Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 523618 12 pageshttpdxdoiorg1011552013523618
Research ArticleNew Recursive Representations for the FavardConstants with Application to Multiple Singular Integrals andSummation of Series
Snezhana Georgieva Gocheva-Ilieva and Ivan Hristov Feschiev
Faculty of Mathematics and Computer Science Paisii Hilendarski University of Plovdiv 24 Tzar Assen Street 4000 Plovdiv Bulgaria
Correspondence should be addressed to Snezhana Georgieva Gocheva-Ilieva snegochevayahoocom
Received 11 February 2013 Accepted 22 April 2013
Academic Editor Josip E Pecaric
Copyright copy 2013 S G Gocheva-Ilieva and I H Feschiev This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
There are obtained integral form and recurrence representations for some Fourier series and connected with themFavard constantsThe method is based on preliminary integration of Fourier series which permits to establish general recursion formulas for Favardconstants This gives the opportunity for effective summation of infinite series and calculation of some classes of multiple singularintegrals by the Favard constants
1 Introduction
The Fourier series and related with them Achieser-Krein-Favard constants often simply called Favard constants havesignificant theoretical and practical roles in many areas [12] These remarkable mathematical constants are introducedfirstly in the theory of Fourier series and approximations offunctions by trigonometric polynomials [3]
The classical definitions of Favard constants are given bythe infinite series [1 4 5]
119870119903=4
120587
infin
sum
]=0
(minus1)](119903+1)
(2] + 1)119903+1 119903 = 0 1 2 (1)
119903=4
120587
infin
sum
]=0
(minus1)]119903
(2] + 1)119903+1 119903 = 1 2 (2)
These constants find wide applications in the approxima-tion theory for exact and asymptotic results on the approx-imation of functions and especially for the best approxi-mations of trigonometric and other classes of functions indifferent spaces and related inequalities [1 5ndash14] In partic-ular many important applications are concerned with theapproximation of Euler cardinal periodic and other type of
splines [15ndash17] It can be noted that Favard constants are con-nected to approximations that are best in a pointwise sensein comparison for instance with the Lebesgue constantswhich are connected to approximations that are best in aleast-squares sense (Fourier series) [5] The Favard constantsplay also an important role in estimating optimal quadratureand cubature formulas calculation of singular integralssome classes of differential integrodifferential and integralequations [18ndash24] and in other areas
Nevertheless widely used as a whole the properties of theFavard constants have not been investigated well enough [14]except for some particular cases
Different methods for their calculation are given forinstance in [2 Ch 52] In general thesemethods are based onthe properties of the well-known constants and special func-tions as gamma function Γ(119911) generalized Riemann zetafunction 120577(119911 119886) the Bernoulli polynomials 119861
119899(119909) and the
Bernoulli numbers 119861119899 and the Euler polynomials 119864
119899(119909) and
the Euler numbers 119864119899 given by the following expressions [2]
Γ (119911) = int
infin
0
119905119911minus1
119890minus119905
119889119905 (Re 119911 gt 0)
Ψ (119911) =Γ1015840
(119911)
Γ (119911)
2 Abstract and Applied Analysis
120577 (119911 119886) =
infin
sum
119896=0
1
(119896 + 119886)119911=
1
Γ (119911)int
infin
0
119905119911minus1
119890minus119886119905
1 minus 119890minus119905119889119905 (Re 119911 gt 1)
119861119899(119909)
119905119890119909119905
119890119905 minus 1=
infin
sum
119896=0
119861119896(119909)
119905119896
119896(|119905| lt 2120587)
119861119899= 119861119899(0) 119861
119899(119909) =
119899
sum
119896=0
(119899
119896)119861119896119909119899minus119896
119864119899(119909)
2119890119909119905
119890119905 + 1=
infin
sum
119896=0
119864119896(119909)
119905119896
119896
(1
cosh (119905)=
infin
sum
119896=0
119864119896
119905119896
119896) (|119905| lt 120587)
(3)
From now on we will use the following notations
119860119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln tan 1205792119889120579 (0 le 119909 le 120587)
119861119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln(2 sin 1205792) 119889120579 (0 le 119909 le 2120587)
119862119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln(2 cos 1205792) 119889120579
(minus120587 le 119909 le 120587) (119899 = 1 2 3 )
(4)
119863119899(119909) =
119909119899
119899 (5)
119879119903=
infin
sum
119899=1
1
119899119903(119903 = 2 3 )
119876119903=
infin
sum
119899=1
(minus1)119899
119899119903(119903 = 1 2 3 )
(6)
where the threemultiple singular integrals (4) contain exactly119899 integral operations
The main purpose of this paper is to establish newrecursive formulas for the Favard constants (1) including onlyfinite number of terms and recursive formulas for (2) Theywill be further used to obtain new integral representations forthe previously stated objects in particular for calculation ofthe multiple singular integrals (4) and summation of series
2 Recursive Representations for119870119903
and Some Fourier Series
We will prove the following
Theorem 1 For the constants119870119903(119903 = 1 2 3 ) the following
recursive representations hold
1198702119904+1
=1
2[(minus1)
119904
1198632119904+1
(120587) +
119904
sum
119901=1
1198702119904+1minus2119901
(minus1)119901minus1
1205872119901
(2119901)]
(1198700= 1 119870
1=120587
2)
(7a)
1198702119904= (minus1)
119904
1198632119904(120587
2) +
119904
sum
119901=1
1198702119904+1minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
(119904 = 1 2 3 )
(7b)
Proof We will use the method of induction and preliminaryintegration of appropriate Fourier series Let us start by thewell-known expansion [2]
4
120587
infin
sum
]=0
sin [(2] + 1) 119909]2] + 1
= 1 (0 lt 119909 lt 120587) (8)
where for 119909 = 1205872we will have1198700= 1 By integration of both
sides of (8) in [0 119909] we get
minus4
120587
infin
sum
]=0
cos [(2] + 1) 119909](2] + 1)2
+4
120587
infin
sum
]=0
1
(2] + 1)2= 119909 = 119863
1(119909)
(0 le 119909 le 120587)
(9)
For 119909 = 120587 (9) gives us 21198701= 120587 = 119863
1(120587) and conse-
quently1198701= 1198631(120587)2 = 1205872 The same results for119870
0and119870
1
can be achieved starting from the equality [2]
infin
sum
119899=1
sin 119899119909119899
=120587
2minus119909
2=120587
2minus1
21198631(119909) (0 lt 119909 lt 2120587)
(10)
For 119909 = 1205872 we find again that 1198700= 1 After integration
of the both sides of (10) we have
minus
infin
sum
119899=1
cos 1198991199091198992
+ 1198792= minus
1199092
4+120587119909
2= minus
1
21198632(119909) +
120587
21198631(119909)
(0 le 119909 le 2120587)
(11)
For 119909 = 120587 we find that 1198701= 1205872 Next after integration
of the both sides of (11) we obtain
minus
infin
sum
119899=1
sin 1198991199091198993
+ 1198792119909 = minus
1
21198633(119909) +
120587
21198632(119909) (0 le 119909 le 2120587)
(12)
which for 119909 = 120587 gives us
1198792= minus
1
21205871198633(120587) +
1
21198632(120587) =
1205872
6 (13)
Abstract and Applied Analysis 3
If we put 119909 = 1205872 in the same equality (12) and make alittle processing we will arrive at the value119870
2= 1205872
8On the other hand after integration of both sides of (9)
we obtain
minus4
120587
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)3
+ 1198701119909 = 119863
2(119909) (0 le 119909 le 120587)
(14)
Here for 119909 = 1205872 we get minus1198702+ (1205872)119870
1= 1198632(1205872) and
consequently
1198702=120587
21198701minus 1198632(120587
2) =
1205872
8 (15)
For the constant 1198703 we must now integrate both sides of
(12) asinfin
sum
119899=1
cos 1198991199091198994
minus 1198794+ 11987921198632(119909) = minus
1
21198634(119909) +
120587
21198633(119909)
(0 le 119909 le 2120587)
(16)
and put 119909 = 120587 here or integrate both sides of (14) as
4
120587
infin
sum
V=1
cos [(2V + 1) 119909](2V + 1)4
minus4
120587
infin
sum
V=1
1
(2V + 1)4+ 11987011198632(119909) = 119863
3(119909)
(0 le 119909 le 120587)
(17)
and put again 119909 = 120587 Then we will have
1198703=1
1205871198634(120587) minus
3
21198633(120587) +
120587
21198632(120587)
=1205872
41198701minus1
21198633(120587) =
1205873
24
(18)
Going on the indicated procedure on the base of induc-tion we easily arrive at the recursive representations (7a) and(7b) and complete the proof
Remark 2 The scheme of this proof is valid for the most ofthe other statements in this paper
In connection with Theorem 1 we would like to noteanother representation of119870
119903(see eg [25]) It can be written
in terms of Lerch transcendent [25] or as it is shown in [2Section 514]
1198702119904minus1
=2
120587 (2119904)(22119904
minus 1) 1205872119904 10038161003816100381610038161198612119904
1003816100381610038161003816
1198702119904=
2
120587 (2119904)(120587
2)
2119904+1
100381610038161003816100381611986421199041003816100381610038161003816 (119904 = 1 2 3 )
(19)
where the Bernoulli and Euler numbers are specified in (3)Data for values of magnitudes of 119870
119903using (7a) and (7b)
are shown in Table 1The equalities (10)ndash(13) outline a procedure for summing
up the numerical series 1198792119904 (119904 = 1 2 3 ) in (6) It leads to
the assertion
Table 1 Exact and approximate values of the Favard constants 119870119903
calculated by the recursive formulas (7a) and (7b) usingMathemat-ica software package
119903 Exact values of119870119903
Approximate values of119870119903
1 1205872 157079632679489661923132169162 120587
2
8 123370055013616982735431137493 120587
3
24 129192819501249250731151312774 5120587
4
384 126834753950524006818281683185 120587
5
240 127508201993867272192808879186 61120587
6
46080 127267232656453061325614987117 17120587
7
40320 127343712480668316338644619008 277120587
8
2064384 127317548065260581363477696719 31120587
9
725760 1273261242472487546381436665610 50521120587
10
3715891200 1273232382729394849508279710811 691120587
11
159667200 1273241945872154096771507790112 540553120587
12
392398110720 12732387471572495304117396905
Corollary 3 The following recursive representation holds
1198792119904= (minus1)
119904
[1
21205871198632119904+1
(120587) minus1
21198632119904(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901
(119904 = 1 2 3 )
(20)
where for 119904 = 1 by definition 1198790= 0
It can be noted that the numbers 1198792119904can be also repre-
sented by the well known formula (see eg [2 512])
1198792119904=22119904minus1
1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (21)
The same procedure applied on the base of the equalityinfin
sum
119899=1
(minus1)119899sin 119899119909119899
= minus1
21198631(119909) (minus120587 lt 119909 lt 120587) (22)
leads to the following
Corollary 4 The following recursive representation holds
1198762119904=(minus1)119904
21205871198632119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901
(119904 = 1 2 3 )
(23)
where for 119904 = 1 by definition 1198760= 0
In this connection we will note the explicit formula for1198762119904represented by the Bernoulli numbers (see [2 521])
1198762119904=
(1 minus 22119904minus1
) 1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (24)
It is easy to see that the constants119870119903satisfy the following
inequalities (see also [1])
1 = 1198700lt 1198702lt 1198704lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt 119870
5lt 1198703lt 1198701=120587
2
(25)
and lim119903rarrinfin
119870119903= 4120587
4 Abstract and Applied Analysis
Theprocedure of getting the representations (11) (12) and(16) with the help of (10) gives us an opportunity to lay downthe following
Theorem 5 The following recursive representations hold
infin
sum
119899=1
sin 1198991199091198992119904minus1
= (minus1)119904
[1
21198632119904minus1
(119909) minus120587
21198632119904minus2
(119909)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
cos 1198991199091198992119904
= (minus1)119904minus1
[1
21198632119904(119909) minus
120587
21198632119904minus1
(119909)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(119909)
(26)
(119904 = 1 2 3 (0 le 119909 le 2120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198790= 0 0 lt 119909 lt 2120587)
At the same time both series in (26) have the well-knownrepresentations [2 542]
infin
sum
119899=1
sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909
2120587)
((0 le 119909 le 2120587) (119904 = 1 2 3 ) 0 lt 119909 lt 2120587 for 119904 = 0)
infin
sum
119899=1
cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909
2120587)
(0 le 119909 le 2120587 119904 = 1 2 3 )
(27)
The previously stated procedure for obtaining (26) cannow be applied on the strength of (22)This leads to the asser-tion
Theorem 6 The following recursive representations hold
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904minus1
=(minus1)119904
21198632119904minus1
(119909)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
21198632119904(119909)
+
119904minus1
sum
119901=0
(minus1)119901
1198762119904minus2119901
1198632119901(119909)
(28)
(119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198760= 0 minus120587 lt 119909 lt 120587)
At the same time both series in (28) have the well-knownrepresentations [2 542]
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909 + 120587
2120587)
(minus120587 lt 119909 le 120587 119904 = 0 1 2 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909 + 120587
2120587)
(minus120587 le 119909 le 120587 119904 = 1 2 3 )
(29)
By analogy with the previous the procedure for obtaining(9) (14) and (17) with the help of (8) leads us to the following
Theorem 7 The following recursive representations hold
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(119909)]
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=120587(minus1)
119904
4[1198632119904minus1
(119909) +
119904
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(119909)]
(30)
119904 = 1 2 3 (0 le 119909 le 120587) for 119904 = 1 by definition 1198630(119909) =
1 (0 lt 119909 lt 120587)
At the same time both series in (30) have the well-knownformulas [2 546]
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587)
(0 lt 119909 lt 120587 119904 = 0 1 2 )
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=(minus1)119904
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587)
(0 le 119909 le 120587 119904 = 1 2 )
(31)
The same procedure applied on the base of the equality
4
120587
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
2V + 1= 1 (minus
120587
2lt 119909 lt
120587
2) (32)
gives us the next theorem
Abstract and Applied Analysis 5
Theorem 8 The following recursive representations holdinfin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(119909)]
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904
=120587(minus1)
119904minus1
4[1198632119904minus1
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(119909)]
(33)
(119904 = 1 2 3 (minus1205872 le 119909 le 1205872) for 119904 = 1 1198700declines
1198630(119909) = 1 (minus1205872 lt 119909 lt 1205872))
At the same time both series in (33) have the well-knownrepresentations ([2 546])infin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587+1
2)
(minus120587
2lt 119909 lt
120587
2 119904 = 0 1 )
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904=(minus1)119904minus1
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587+1
2)
(minus120587
2le 119909 le
120587
2 119904 = 1 2 )
(34)
Meanwhile it is important to note that the number ofaddends in our recurrence representations (26) (28) (30)and (33) is two times less than the number of the addendsin the corresponding cited formulas from [2] So our methodappears to be more economic and effective
Moreover one can get many other representations of theconstants 119870
119903(119903 = 1 2 3 ) and numerical series 119876
2119904(119904 =
1 2 ) from Theorems 5ndash8 putting in particular 119909 = 1205872
or 119909 = 120587 For completeness we will note the main resultsFromTheorem 5 for 119909 = 1205872 and 119909 = 120587 immediately fol-
lows the following
Corollary 9 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)
119904
[1
21198632119904minus1
(120587
2) minus
120587
21198632119904minus2
(120587
2)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)
119904
[1
21198632119904(120587) minus
120587
21198632119904minus1
(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901(120587)
(35)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119879
0= 0)
For 119909 = 120587 one can get (20) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 we obtain the next corollary
Corollary 10 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904= 4119904
(minus1)119904minus1
[1
21198632119904(120587
2) minus
120587
21198632119904minus1
(120587
2)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(120587
2)
(36)
(119904 = 1 2 for 119904 = 11198630(119909) = 1)
Similarly from Theorem 6 for 119909 = 1205872 and 119909 = 120587 weobtain respectively the following
Corollary 11 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)119904minus1
21198632119904minus1
(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)119904minus1
21198632119904(120587)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901(120587)
(37)
where 119904 = 1 2 and for 119904 = 11198630(119909) = 1 119876
0= 0
For 119909 = 120587 one can get (23) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 from Theorem 6 (the second formula in(28)) we will have also the next analogous corollary
Corollary 12 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904=
4119904
4119904 minus 1(minus1)119904
21198632119904(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901(120587
2)
(38)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119876
0= 0)
By the same manner from Theorem 7 for 119909 = 120587 and119909 = 1205872 we obtain respectively the formulas for 119870
119903(119903 =
1 2 3 ) different from these inTheorem 1
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
120577 (119911 119886) =
infin
sum
119896=0
1
(119896 + 119886)119911=
1
Γ (119911)int
infin
0
119905119911minus1
119890minus119886119905
1 minus 119890minus119905119889119905 (Re 119911 gt 1)
119861119899(119909)
119905119890119909119905
119890119905 minus 1=
infin
sum
119896=0
119861119896(119909)
119905119896
119896(|119905| lt 2120587)
119861119899= 119861119899(0) 119861
119899(119909) =
119899
sum
119896=0
(119899
119896)119861119896119909119899minus119896
119864119899(119909)
2119890119909119905
119890119905 + 1=
infin
sum
119896=0
119864119896(119909)
119905119896
119896
(1
cosh (119905)=
infin
sum
119896=0
119864119896
119905119896
119896) (|119905| lt 120587)
(3)
From now on we will use the following notations
119860119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln tan 1205792119889120579 (0 le 119909 le 120587)
119861119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln(2 sin 1205792) 119889120579 (0 le 119909 le 2120587)
119862119899(119909) = int
119909
0
119889119909 sdot sdot sdot int
119909
0
ln(2 cos 1205792) 119889120579
(minus120587 le 119909 le 120587) (119899 = 1 2 3 )
(4)
119863119899(119909) =
119909119899
119899 (5)
119879119903=
infin
sum
119899=1
1
119899119903(119903 = 2 3 )
119876119903=
infin
sum
119899=1
(minus1)119899
119899119903(119903 = 1 2 3 )
(6)
where the threemultiple singular integrals (4) contain exactly119899 integral operations
The main purpose of this paper is to establish newrecursive formulas for the Favard constants (1) including onlyfinite number of terms and recursive formulas for (2) Theywill be further used to obtain new integral representations forthe previously stated objects in particular for calculation ofthe multiple singular integrals (4) and summation of series
2 Recursive Representations for119870119903
and Some Fourier Series
We will prove the following
Theorem 1 For the constants119870119903(119903 = 1 2 3 ) the following
recursive representations hold
1198702119904+1
=1
2[(minus1)
119904
1198632119904+1
(120587) +
119904
sum
119901=1
1198702119904+1minus2119901
(minus1)119901minus1
1205872119901
(2119901)]
(1198700= 1 119870
1=120587
2)
(7a)
1198702119904= (minus1)
119904
1198632119904(120587
2) +
119904
sum
119901=1
1198702119904+1minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
(119904 = 1 2 3 )
(7b)
Proof We will use the method of induction and preliminaryintegration of appropriate Fourier series Let us start by thewell-known expansion [2]
4
120587
infin
sum
]=0
sin [(2] + 1) 119909]2] + 1
= 1 (0 lt 119909 lt 120587) (8)
where for 119909 = 1205872we will have1198700= 1 By integration of both
sides of (8) in [0 119909] we get
minus4
120587
infin
sum
]=0
cos [(2] + 1) 119909](2] + 1)2
+4
120587
infin
sum
]=0
1
(2] + 1)2= 119909 = 119863
1(119909)
(0 le 119909 le 120587)
(9)
For 119909 = 120587 (9) gives us 21198701= 120587 = 119863
1(120587) and conse-
quently1198701= 1198631(120587)2 = 1205872 The same results for119870
0and119870
1
can be achieved starting from the equality [2]
infin
sum
119899=1
sin 119899119909119899
=120587
2minus119909
2=120587
2minus1
21198631(119909) (0 lt 119909 lt 2120587)
(10)
For 119909 = 1205872 we find again that 1198700= 1 After integration
of the both sides of (10) we have
minus
infin
sum
119899=1
cos 1198991199091198992
+ 1198792= minus
1199092
4+120587119909
2= minus
1
21198632(119909) +
120587
21198631(119909)
(0 le 119909 le 2120587)
(11)
For 119909 = 120587 we find that 1198701= 1205872 Next after integration
of the both sides of (11) we obtain
minus
infin
sum
119899=1
sin 1198991199091198993
+ 1198792119909 = minus
1
21198633(119909) +
120587
21198632(119909) (0 le 119909 le 2120587)
(12)
which for 119909 = 120587 gives us
1198792= minus
1
21205871198633(120587) +
1
21198632(120587) =
1205872
6 (13)
Abstract and Applied Analysis 3
If we put 119909 = 1205872 in the same equality (12) and make alittle processing we will arrive at the value119870
2= 1205872
8On the other hand after integration of both sides of (9)
we obtain
minus4
120587
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)3
+ 1198701119909 = 119863
2(119909) (0 le 119909 le 120587)
(14)
Here for 119909 = 1205872 we get minus1198702+ (1205872)119870
1= 1198632(1205872) and
consequently
1198702=120587
21198701minus 1198632(120587
2) =
1205872
8 (15)
For the constant 1198703 we must now integrate both sides of
(12) asinfin
sum
119899=1
cos 1198991199091198994
minus 1198794+ 11987921198632(119909) = minus
1
21198634(119909) +
120587
21198633(119909)
(0 le 119909 le 2120587)
(16)
and put 119909 = 120587 here or integrate both sides of (14) as
4
120587
infin
sum
V=1
cos [(2V + 1) 119909](2V + 1)4
minus4
120587
infin
sum
V=1
1
(2V + 1)4+ 11987011198632(119909) = 119863
3(119909)
(0 le 119909 le 120587)
(17)
and put again 119909 = 120587 Then we will have
1198703=1
1205871198634(120587) minus
3
21198633(120587) +
120587
21198632(120587)
=1205872
41198701minus1
21198633(120587) =
1205873
24
(18)
Going on the indicated procedure on the base of induc-tion we easily arrive at the recursive representations (7a) and(7b) and complete the proof
Remark 2 The scheme of this proof is valid for the most ofthe other statements in this paper
In connection with Theorem 1 we would like to noteanother representation of119870
119903(see eg [25]) It can be written
in terms of Lerch transcendent [25] or as it is shown in [2Section 514]
1198702119904minus1
=2
120587 (2119904)(22119904
minus 1) 1205872119904 10038161003816100381610038161198612119904
1003816100381610038161003816
1198702119904=
2
120587 (2119904)(120587
2)
2119904+1
100381610038161003816100381611986421199041003816100381610038161003816 (119904 = 1 2 3 )
(19)
where the Bernoulli and Euler numbers are specified in (3)Data for values of magnitudes of 119870
119903using (7a) and (7b)
are shown in Table 1The equalities (10)ndash(13) outline a procedure for summing
up the numerical series 1198792119904 (119904 = 1 2 3 ) in (6) It leads to
the assertion
Table 1 Exact and approximate values of the Favard constants 119870119903
calculated by the recursive formulas (7a) and (7b) usingMathemat-ica software package
119903 Exact values of119870119903
Approximate values of119870119903
1 1205872 157079632679489661923132169162 120587
2
8 123370055013616982735431137493 120587
3
24 129192819501249250731151312774 5120587
4
384 126834753950524006818281683185 120587
5
240 127508201993867272192808879186 61120587
6
46080 127267232656453061325614987117 17120587
7
40320 127343712480668316338644619008 277120587
8
2064384 127317548065260581363477696719 31120587
9
725760 1273261242472487546381436665610 50521120587
10
3715891200 1273232382729394849508279710811 691120587
11
159667200 1273241945872154096771507790112 540553120587
12
392398110720 12732387471572495304117396905
Corollary 3 The following recursive representation holds
1198792119904= (minus1)
119904
[1
21205871198632119904+1
(120587) minus1
21198632119904(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901
(119904 = 1 2 3 )
(20)
where for 119904 = 1 by definition 1198790= 0
It can be noted that the numbers 1198792119904can be also repre-
sented by the well known formula (see eg [2 512])
1198792119904=22119904minus1
1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (21)
The same procedure applied on the base of the equalityinfin
sum
119899=1
(minus1)119899sin 119899119909119899
= minus1
21198631(119909) (minus120587 lt 119909 lt 120587) (22)
leads to the following
Corollary 4 The following recursive representation holds
1198762119904=(minus1)119904
21205871198632119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901
(119904 = 1 2 3 )
(23)
where for 119904 = 1 by definition 1198760= 0
In this connection we will note the explicit formula for1198762119904represented by the Bernoulli numbers (see [2 521])
1198762119904=
(1 minus 22119904minus1
) 1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (24)
It is easy to see that the constants119870119903satisfy the following
inequalities (see also [1])
1 = 1198700lt 1198702lt 1198704lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt 119870
5lt 1198703lt 1198701=120587
2
(25)
and lim119903rarrinfin
119870119903= 4120587
4 Abstract and Applied Analysis
Theprocedure of getting the representations (11) (12) and(16) with the help of (10) gives us an opportunity to lay downthe following
Theorem 5 The following recursive representations hold
infin
sum
119899=1
sin 1198991199091198992119904minus1
= (minus1)119904
[1
21198632119904minus1
(119909) minus120587
21198632119904minus2
(119909)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
cos 1198991199091198992119904
= (minus1)119904minus1
[1
21198632119904(119909) minus
120587
21198632119904minus1
(119909)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(119909)
(26)
(119904 = 1 2 3 (0 le 119909 le 2120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198790= 0 0 lt 119909 lt 2120587)
At the same time both series in (26) have the well-knownrepresentations [2 542]
infin
sum
119899=1
sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909
2120587)
((0 le 119909 le 2120587) (119904 = 1 2 3 ) 0 lt 119909 lt 2120587 for 119904 = 0)
infin
sum
119899=1
cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909
2120587)
(0 le 119909 le 2120587 119904 = 1 2 3 )
(27)
The previously stated procedure for obtaining (26) cannow be applied on the strength of (22)This leads to the asser-tion
Theorem 6 The following recursive representations hold
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904minus1
=(minus1)119904
21198632119904minus1
(119909)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
21198632119904(119909)
+
119904minus1
sum
119901=0
(minus1)119901
1198762119904minus2119901
1198632119901(119909)
(28)
(119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198760= 0 minus120587 lt 119909 lt 120587)
At the same time both series in (28) have the well-knownrepresentations [2 542]
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909 + 120587
2120587)
(minus120587 lt 119909 le 120587 119904 = 0 1 2 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909 + 120587
2120587)
(minus120587 le 119909 le 120587 119904 = 1 2 3 )
(29)
By analogy with the previous the procedure for obtaining(9) (14) and (17) with the help of (8) leads us to the following
Theorem 7 The following recursive representations hold
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(119909)]
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=120587(minus1)
119904
4[1198632119904minus1
(119909) +
119904
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(119909)]
(30)
119904 = 1 2 3 (0 le 119909 le 120587) for 119904 = 1 by definition 1198630(119909) =
1 (0 lt 119909 lt 120587)
At the same time both series in (30) have the well-knownformulas [2 546]
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587)
(0 lt 119909 lt 120587 119904 = 0 1 2 )
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=(minus1)119904
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587)
(0 le 119909 le 120587 119904 = 1 2 )
(31)
The same procedure applied on the base of the equality
4
120587
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
2V + 1= 1 (minus
120587
2lt 119909 lt
120587
2) (32)
gives us the next theorem
Abstract and Applied Analysis 5
Theorem 8 The following recursive representations holdinfin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(119909)]
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904
=120587(minus1)
119904minus1
4[1198632119904minus1
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(119909)]
(33)
(119904 = 1 2 3 (minus1205872 le 119909 le 1205872) for 119904 = 1 1198700declines
1198630(119909) = 1 (minus1205872 lt 119909 lt 1205872))
At the same time both series in (33) have the well-knownrepresentations ([2 546])infin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587+1
2)
(minus120587
2lt 119909 lt
120587
2 119904 = 0 1 )
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904=(minus1)119904minus1
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587+1
2)
(minus120587
2le 119909 le
120587
2 119904 = 1 2 )
(34)
Meanwhile it is important to note that the number ofaddends in our recurrence representations (26) (28) (30)and (33) is two times less than the number of the addendsin the corresponding cited formulas from [2] So our methodappears to be more economic and effective
Moreover one can get many other representations of theconstants 119870
119903(119903 = 1 2 3 ) and numerical series 119876
2119904(119904 =
1 2 ) from Theorems 5ndash8 putting in particular 119909 = 1205872
or 119909 = 120587 For completeness we will note the main resultsFromTheorem 5 for 119909 = 1205872 and 119909 = 120587 immediately fol-
lows the following
Corollary 9 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)
119904
[1
21198632119904minus1
(120587
2) minus
120587
21198632119904minus2
(120587
2)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)
119904
[1
21198632119904(120587) minus
120587
21198632119904minus1
(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901(120587)
(35)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119879
0= 0)
For 119909 = 120587 one can get (20) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 we obtain the next corollary
Corollary 10 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904= 4119904
(minus1)119904minus1
[1
21198632119904(120587
2) minus
120587
21198632119904minus1
(120587
2)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(120587
2)
(36)
(119904 = 1 2 for 119904 = 11198630(119909) = 1)
Similarly from Theorem 6 for 119909 = 1205872 and 119909 = 120587 weobtain respectively the following
Corollary 11 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)119904minus1
21198632119904minus1
(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)119904minus1
21198632119904(120587)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901(120587)
(37)
where 119904 = 1 2 and for 119904 = 11198630(119909) = 1 119876
0= 0
For 119909 = 120587 one can get (23) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 from Theorem 6 (the second formula in(28)) we will have also the next analogous corollary
Corollary 12 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904=
4119904
4119904 minus 1(minus1)119904
21198632119904(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901(120587
2)
(38)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119876
0= 0)
By the same manner from Theorem 7 for 119909 = 120587 and119909 = 1205872 we obtain respectively the formulas for 119870
119903(119903 =
1 2 3 ) different from these inTheorem 1
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
If we put 119909 = 1205872 in the same equality (12) and make alittle processing we will arrive at the value119870
2= 1205872
8On the other hand after integration of both sides of (9)
we obtain
minus4
120587
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)3
+ 1198701119909 = 119863
2(119909) (0 le 119909 le 120587)
(14)
Here for 119909 = 1205872 we get minus1198702+ (1205872)119870
1= 1198632(1205872) and
consequently
1198702=120587
21198701minus 1198632(120587
2) =
1205872
8 (15)
For the constant 1198703 we must now integrate both sides of
(12) asinfin
sum
119899=1
cos 1198991199091198994
minus 1198794+ 11987921198632(119909) = minus
1
21198634(119909) +
120587
21198633(119909)
(0 le 119909 le 2120587)
(16)
and put 119909 = 120587 here or integrate both sides of (14) as
4
120587
infin
sum
V=1
cos [(2V + 1) 119909](2V + 1)4
minus4
120587
infin
sum
V=1
1
(2V + 1)4+ 11987011198632(119909) = 119863
3(119909)
(0 le 119909 le 120587)
(17)
and put again 119909 = 120587 Then we will have
1198703=1
1205871198634(120587) minus
3
21198633(120587) +
120587
21198632(120587)
=1205872
41198701minus1
21198633(120587) =
1205873
24
(18)
Going on the indicated procedure on the base of induc-tion we easily arrive at the recursive representations (7a) and(7b) and complete the proof
Remark 2 The scheme of this proof is valid for the most ofthe other statements in this paper
In connection with Theorem 1 we would like to noteanother representation of119870
119903(see eg [25]) It can be written
in terms of Lerch transcendent [25] or as it is shown in [2Section 514]
1198702119904minus1
=2
120587 (2119904)(22119904
minus 1) 1205872119904 10038161003816100381610038161198612119904
1003816100381610038161003816
1198702119904=
2
120587 (2119904)(120587
2)
2119904+1
100381610038161003816100381611986421199041003816100381610038161003816 (119904 = 1 2 3 )
(19)
where the Bernoulli and Euler numbers are specified in (3)Data for values of magnitudes of 119870
119903using (7a) and (7b)
are shown in Table 1The equalities (10)ndash(13) outline a procedure for summing
up the numerical series 1198792119904 (119904 = 1 2 3 ) in (6) It leads to
the assertion
Table 1 Exact and approximate values of the Favard constants 119870119903
calculated by the recursive formulas (7a) and (7b) usingMathemat-ica software package
119903 Exact values of119870119903
Approximate values of119870119903
1 1205872 157079632679489661923132169162 120587
2
8 123370055013616982735431137493 120587
3
24 129192819501249250731151312774 5120587
4
384 126834753950524006818281683185 120587
5
240 127508201993867272192808879186 61120587
6
46080 127267232656453061325614987117 17120587
7
40320 127343712480668316338644619008 277120587
8
2064384 127317548065260581363477696719 31120587
9
725760 1273261242472487546381436665610 50521120587
10
3715891200 1273232382729394849508279710811 691120587
11
159667200 1273241945872154096771507790112 540553120587
12
392398110720 12732387471572495304117396905
Corollary 3 The following recursive representation holds
1198792119904= (minus1)
119904
[1
21205871198632119904+1
(120587) minus1
21198632119904(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901
(119904 = 1 2 3 )
(20)
where for 119904 = 1 by definition 1198790= 0
It can be noted that the numbers 1198792119904can be also repre-
sented by the well known formula (see eg [2 512])
1198792119904=22119904minus1
1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (21)
The same procedure applied on the base of the equalityinfin
sum
119899=1
(minus1)119899sin 119899119909119899
= minus1
21198631(119909) (minus120587 lt 119909 lt 120587) (22)
leads to the following
Corollary 4 The following recursive representation holds
1198762119904=(minus1)119904
21205871198632119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901
(119904 = 1 2 3 )
(23)
where for 119904 = 1 by definition 1198760= 0
In this connection we will note the explicit formula for1198762119904represented by the Bernoulli numbers (see [2 521])
1198762119904=
(1 minus 22119904minus1
) 1205872119904
(2119904)
100381610038161003816100381611986121199041003816100381610038161003816 (119904 = 1 2 3 ) (24)
It is easy to see that the constants119870119903satisfy the following
inequalities (see also [1])
1 = 1198700lt 1198702lt 1198704lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt 119870
5lt 1198703lt 1198701=120587
2
(25)
and lim119903rarrinfin
119870119903= 4120587
4 Abstract and Applied Analysis
Theprocedure of getting the representations (11) (12) and(16) with the help of (10) gives us an opportunity to lay downthe following
Theorem 5 The following recursive representations hold
infin
sum
119899=1
sin 1198991199091198992119904minus1
= (minus1)119904
[1
21198632119904minus1
(119909) minus120587
21198632119904minus2
(119909)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
cos 1198991199091198992119904
= (minus1)119904minus1
[1
21198632119904(119909) minus
120587
21198632119904minus1
(119909)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(119909)
(26)
(119904 = 1 2 3 (0 le 119909 le 2120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198790= 0 0 lt 119909 lt 2120587)
At the same time both series in (26) have the well-knownrepresentations [2 542]
infin
sum
119899=1
sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909
2120587)
((0 le 119909 le 2120587) (119904 = 1 2 3 ) 0 lt 119909 lt 2120587 for 119904 = 0)
infin
sum
119899=1
cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909
2120587)
(0 le 119909 le 2120587 119904 = 1 2 3 )
(27)
The previously stated procedure for obtaining (26) cannow be applied on the strength of (22)This leads to the asser-tion
Theorem 6 The following recursive representations hold
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904minus1
=(minus1)119904
21198632119904minus1
(119909)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
21198632119904(119909)
+
119904minus1
sum
119901=0
(minus1)119901
1198762119904minus2119901
1198632119901(119909)
(28)
(119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198760= 0 minus120587 lt 119909 lt 120587)
At the same time both series in (28) have the well-knownrepresentations [2 542]
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909 + 120587
2120587)
(minus120587 lt 119909 le 120587 119904 = 0 1 2 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909 + 120587
2120587)
(minus120587 le 119909 le 120587 119904 = 1 2 3 )
(29)
By analogy with the previous the procedure for obtaining(9) (14) and (17) with the help of (8) leads us to the following
Theorem 7 The following recursive representations hold
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(119909)]
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=120587(minus1)
119904
4[1198632119904minus1
(119909) +
119904
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(119909)]
(30)
119904 = 1 2 3 (0 le 119909 le 120587) for 119904 = 1 by definition 1198630(119909) =
1 (0 lt 119909 lt 120587)
At the same time both series in (30) have the well-knownformulas [2 546]
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587)
(0 lt 119909 lt 120587 119904 = 0 1 2 )
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=(minus1)119904
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587)
(0 le 119909 le 120587 119904 = 1 2 )
(31)
The same procedure applied on the base of the equality
4
120587
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
2V + 1= 1 (minus
120587
2lt 119909 lt
120587
2) (32)
gives us the next theorem
Abstract and Applied Analysis 5
Theorem 8 The following recursive representations holdinfin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(119909)]
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904
=120587(minus1)
119904minus1
4[1198632119904minus1
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(119909)]
(33)
(119904 = 1 2 3 (minus1205872 le 119909 le 1205872) for 119904 = 1 1198700declines
1198630(119909) = 1 (minus1205872 lt 119909 lt 1205872))
At the same time both series in (33) have the well-knownrepresentations ([2 546])infin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587+1
2)
(minus120587
2lt 119909 lt
120587
2 119904 = 0 1 )
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904=(minus1)119904minus1
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587+1
2)
(minus120587
2le 119909 le
120587
2 119904 = 1 2 )
(34)
Meanwhile it is important to note that the number ofaddends in our recurrence representations (26) (28) (30)and (33) is two times less than the number of the addendsin the corresponding cited formulas from [2] So our methodappears to be more economic and effective
Moreover one can get many other representations of theconstants 119870
119903(119903 = 1 2 3 ) and numerical series 119876
2119904(119904 =
1 2 ) from Theorems 5ndash8 putting in particular 119909 = 1205872
or 119909 = 120587 For completeness we will note the main resultsFromTheorem 5 for 119909 = 1205872 and 119909 = 120587 immediately fol-
lows the following
Corollary 9 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)
119904
[1
21198632119904minus1
(120587
2) minus
120587
21198632119904minus2
(120587
2)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)
119904
[1
21198632119904(120587) minus
120587
21198632119904minus1
(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901(120587)
(35)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119879
0= 0)
For 119909 = 120587 one can get (20) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 we obtain the next corollary
Corollary 10 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904= 4119904
(minus1)119904minus1
[1
21198632119904(120587
2) minus
120587
21198632119904minus1
(120587
2)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(120587
2)
(36)
(119904 = 1 2 for 119904 = 11198630(119909) = 1)
Similarly from Theorem 6 for 119909 = 1205872 and 119909 = 120587 weobtain respectively the following
Corollary 11 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)119904minus1
21198632119904minus1
(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)119904minus1
21198632119904(120587)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901(120587)
(37)
where 119904 = 1 2 and for 119904 = 11198630(119909) = 1 119876
0= 0
For 119909 = 120587 one can get (23) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 from Theorem 6 (the second formula in(28)) we will have also the next analogous corollary
Corollary 12 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904=
4119904
4119904 minus 1(minus1)119904
21198632119904(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901(120587
2)
(38)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119876
0= 0)
By the same manner from Theorem 7 for 119909 = 120587 and119909 = 1205872 we obtain respectively the formulas for 119870
119903(119903 =
1 2 3 ) different from these inTheorem 1
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
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Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
Theprocedure of getting the representations (11) (12) and(16) with the help of (10) gives us an opportunity to lay downthe following
Theorem 5 The following recursive representations hold
infin
sum
119899=1
sin 1198991199091198992119904minus1
= (minus1)119904
[1
21198632119904minus1
(119909) minus120587
21198632119904minus2
(119909)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
cos 1198991199091198992119904
= (minus1)119904minus1
[1
21198632119904(119909) minus
120587
21198632119904minus1
(119909)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(119909)
(26)
(119904 = 1 2 3 (0 le 119909 le 2120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198790= 0 0 lt 119909 lt 2120587)
At the same time both series in (26) have the well-knownrepresentations [2 542]
infin
sum
119899=1
sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909
2120587)
((0 le 119909 le 2120587) (119904 = 1 2 3 ) 0 lt 119909 lt 2120587 for 119904 = 0)
infin
sum
119899=1
cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909
2120587)
(0 le 119909 le 2120587 119904 = 1 2 3 )
(27)
The previously stated procedure for obtaining (26) cannow be applied on the strength of (22)This leads to the asser-tion
Theorem 6 The following recursive representations hold
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904minus1
=(minus1)119904
21198632119904minus1
(119909)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901minus1
(119909)
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
21198632119904(119909)
+
119904minus1
sum
119901=0
(minus1)119901
1198762119904minus2119901
1198632119901(119909)
(28)
(119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 1 by definition1198630(119909) = 1
and 1198760= 0 minus120587 lt 119909 lt 120587)
At the same time both series in (28) have the well-knownrepresentations [2 542]
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904+1
=(minus1)119904minus1
2 (2119904 + 1)(2120587)2119904+1
1198612119904+1
(119909 + 120587
2120587)
(minus120587 lt 119909 le 120587 119904 = 0 1 2 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904
=(minus1)119904minus1
2 (2119904)(2120587)2119904
1198612119904(119909 + 120587
2120587)
(minus120587 le 119909 le 120587 119904 = 1 2 3 )
(29)
By analogy with the previous the procedure for obtaining(9) (14) and (17) with the help of (8) leads us to the following
Theorem 7 The following recursive representations hold
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(119909)]
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=120587(minus1)
119904
4[1198632119904minus1
(119909) +
119904
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(119909)]
(30)
119904 = 1 2 3 (0 le 119909 le 120587) for 119904 = 1 by definition 1198630(119909) =
1 (0 lt 119909 lt 120587)
At the same time both series in (30) have the well-knownformulas [2 546]
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587)
(0 lt 119909 lt 120587 119904 = 0 1 2 )
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904
=(minus1)119904
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587)
(0 le 119909 le 120587 119904 = 1 2 )
(31)
The same procedure applied on the base of the equality
4
120587
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
2V + 1= 1 (minus
120587
2lt 119909 lt
120587
2) (32)
gives us the next theorem
Abstract and Applied Analysis 5
Theorem 8 The following recursive representations holdinfin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(119909)]
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904
=120587(minus1)
119904minus1
4[1198632119904minus1
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(119909)]
(33)
(119904 = 1 2 3 (minus1205872 le 119909 le 1205872) for 119904 = 1 1198700declines
1198630(119909) = 1 (minus1205872 lt 119909 lt 1205872))
At the same time both series in (33) have the well-knownrepresentations ([2 546])infin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587+1
2)
(minus120587
2lt 119909 lt
120587
2 119904 = 0 1 )
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904=(minus1)119904minus1
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587+1
2)
(minus120587
2le 119909 le
120587
2 119904 = 1 2 )
(34)
Meanwhile it is important to note that the number ofaddends in our recurrence representations (26) (28) (30)and (33) is two times less than the number of the addendsin the corresponding cited formulas from [2] So our methodappears to be more economic and effective
Moreover one can get many other representations of theconstants 119870
119903(119903 = 1 2 3 ) and numerical series 119876
2119904(119904 =
1 2 ) from Theorems 5ndash8 putting in particular 119909 = 1205872
or 119909 = 120587 For completeness we will note the main resultsFromTheorem 5 for 119909 = 1205872 and 119909 = 120587 immediately fol-
lows the following
Corollary 9 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)
119904
[1
21198632119904minus1
(120587
2) minus
120587
21198632119904minus2
(120587
2)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)
119904
[1
21198632119904(120587) minus
120587
21198632119904minus1
(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901(120587)
(35)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119879
0= 0)
For 119909 = 120587 one can get (20) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 we obtain the next corollary
Corollary 10 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904= 4119904
(minus1)119904minus1
[1
21198632119904(120587
2) minus
120587
21198632119904minus1
(120587
2)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(120587
2)
(36)
(119904 = 1 2 for 119904 = 11198630(119909) = 1)
Similarly from Theorem 6 for 119909 = 1205872 and 119909 = 120587 weobtain respectively the following
Corollary 11 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)119904minus1
21198632119904minus1
(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)119904minus1
21198632119904(120587)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901(120587)
(37)
where 119904 = 1 2 and for 119904 = 11198630(119909) = 1 119876
0= 0
For 119909 = 120587 one can get (23) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 from Theorem 6 (the second formula in(28)) we will have also the next analogous corollary
Corollary 12 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904=
4119904
4119904 minus 1(minus1)119904
21198632119904(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901(120587
2)
(38)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119876
0= 0)
By the same manner from Theorem 7 for 119909 = 120587 and119909 = 1205872 we obtain respectively the formulas for 119870
119903(119903 =
1 2 3 ) different from these inTheorem 1
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Theorem 8 The following recursive representations holdinfin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904minus1
=120587(minus1)
119904minus1
4[1198632119904minus2
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(119909)]
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904
=120587(minus1)
119904minus1
4[1198632119904minus1
(119909) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(119909)]
(33)
(119904 = 1 2 3 (minus1205872 le 119909 le 1205872) for 119904 = 1 1198700declines
1198630(119909) = 1 (minus1205872 lt 119909 lt 1205872))
At the same time both series in (33) have the well-knownrepresentations ([2 546])infin
sum
V=0
(minus1)V cos [(2V + 1) 119909](2V + 1)2119904+1
=(minus1)119904
1205872119904+1
4 (2119904)1198642119904(119909
120587+1
2)
(minus120587
2lt 119909 lt
120587
2 119904 = 0 1 )
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
(2V + 1)2119904=(minus1)119904minus1
1205872119904
4 (2119904 minus 1)1198642119904minus1
(119909
120587+1
2)
(minus120587
2le 119909 le
120587
2 119904 = 1 2 )
(34)
Meanwhile it is important to note that the number ofaddends in our recurrence representations (26) (28) (30)and (33) is two times less than the number of the addendsin the corresponding cited formulas from [2] So our methodappears to be more economic and effective
Moreover one can get many other representations of theconstants 119870
119903(119903 = 1 2 3 ) and numerical series 119876
2119904(119904 =
1 2 ) from Theorems 5ndash8 putting in particular 119909 = 1205872
or 119909 = 120587 For completeness we will note the main resultsFromTheorem 5 for 119909 = 1205872 and 119909 = 120587 immediately fol-
lows the following
Corollary 9 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)
119904
[1
21198632119904minus1
(120587
2) minus
120587
21198632119904minus2
(120587
2)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)
119904
[1
21198632119904(120587) minus
120587
21198632119904minus1
(120587)]
+
119904minus1
sum
119901=1
(minus1)119901+1
1198792119904minus2119901
1198632119901(120587)
(35)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119879
0= 0)
For 119909 = 120587 one can get (20) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 we obtain the next corollary
Corollary 10 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904= 4119904
(minus1)119904minus1
[1
21198632119904(120587
2) minus
120587
21198632119904minus1
(120587
2)]
+
119904minus1
sum
119901=0
(minus1)119901
1198792119904minus2119901
1198632119901(120587
2)
(36)
(119904 = 1 2 for 119904 = 11198630(119909) = 1)
Similarly from Theorem 6 for 119909 = 1205872 and 119909 = 120587 weobtain respectively the following
Corollary 11 For the Favard constants119870119903 the following recur-
sive representations hold
1198702119904minus2
=4
120587(minus1)119904minus1
21198632119904minus1
(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901minus1
(120587
2)
1198702119904minus1
=2
120587(minus1)119904minus1
21198632119904(120587)
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901
1198632119901(120587)
(37)
where 119904 = 1 2 and for 119904 = 11198630(119909) = 1 119876
0= 0
For 119909 = 120587 one can get (23) too by replacing previously 119904by 119904 + 1
For 119909 = 1205872 from Theorem 6 (the second formula in(28)) we will have also the next analogous corollary
Corollary 12 For numbers 1198762119904 the following recursive repre-
sentations hold
1198762119904=
4119904
4119904 minus 1(minus1)119904
21198632119904(120587
2)
+
119904minus1
sum
119901=1
(minus1)119901+1
1198762119904minus2119901
1198632119901(120587
2)
(38)
(119904 = 1 2 for 119904 = 11198630(119909) = 1 119876
0= 0)
By the same manner from Theorem 7 for 119909 = 120587 and119909 = 1205872 we obtain respectively the formulas for 119870
119903(119903 =
1 2 3 ) different from these inTheorem 1
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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OptimizationJournal of
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International Journal of
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Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
Corollary 13 For the Favard constants 1198702119904minus3
and 1198702119904minus1
thefollowing recursive representations hold
1198702119904minus3
=(minus1)119904
1205871198632119904minus2
(120587) +
119904minus2
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901minus1
(120587)
(39)
(119904 = 2 3 for 119904 = 2 11987011198631(120587) must be canceled) and
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
1198702119901minus1
1198632119904minus2119901
(120587
2)
(40)
(119904 = 1 2 3 for 119904 = 1 11987011198630(1205872)must be canceled)
The remaining cases for 119909 = 1205872 and 119909 = 120587 immediatelylead toTheorem 1 after replacing 119904 by 119904 + 1
From Theorem 8 for 119909 = 1205872 one can get respectivelyother representations for 119870
119903(119903 = 1 2 3 ) different from
these inTheorem 1
Corollary 14 For the Favard constants 119870119903 the following re-
cursive representations hold
1198702119904minus2
= (minus1)119904
1198632119904minus2
(120587
2) +
119904minus2
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus2
(120587
2)
1198702119904minus1
= (minus1)119904minus1
1198632119904minus1
(120587
2) +
119904minus1
sum
119901=1
(minus1)119901
11987021199011198632119904minus2119901minus1
(120587
2)
(41)
where 119904 = 1 2 3 for 119904 = 1 1198630(1205872) = minus1 and for 119904 = 2
11987021198630(1205872) must be canceled
Corollary 15 From the difference 1198792119904minus1198762119904= (1205872)119870
2119904minus1(119904 =
1 2 ) and after replacing 119904 by 119904+1 in the obtained expressionone gets the following formula
1198702119904+1
=(minus1)119904
1205871198632119904+2
(120587) +
119904
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198702119904minus2119901+1
(119904 = 0 1 )
(42)
This is somewhat better than the corresponding formula inTheorem 1 because (2119904 + 1) gt 2(2119904) for 119904 = 1 2
3 Recursive Representations for119903
and Some Fourier Series
Here we will get down to the integral representation andrecursive formulas for the constants
119903 defined in (2) As
one can see they are closely linked with the approximationof the conjugate classes of functions obtained on the baseof the Hilbert transform [1 26] First of all let us note theirrepresentations easily obtained by means of special functions
in (3) as it is shown in [2 514] for the Catalan constant asfollows
119903=
4
120587(1 minus 2
minus(2119904+1)
) 120577 (2119904 + 1) 119903=2119904 119904 = 1 2 3
22minus4119904
120587(120577 (2119904
1
4)minus120577 (2119904
3
4))
=2
120587Γ (2119904)int
infin
0
1199052119904minus1
ch 119905119889119905 119903=2119904minus1 119904=1 2 3
(43)
In the beginning we will prove the following assertion
Theorem 16 For the constants 119903(119903 = 1 2 ) the following
recursive representations hold
2119904minus1
=2(minus1)119904
1205871198602119904minus1
(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901 minus 1)(120587
2)
2119901minus1
2119904=2(minus1)119904
1205871198602119904(120587
2) +
119904minus1
sum
119901=1
2119904minus2119901
(minus1)119901minus1
(2119901)(120587
2)
2119901
(119904 = 1 2 3 0
def= 0)
(44)
Proof The proof of this theorem is based on induction againHowever for completeness we must give somewhat moredetailed considerations at first steps which underline furtherdiscussions
Let us start by the well-known Fourier expansions (seeeg [2 54])
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2
= minus1
21198601(119909) (0 le 119909 le 120587)
infin
sum
119899=1
sin 1198991199091198992
= minus1198611(119909) (0 le 119909 le 2120587)
(45)
For 119909 = 1205872 the first equality gives us immediately (1205874)1= minus(12)119860
1(1205872) At the same time the second equality
in (45) leads to (1205874)1= minus1198611(1205872) Sowe obtain the integral
representation of 1in the form
1= minus
4
1205871198611(120587
2) = minus
2
1205871198601(120587
2) = 1166243616123275 sdot sdot sdot
(46)
We have another integral representation of the same con-stant
1in our paper [8]
Further by integration of both sides of the first equality in(45) we getinfin
sum
V=0
cos [(2V + 1) 119909](2V + 1)3
=1
21198602(119909) +
120587
42
(0 le 119909 le 120587)
(47)
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
from where for 119909 = 1205872 and 119909 = 120587 we have simultaneously0 = (12)119860
2(1205872) + (1205874)
2and minus(1205874)
2= (12)119860
2(120587) +
(1205874)2 and consequently
2= minus
2
1205871198602(120587
2) = minus
1
1205871198602(120587) (48)
Then after the integration of both sides of the secondequality in (45) we obtain
infin
sum
119899=1
cos 1198991199091198993
= 1198612(119909) + 119879
3(0 le 119909 le 2120587) (49)
from where for 119909 = 120587 we have (1205872)2= minus1198612(120587) If we cor-
respond this with (48) we find
2= minus
2
1205871198612(120587) = minus
2
1205871198602(120587
2)
= minus1
1205871198602(120587) = 1339193086109090 sdot sdot sdot
(50)
In order to obtain the integral representation of 3 we
must at first integrate the both sides of (47) as
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)4
=1
21198603(119909) +
120587
421198631(119909) (0 le 119909 le 120587)
(51)
Next it remains to put 119909 = 1205872 as
3=2
1205871198603(120587
2) +
120587
22=2
1205871198603(120587
2) minus 119860
2(120587
2) (52)
As another integral representation of 3 we can get after
integration of both sides of (49)
infin
sum
119899=1
sin 1198991199091198994
= 1198613(119909) + 119879
31198631(119909) (0 le 119909 le 2120587) (53)
On one hand for 119909 = 120587 (53) gives
1198793= minus
1
1205871198613(120587) = 120205690 sdot sdot sdot (54)
On the other hand the same equality (53) for 119909 = 1205872
givessuminfinV=0 sin[(2V+1)1205872](2V+1)4
= 1198613(1205872)minus (12)119861
3(120587)
Then
3=4
1205871198613(120587
2) minus
2
1205871198613(120587) =
2
1205871198603(120587
2) minus 119860
2(120587
2)
= 1259163310827165 sdot sdot sdot
(55)
For the constant 4there are also different ways to receive
its integral representations One of them is based on theintegration of both sides of (53) as
minus
infin
sum
119899=1
cos 1198991199091198995
+ 1198795= 1198614(119909) + 119879
31198632(119909) (0 le 119909 le 2120587)
(56)
For 119909 = 120587 (56) gives (1205872)4= 1198614(120587)+(120587
2
2)1198793 Admit-
ting (54) we will have
4=2
1205871198614(120587) minus 119861
3(120587) = 1278999378416936 sdot sdot sdot
(57)
Another integral representation of 4can be obtained by
integration of both sides of (51) as
minus
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)5
+120587
44=1
21198604(119909) +
120587
421198632(119909)
(0 le 119909 le 120587)
(58)
For 119909 = 1205872 and 119909 = 120587 we get respectively
4=2
1205871198604(120587
2) minus
120587
41198602(120587
2) =
2
1205871198604(120587
2) +
1205872
82
=1
1205871198604(120587) +
1205872
42
(59)
Further after integration of both sides of (56) we get
minus
infin
sum
V=0
sin 1198991199091198996
+ 11987951198631(119909) = 119861
5(119909) + 119879
31198633(119909) (0 le 119909 le 2120587)
(60)
On one hand if we put here 119909 = 120587 and admit (54) we willhave
1198795=1
1205871198615(120587) +
1205872
31198793=1
1205871198615(120587) minus
120587
61198613(120587)
= 103692776 sdot sdot sdot
(61)
On the other hand the same formula (60) for 119909 = 1205872
gives us minus(1205874)5+ (1205872)119879
5= 1198615(1205872) + (120587
3
23
3)1198793 Then
admitting (54) and (61) we obtain
5= minus
120587
41198615(120587
2) +
2
1205871198615(120587) minus
120587
41198613(120587)
= 1271565517671139 sdot sdot sdot
(62)
Next in order to receive another integral representationof 5 we must integrate both sides of (58) So we will have
minus
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)6
+120587
441198631(119909) =
1
21198602(119909) +
120587
421198633(119909)
(0 le 119909 le 120587)
(63)
Putting 119909 = 1205872 here we get
5= minus
2
1205871198605(120587
2) +
120587
24minus1205873
2332
= minus2
1205871198605(120587
2) + 119860
4(120587
2) minus
1205872
22 sdot 31198602(120587
2)
(64)
Now after this preparatory work we can go on the indi-cated procedure which leads us to the general recursive rep-resentations (43) and so complete the proof
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
The previously stated procedure gives us the opportunityto obtain recursive formulas for the multiple integrals119860
119903 119861119903
and 119862119903in (4) for a given 119909
First of all we will note that on the base of induction andwith the help of the first formula in (45) (47) (51) (58) and(63) we can lay down the following
Theorem 17 For themultiple singular integrals119860119903 the follow-
ing recursive representations hold
1198602119904minus1
(119909) = 2(minus1)119904
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
+120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(119909)
1198602119904(119909) = minus 2(minus1)
119904
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(119909)
(65)
119904 = 1 2 3 (0 le 119909 le 120587) 1198630= 1 for 119904 = 0119860
0(119909) =
ln(tan(1199092)) (0 lt 119909 lt 120587) and 0= 0 where
infin
sum
V=0
sin [(2V + 1) 119909](2V + 1)2119904
=1
2
infin
sum
119899=1
sin 1198991199091198992119904
minus
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) sin119909
(1198902119905 + 1)2
minus 41198902119905cos2119909119889119905
infin
sum
V=0
cos [(2V + 1) 119909](2V + 1)2119904+1
=1
2
infin
sum
119899=1
cos 1198991199091198992119904+1
minus
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904+1
=1
Γ (2119904+1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) cos119909
(1198902119905+1)2
minus41198902119905cos2119909119889119905
(66)
By analogy with the previous and by means of the secondformula in (45) (49) (53) (56) and (60) one can get thefollowing
Theorem 18 For the multiple singular integrals 119861119903 the follow-
ing recursive representations hold
1198612119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus1
(119909)
1198612119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198792119904minus2119901+1
1198632119901minus2
(119909)
(67)
119904 = 1 2 3 (0 le 119909 le 2120587) 1198630= 1 for 119904 = 1119879
1119863119896(119896 = 0 1)
declines by definition 1198610(119909) = ln(2 sin(1199092)) (0 lt 119909 lt 2120587)
where (see [2])infin
sum
119899=1
sin 1198991199091198992119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
infin
sum
119899=1
cos 1198991199091198992119904minus1
=1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 minus 1)
1 minus 2119890119905 cos119909 + 1198902119905119889119905
(119909 = 0 119904 = 1 2 3 )
(68)
As similar representations of 119862119903(119909) one can get on the
base of the expansion (see [2 542])infin
sum
119899=1
(minus1)119899cos 119899119909119899
= minus ln(2 cos(1199092)) (minus120587 lt 119909 lt 120587)
1198761= minus ln 2
(69)
Theorem19 For themultiple singular integrals 119862119903 the follow-
ing recursive representations hold
1198622119904minus1
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus1
(119909)
1198622119904minus2
(119909)
= (minus1)119904
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
+
119904minus1
sum
119901=1
(minus1)119901
1198762119904minus2119901+1
1198632119901minus2
(119909)
(70)
119904 = 1 2 3 (minus120587 le 119909 le 120587) for 119904 = 11198620(119909) = ln (2 cos(119909
2)) (minus120587 lt 119909 lt 120587) 1198761119863119896(119896 = 0 1) declines where (see [2])
infin
sum
119899=1
(minus1)119899sin 1198991199091198992119904
=minus1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905 sin119909
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 3 )
infin
sum
119899=1
(minus1)119899cos 1198991199091198992119904minus1
=minus1
Γ (2119904 minus 1)int
infin
0
1199052119904minus2
(119890119905 cos119909 + 1)
1 + 2119890119905 cos119909 + 1198902119905119889119905
(119904 = 1 2 )
(71)
By analogy with the previous if we start from the expan-sion [2 546]infin
sum
V=0
(minus1)V sin [(2V + 1) 119909]
2V + 1=minus1
2ln(tan(120587
4minus119909
2))
(minus120587
2lt 119909 lt
120587
2)
(72)
we can obtain the next theorem
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
Theorem 20 The following recursive formulas hold
int
1205872minus119909
0
1198602119904minus1
(119906) 119889119906 = minus 2(minus1)119904
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
+120587
2
119904
sum
119901=1
(minus1)119901
21199011198632119904minus2119901
(120587
2minus 119909)
int
1205872minus119909
0
1198602119904minus2
(119906) 119889119906 = 2(minus1)119904
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
minus120587
2
119904minus1
sum
119901=1
(minus1)119901
21199011198632119904minus2119901minus1
(120587
2minus 119909)
(73)
119904 = 1 2 3 (minus1205872 le 119909 le 1205872) 1198630(119909) = 1 for 119904 = 1
1198600(119906) = ln tan(1199062) (minus1205872 lt 119906 lt 1205872) where
infin
sum
V=0
(minus1)V sin [(2V + 1) 119909](2V + 1)2119904+1
=1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905
(1198902119905
minus 1) sin119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
infin
sum
V=0
(minus1)V cos [(2V + 1) 119909]
(2V + 1)2119904
=1
Γ (2119904)int
infin
0
1199052119904minus1
119890119905
(1198902119905
+ 1) cos119909
(1198902119905 + 1)2
minus 41198902119905sin2119909119889119905
(119904 = 1 2 )
(74)
The proof of this theorem requires to take integrationfrom 1205872 to 119909 and then to make the substitution 119905 = 1205872 minus 119909
From Table 2 one can see that119860119903(119909) = 119861
119903(119909)minus119862
119903(119909) (0 le
119909 le 120587) 119903 = 1 2 The constants 119903satisfy the following
inequalities [1]
1 lt 1lt 3lt 5lt sdot sdot sdot lt
4
120587lt sdot sdot sdot lt
6lt 4lt 2lt120587
2
(75)
Remark 21 The equalities (49) (53) (56) and (60) outline aprocedure for summing up the numerical series 119879
2119904+1(119904 =
1 2 ) in (2) It leads us to the following
Corollary 22 The following recursive representation holds
1198792119904+1
=(minus1)119904
1205871198612119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198792119904minus2119901+1
(76)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198791declines
For comparisonwewill note also the well-known formula[2 512]
1198792119904+1
= 120577 (2119904 + 1) =1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 minus 1119889119905
=minus1
(2119904)Ψ(2119904)
(1) (119904 = 1 2 )
(77)
The same procedure based on induction and preliminarymultiple integration of both sides of (69) leads us to the next
Corollary 23 The following recursive representation holds
1198762119904+1
=(minus1)119904
1205871198622119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)1198762119904minus2119901+1
(78)
119904 = 1 2 for 119904 = 1 the term (1205872
6) 1198761declines
As in previous let us give the alternate formula [2 513]
1198762119904+1
= (2minus2119904
minus 1) 120577 (2119904 + 1) =minus1
Γ (2119904 + 1)int
infin
0
1199052119904
119890119905 + 1119889119905
(119904 = 1 2 )
(79)
At the end we would like to note also that one can getmany other representations (through multiple integrals) ofthe constants
119903(119903 = 1 2 ) fromTheorems 17ndash20 setting
in particular 119909 = 1205872 or 119909 = 120587 as it is made for the constants119870119903From the difference 119879
2119904+1minus 1198762119904+1
= (1205872) 2119904(119904 = 1 2
) one can get the following formula
2119904=2(minus1)119904
12058721198602119904+1
(120587) +
119904minus1
sum
119901=1
(minus1)119901+1
1205872119901
(2119901 + 1)2119904minus2119901
(119904 = 1 2 )
(80)
which is somewhat inferior to the similar recursion formulainTheorem 16 because (2119904minus1) le 22119904minus2(2119904minus2) for 119904 = 1 2
As an exception we will give the inverse formula of (80)as
1198602119904+1
(120587) = (minus1)119904
119904
sum
119901=1
(minus1)119901+1
1205872119901
2 (2119901 minus 1)2119904minus2119901+2
(119904 = 1 2 )
(81)
If we by definition lay 0= 0 then the equality (80) is
valid for 119904 = 0 too
4 Some Notes on Numerical and ComputerImplementations of the Derived Formulas
We will consider some aspects of numerical and symboliccalculations of the Favard constants119870
119903 singular integrals (4)
and summation of series
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
Table 2 Calculated values of the magnitudes 119860119903(119909) 119861
119903(119909) and 119862
119903(119909) for 119909 = 1205872 120587 by using formulas (65) (67) and (70) respectively
119903 119860119903(1205872) 119861
119903(1205872) 119862
119903(1205872)
0 0346573590279972654709 0346573590279972654711 minus1831931188354438030 minus091596559417721901505 0915965594177219015062 minus2103599580529289999 minus131474973783080624966 0788849842698483749793 minus1326437390660483389 minus089924201634043412749 0427195374320049261594 minus0586164434174921923 minus041567176475313624651 0170492669421785676365 minus0200415775434410589 minus014636851588819525831 0054047259546215330636 minus0055999130148030908 minus004177075772178286763 0014228372426248040427 minus0013242890305882861 minus001003832077074076074 0003204569535142100008 minus0002716221784223860 minus000208540018388810697 0000630821600335753299 minus0000492033289602442 minus000038173093475665008 00001103023548457920610 minus0000079824623761109 minus000006247448634651347 00000173501374145951811 minus0000011727851308441 minus000000924763886476940 00000024802124436716712 minus0000001574609462951 minus000000124968150024699 000000032492796270360r 119860
119903(120587) 119861
119903(120587) 119862
119903(120587)
0 06931471805599453094171 0 0 02 minus42071991610585799989 minus210359958052928999945 2103599580529289999453 minus66086529882853881226 minus377637313616307892720 2832279852122309195404 minus63627527878802461560 minus392286552530160912232 2439887262578637033685 minus45591894893017946564 minus295428020294335768590 1604909286358436970496 minus26255391709472096141 minus176271265891422894113 0862826512032980672957 minus12684915655900274347 minus087472015815662857450 0393771407433398860168 minus05287656648068309189 minus037237211738940254175 0156393547417428377149 minus01940031559506448617 minus013897143291435930434 00550317230362855573910 minus00635985732918145806 minus004620733330628432785 00173912399855302527711 minus00188489390019113804 minus001385968931040814707 00049892496915032333612 minus00050985327885974909 minus000378779048977775268 000131074229881973820
Let us take only the first119898minus1119898 le 119904+1 terms in the finalsums in the right-hand side of the formulas (7a) and (7b) anddenote the remaining truncation sums by
1198781119898
=
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
1198782119898
=
119904
sum
119895=119898
(minus1)119895minus1
(2119895 minus 1)(120587
2)
2119895minus1
1198702119904minus2119895+1
(82)
respectively
Theorem 24 For 119898 ge 2 and any 119904 gt 119898 the following esti-mates for the truncation errors (82) hold
100381610038161003816100381611987811198981003816100381610038161003816 = 119874(
1205872119898
(2119898))
100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(83)
where Landau big119874 notation is used For the Landau notationsee [27]
Proof By means of the inequalities (25) it is easy to obtain
100381610038161003816100381611987811198981003816100381610038161003816 =
10038161003816100381610038161003816100381610038161003816100381610038161003816
119904
sum
119895=119898
(minus1)119895minus1
120587119895119901
(2119895)1198702119904minus2119895+1
10038161003816100381610038161003816100381610038161003816100381610038161003816
le120587
2
119904
sum
119895=119898
1205872119895
(2119895)=120587
2
1205872119898
(2119898)
times (1 +1205872
(2119898 + 1) (2119898 + 2)+ sdot sdot sdot +
1205872119904minus2119898
(2119898 + 1) sdot sdot sdot (2119904))
le1205872119898+1
2 (2119898)(1 +
10
56+102
5262+ sdot sdot sdot )
le1205872119898+1
2 (2119898)(1 +
1
3+1
32+ sdot sdot sdot )
=1205872119898+1
2 (2119898)
3
2le31205872119898
(2119898)
(84)
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
m = 12 Array [K m] K1=120587
2 d [n
minus xminus] =
xn
n
For [s = 1 s le m2 s + +
K2s+1 =
1
2((minus1)
sd [2s + 1 120587] +s
sum
p=1
K2s+1minus2p
(minus1)pminus11205872p
(2p))
K2s = (minus1)
sd [2s120587
2] +
s
sum
p=1
K2s+1minus2p
(minus1)pminus1(120587
2)
2pminus1
(2p minus 1)
]
For [s = 1 s le m s ++ Print [Ks N[Ks 30]]]
Algorithm 1Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constantsby recursive representations (7a) and (7b)
Consequently |1198781119898| = 119874(120587
2119898
(2119898)) In the sameway forthe second truncation sum we have
100381610038161003816100381611987821198981003816100381610038161003816 le 2(
120587
2)
2119898minus11
(2119898 minus 1)
so 100381610038161003816100381611987821198981003816100381610038161003816 = 119874((
120587
2)
2119898minus11
(2119898 minus 1))
(85)
By the details of the proof it follows that the obtainedrepresentations did not depend on 119904 for any 119898 lt 119904 The apriori estimates (83) can be used for calculation of 119870
119903with a
given numerical precision 120576 in order to decrease the numberof addends in the sums for larger 119904 at the condition119898 lt 119904
Remark 25 The error estimates similar to (83) can be estab-lished for other recurrence representations in this paper too
In Algorithm 1 we provide the simplest Mathematicacode for exact symbolic and numerical calculation withoptional 30 digits accuracy of the Favard constants 119870
119903for
119903 = 1 2 3 119898 based on recursive representations (7a) and(7b) for a given arbitrary integer119898 gt 0 The obtained resultsare given in Table 1 We have to note that this code is not themost economic It can be seen that the thrifty code will takeabout 21198982 + 8119898 or 119874(1198982) arithmetic operations in (7a) and(7b)
The basic advantage of using formulas (7a) and (7b) is thatthey contain only finite number of terms (ie finite number ofarithmetic operations) in comparisonwith the initial formula119870119903= (4120587)sum
infin
]=0((minus1)](119903+1)
(2] + 1)119903+1
) (119903 = 0 1 2 )which needs the calculation of the slowly convergent infinitesum It must be also mentioned that inMathematicaMapleand other powerful mathematical software packages theFavard constants are represented by sums of Zeta and relatedfunctions which are calculated by the use of Euler-Maclaurinsummation and functional equations Near the critical stripthey also use the Riemann-Siegel formula (see eg [28A94])
Finally we will note that the direct numerical integrationin (4) is very difficult This way a big opportunity is given
by Theorems 17 18 and 19 (formulas (65) (67) and (70))for effective numerical calculation of the classes of multiplesingular integrals 119860
119903(119909) 119861
119903(119909) and 119862
119903(119909) for any 119909 Their
computation is reduced to find the finite number of theFavard constants V for all V le 119903 and the calculation of oneadditional numerical sum Numerical values for some of thesingular integrals are presented in Table 2
5 Concluding Remarks
As it became clear there are many different ways and well-developed computer programs at present for calculation ofthe significant constants inmathematics (in particular Favardconstants) and for summation of important numerical seriesMost of them are based on using of generalized functions asone can see in [2ndash5]Nonetheless we hope that our previouslystated approach through integration of Fourier series appearsto be more convenient and has its theoretical and practicalmeanings in the scope of applications in particular forcomputing of the pointed special types of multiple singularintegrals The basic result with respect to multiple integralsreduces their calculation to finite number of numerical sums
Acknowledgments
This paper is supported by the Bulgarian Ministry of Edu-cation Youth and Science Grant BG051PO00133-05-0001ldquoScience and businessrdquo financed under operational programldquoHuman Resources Developmentrdquo by the European SocialFund
References
[1] N Korneıchuk Exact Constants in Approximation Theory vol38 chapter 3 4 Cambridge University Press New York NYUSA 1991
[2] A P Prudnikov A Yu Brychkov and O I Marichev Integralsand Series Elementary Functions chapter 5 CRC Press BocaRaton Fla USA 1998
[3] J Favard ldquoSur les meilleurs precedes drsquoapproximation de cer-taines classes de fonctions par des polynomes trigonometri-quesrdquo Bulletin des Sciences Mathematiques vol 61 pp 209ndash2241937
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Abstract and Applied Analysis
[4] S R Finch Mathematical Constants Cambridge UniversityPress New York NY USA 2003
[5] J Bustamante Algebraic Approximation A Guide to Past andCurrent Solutions Springer Basel AG Basel Switzerland 2012
[6] S Foucart Y Kryakin and A Shadrin ldquoOn the exact constantin the Jackson-Stechkin inequality for the uniform metricrdquoConstructive Approximation vol 29 no 2 pp 157ndash179 2009
[7] Yu N Subbotin and S A Telyakovskiı ldquoOn the equality ofKolmogorov and relative widths of classes of differentiablefunctionsrdquoMatematicheskie Zametki vol 86 no 3 pp 432ndash4392009
[8] I H Feschiev and S G Gocheva-Ilieva ldquoOn the extension ofa theorem of Stein and Weiss and its applicationrdquo ComplexVariables vol 49 no 10 pp 711ndash730 2004
[9] R A DeVore and G G Lorentz Constructive Approximationvol 303 Springer Berlin Germany 1993
[10] V P Motornyı ldquoOn sharp estimates for the pointwise approx-imation of the classes 119882
119903
119867120596 by algebraic polynomialsrdquo
Ukrainian Mathematical Journal vol 53 no 6 pp 916ndash9372001 Translation fromUkrainsrsquokyiMatematychnyi Zhurnal vol53 no 6 pp 783ndash799 2001
[11] W Xiao ldquoRelative infinite-dimensional width of Sobolev classes119882119903
119901(119877)rdquo Journal of Mathematical Analysis and Applications vol
369 no 2 pp 575ndash582 2010[12] P C Nitiema ldquoOn the best one-sided approximation of func-
tions in themeanrdquo Far East Journal of AppliedMathematics vol49 no 2 pp 139ndash150 2010
[13] O L Vinogradov ldquoSharp inequalities for approximations ofclasses of periodic convolutions by subspaces of shifts of odddimensionrdquo Mathematical Notes vol 85 no 4 pp 544ndash5572009 Translation fromMatematicheskie Zametki vol 85 no 4pp 569ndash584 2009
[14] A V Mironenko ldquoOn the Jackson-Stechkin inequality foralgebraic polynomialsrdquo Proceedings of Institute of Mathematicsand Mechanics vol 273 supplement 1 pp S116ndashS123 2011
[15] V F Babenko and V A Zontov ldquoBernstein-type inequalitiesfor splines defined on the real axisrdquo Ukrainian MathematicalJournal vol 63 pp 699ndash708 2011
[16] G Vainikko ldquoError estimates for the cardinal spline interpola-tionrdquo Journal of Analysis and its Applications vol 28 no 2 pp205ndash222 2009
[17] O L Vinogradov ldquoAnalog of the Akhiezer-Krein-Favard sumsfor periodic splines of minimal defectrdquo Journal of MathematicalSciences vol 114 no 5 pp 1608ndash1627 2003
[18] L A Apaıcheva ldquoOptimal quadrature and cubature formulasfor singular integrals with Hilbert kernelsrdquo Izvestiya VysshikhUchebnykh Zavedeniı no 4 pp 14ndash25 2004
[19] F D Gakhov and I Kh Feschiev ldquoApproximate calculationof singular integralsrdquo Izvestiya Akademii Nauk BSSR SeriyaFiziko-Matematicheskikh Nauk vol 4 pp 5ndash12 1977
[20] F D Gakhov and I Kh Feschiev ldquoInterpolation of singularintegrals and approximate solution of the Riemann boundaryvalue problemrdquo Vestsı Akademıı Navuk BSSR Seryya Fızıka-Matematychnykh Navuk no 5 pp 3ndash13 1982
[21] B G Gabdulkhaev ldquoFinite-dimensional approximations of sin-gular integrals and direct methods of solution of singular inte-gral and integro-differential equationsrdquo Journal of Soviet Math-ematics vol 18 pp 593ndash627 1982
[22] V P Motornyı ldquoApproximation of some classes of singularintegrals by algebraic polynomialsrdquo Ukrainian MathematicalJournal vol 53 no 3 pp 377ndash394 2001
[23] E Vainikko and G Vainikko ldquoProduct quasi-interpolationin logarithmically singular integral equationsrdquo MathematicalModelling and Analysis vol 17 no 5 pp 696ndash714 2012
[24] H Brass and K Petras Quadrature Theory The Theory ofNumerical Integration on a Compact Interval vol 178 AmericanMathematical Society Providence RI USA 2011
[25] E W Weisstein ldquoFavard constantsrdquo httpmathworldwolframcomFavardConstantshtml
[26] A G ZygmundTrigonometric Series vol 1 Cambridge Univer-sity Press 2nd edition 1959
[27] D E Knuth The Art of Computer Programming Vol 1 Funda-mental Algorithms Section 1211 Asymptotic RepresentationsAddison-Wesley 3rd edition 1997
[28] S Wolfram The Mathematica Book Wolfram Media Cham-paign Ill USA 5th edition 2003
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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