On Stirling's Numbers(1), by Charles JORDAN, Budapest. 1 ...

On Stirling's Numbers(1),

by

Charles JORDAN, Budapest.

1. The numbers introduced by Stirling in 1730 (Methodus differentialis) and subsequently called "Stirling's Numbers ", are of the greatest utility in mathematical calculus. This however has not

been fully recognised ; the numbers have been neglected, and are very seldom used. This is especially due to the fact that different authors have reintroduced them under different notations, often not knowing, or not mentioning, that they dealt with the same numbers. Stirling's Numbers are as important or even more so than Bernoulli's Num-

bers ; they should be placed in the centre of the Calculus of Finite Differences.

In this paper different known formulae in which they occur are assembled, many new formulae deduced, and two small tables given. 2. Let us resume the different notations used here:-

Abbreviations :

Factorial function of the n-th degree :

Binomial coefficient of the n-th degree :

Stm:

According to the principles of the Calculus of Differences, the value of f(x) at the upper limit, say f(n), is not included in the sum.

(1) As this paper is a monograph on St it linq's Numbers, it necessarily contains some known results, but most of the formulae are new. The demonstrations of the old formulae are greatly shortened by my new methods. For instance the demon-stration of the formulae (74) and (76), to which Schlomi1ch devotes several pages are given in a few lines.

ON STIRLING'S NUMBERS. 255

3. Let (x)n be expanded into Maclaurin's series,

(1)

Let us call the number in brackets above Stirling's Number of the First Kind, and let it be denoted by S1n, so that we shall have

(2) From equation (1) it results that, S0n=0; Sn+mn=0 ; and S11=1.

The other numbers also could be calculated by (1), but we will give a shorter way. Leibniz's formula giving higher derivatives of a

product, is the following :

Let us write then

we have

The division by v! will give, if we put into the result x=0,

(3) This is an equation of partial differences of the first order. It

is true that the solution of this equation is unknown, but starting

from S01=0 and S11=1 we can easily calculate, step by step, the num-

bers Svn. This will give the following Table.

Table of Stirling's Numbers of the First Kind Svn.

The equation of differences (3) shows that the sign of Svn is iden-tical with that of If we put x=1 into (2), we get

256 C.JORDAN :

(4)

Viz. the sum of the numbers in each line of the Table is equal to zero. This can servo to check the Table. To have a second check, we put x=-1 into equation (2). It results that

(5)

i.e., the sum of the absolute values of the numbers in the lines n of

the Table S1n is equal to n!.

Difference equation (3) cat be solved in some particular cases.

1. Putting ƒË=1, equation (3) will be

taking account of S11=1, the solution of this is

2. Putting ƒË=n+1, equation (3) gives

3. Putting ƒË=n, we have

Since S12=-l, therefore the solution of the above equation is

Schlomilch has given the following general solution of equa. tion (3)

(6)

where Gun denotes the Stirling Number of the second kind ; the latter is expressed by a single sum, therefore Schlomilch's solution contains a double sum ; those are difficult to manage, but there is no known solution of (3) expressible by a single sum.

4. Generating-Function. From (2) it follows that

(7)is the generating-function of the numbers S1n; that is, in the expan-

sion of (7), in powers of x, the coefficient of zƒË is equal to SƒËn. This

coefficient may be expressed in two different manners. First :

(8)


The sum figuring in the second member is extended to all

combinations, without repetition, of order n-ƒË, of the numbers

1,2,3,...., (n-1).

Second :

(9)

The sum in the second member should be extended to all

combinations, without repetition, of order ƒË-1, of the numbers

1,2,3, .... , (n-1).

We will show later (76) that

(10)

On the other hand, starting from the known series of log (1+t) and applying m times Cauchy's rule of multiplication of infinite series, we get

(11)

From (10) and (11) it results that

(12)

In the sum contained in the second member, r1 takes every value

of 1, 2,3,...., with repetition, but satisfying the condition that

r1 +r2+ .... +rm=n. It will be shown that lira , moreover as the series in

(10) is alternate, therefore this series is convergent for all values of

t•…1. Let us put t=1; then we have

(12')

5. Multiplying the difference equation (3) by(-1)m+ƒË+1/n! we get

From this we conclude

(13)

Starting from the generating function. (7), we get by integration:

268 C. JORDAN :

It has been shown that this integral is also equal to n! bn, where

bn is a coefficient of the Bernoulli polynomial of the second kind(1) Therefore

(14)

If we denote the power-moments of order ƒË by MƒË and the

factorial-moments of order n by Mn, so that

Then the latter can be expressed by power-moments and Stirling

numbers

(15)

6. Derivatives expressed by differences. The expansion of f(x) in Newton series gives,

But since

we have

(16)

7. It can be shown that the Stirling numbers Sn-mn are poly-nomials of n, of degree 2m, so that they can be written in the following manner :

(17)

The numbers Cml are given by the difference equation :

(18)

(1) Char1es Jordan, Sur les polynomes analogues aux polynomes de B er. nou11i et sur des formules de sommation analogues a celle de Mac1aurin-Eu1cr. Acta Scientiarum Mathematiearum, Szeged 1929.


This equation can be solved in some particular cases, from

obtained above ; it follows that C10= -1, C11=0, C12=0, etc. Starting

from these numbers we get, step by step,

The solution of (18), for s=m-1 is,

and for s=0

A table of the other numbers Cmlt can be rapidly calculated by aid of the difference equation (18).

Table of Cml.

The Table can be checked by

From (17) it follows that

(19)

From this we can deduce an asymptotic value of Sn-mn.

If we denote the Stirling polynomial(1) of degree m by ƒÕm(x),

this can be expressed in the following manner for x>O

(20)

and putting into it our value of the Stir1ing number given by (17), we obtain the expansions of the Stirling polynomial in factorials (2):

(1) N. Nielsen, Gamma Funktionen, Leipzig, 1906, p. 71. (2) It might be useful to remark, that the expansions of a function in fac-

torials and in Newton-series are often preferable to the expansion in power- series, especially if sums or differences are to be calculated.

260 C.JORDAN:

(21)

8, Stirling's numbers can be obtained directly from their de-finition (1) by carrying out the indicated derivations. It is easy to

see that we get

(22)

The sum in this expansion must be extended to all combinations of

ul=1, 2, 3....,n, without repetition. These sums may be expressed

by other stuns with no restriction. Let us denote by

Then we may write the sum above in the following manner

(23)

where the expresions auƒÊƒÉƒÌ are numbers still to be determined ;ƒ°ƒÉ;

takes every value of l , 2, 3,...., but so that ƒ°iƒÉl=m-1. If ƒ°ƒÉi is

given we arrange the terms in order of magnitude of ƒÉ1, the first

being that in which ƒÉ1 is the greatest ; in this term we put u=1, in

the following ƒÊ=2, and so on. If there are several terms in which

ƒÉ1 has the same value, we arrange these in order of magnitude of

ƒÉ2, and so on.

The number of the terms of (23) in which ƒ°ƒÉi is the same, is

equal to the number of partitions of m-1, with repetition, into

ƒ°ƒÉi, parts. The total number of the terms in (23) is equal to the

number of all partitions of m-1, with repetition.

It can be shown that

(24)

and therefore,

the other aƒÊƒ°ƒÉi are obtained in a more complicated manner.

Moreover according to (4) the sum of all the coefficients aƒÊƒ°ƒÉi in

(23) is equal to zero.

9. Limits of certain expressions of stirling's numbers for n=•‡.

Considering that


where the logarithm is a Napier's logarithm, and C the Euler's constant 0,577.... ; it results from (23) that

(25)

This gives an asymptotic value for Smn. In the Table of Stirling's mumbers we see that in each row the

numbors first increase, and having attained their largest value, they decrease again. From (23) it follows that

(26)

Hence the corresponding asymptotic value of the relation is

(27)

Therefore Sxn is approximately the greatest term of the n-th row ; if for m=x-1 this relation is greater than one, and if for m=X it is smaller. Consequently x is equal to the greatest integer con-tnined in (log n+C+1).

Notwithstanding the small numbers contained in our Table, this rule is sufficiently verified. In the row n=100, the greatest term will be S6100. From (23) we can deduce moreover

(28)

which gives an asymptotic value of this relation. 10. Different formulae containing Stirling's numbers of the

first kind. Let us write

It follows that

The coefficients, of xm in both expressions being identical, it

results that

(29)

262 C.JORDAN:

As the expression

is equal to zero if k is different from ƒË, and equal to (-1)k if ƒË=k,

it can be useful for inverting certain sums. For instance if Ive

multiply both members of (29) by and sum the result, m

varying from k to ƒË+1, all the terms of the second member will

vanish except the term in which ƒË=k, so that we shall have

It is easy to see that if we put m+1 instead of m in the first,

term of the first member, its value will not change, and then the

two terms can be reunited. Moreover if we write in the result ƒË

instead of k+l and n instead of n-1, it follows that

(30)

Let us repeat the above operation, multiplying (30) by (-1)ƒË(ƒËk)

and summing, ƒË varying from k to n+1 ; if we write moreover in

the result ƒË instead of ƒË+ 1 and n instead of n+1, we have

(31)

Formulae (29) and (30) give interesting particular cases, for instance

(29')

(30')

11. Stirling's numbers of the second kind. Let xn be expanded into a Newton series :

(32)

We will call the number in the brackets, Stirling's number of the

second kind, and denote it by GƒËn; hence we have :

(33) This equation gives immediately G0n=0 and Gn+mn=0. Moreover,

if we put into it x=1, we get


The formulae (32) would also give the other numbers GƒËn, but

we will calculate thein in a shorter way.

The well known expression giving the higher differences of a

product is the following :

dividing both members of this equation by ƒË! and putting ,x=O, we

have, using the above notation

(34)

Starting from the values G01=O and G11=1 this difference equa=

tion gives immediately the following table :

Table of Stirling's Numbers of the Second Kind Glu.

12. From the difference equation it follows that GƒËn>0. If we

put in (33) x=-1 we get the expression :

(35)

which can be used to check the Table. It is possible to obtain an analogous expression starting from

the difference equation (34),

(36)

Solution of the differenee equation. For ƒË=n+1 it gives Gnn=1,

and for ƒË=n we get Gn-1n=(n2)

The general solution of this equation is

264. C.JORDAN:

(37)

This result can be deduced immediately from the definition (32)

of the numbers GƒËn. After a well known formula the m-th difference

of f(x) for x=0 is given by

This will lead to formula (37). Limit of certain expressions of the Stirling's numbers. From (37) we deduce

(38)

and

(39)

and moreover

(40)

The above three formulae give corresponding asymptotic values. 13. It can be shown that the Stirling's numbers such as Gn-mn

are polynomials of n of degree 2m; hence they can be written in the following way:.

(41)

whore the numbers Cms are given by the differonu equation

(42)

From it results that

Starting from these values equation (42) can be solved in some particular cases


A table of the other values can easily be calculated.

Table of Cmt

Equation

(43)

may serve for checking the results. From (41) we conclude that

(44)

The values of the Stirling polynomial for negative integer argu-

ments can be given by the Stirling numbers of the second kind

(44')

From this formula and from (20) limits of the Stitling poly-nomial can be deduced by using the limits obtained for the Stirling numbers ; e.g. for x>0 we get

14. Relation between the Stirling numbers of the two kinds. Let us write

As the coefficients of (x)n in both members must be identical, it results that

(45)

if i is different from n, and that this expression is equal to one if

i=n. Starting from xn we get in the same way a similar relation

(46)

if i is different from n ; and equal to one if i=u.

266 C.JORDAN:

The latter two equations permit the inversion of certain sums.

For instance if we multiply both sides of (12') by Gkm, and sum from

m =k to m=•‡, it results, taking account of (45), that

(47)

In the same mannor we get front (14)

(48)

From (15) we may obtain in the same way the power-moments expressed in function of factorial-moments

(49)

As the factorial-moments are more rapidly calculated than the

power-moments(1), this formula becomes useful. By this method we obtain also the expression of differences in function of the derivatives ; starting from (16) we have

(50)

If we write formula (30) in the following manner :

multi plying it by Gxm and summing from x=m-1 to x=n+1 it

results that

(51)

Equation (31)

multiplied. by Gxn and summed from x=m +1 to x=n+1 gives

(52)

Writing equation (36) in the following way :

(1) Ch. Jordan, Sur la determination de la tendance seculaire, Journal de la Societe Hongroise de Statistique, 1929.


multiplying it by Sx+1n and summing from x=m to x=n we get

(53)

Let us start now from

Since the coefficients of (x)l in both members must be identical,

therefore if i is different from n and from n+l, it results that.

(54)

If i=n, the sum (54) is equal to n, and if i=n+1, it is equal to one.

15. Generating-function of the Stirling number of the second

kind, with respect to n. If this function is denoted by u=u(ƒË, t), then,

To determine the function u, we shall use Laplace's method,

which starts from the difference equation of the numbers. The pro-

ceeding states that : if u=u(ƒË, t) is the generating function of GƒËn,

then u=u(ƒË+1, t) will be generating function of GƒË+1n, and finally 1/t

. u(ƒË+1, t) will be that of GƒË+1n+1.

Since the generating-function of a quantity satisfies the dif-ference equation, which determines the quantity itself ; therefore let us put the values obtained above into equation (34); it results that

This is a linear difference equation of the first order with a single

variable ƒË, as t is. only a parameter, whereas (34) contained two vari-

ables ƒË and n. This equation is easily solved, and we find

(55)

where ƒÖ(t) is an arbitrary function of t, determined by the initial

values. As G1n=1, consequently the generating-function for ƒË=1 is

u(1, t)=t/(1-t), but (55) gives u(1, t)= t.ƒÖ(t)/(1-t), therefore w(t)=1,

so that the generating function which was to be determined is :

268 C.JORDAN:

(56)

The second member of (56) can be Written as follows :

tƒË(1+t+t2+t3+ ....)(1+2t+4t2+8t3+.... )....(1+ƒËt+ƒË2t2+ƒË2t2+ƒË3t3+....).

Hence GƒËn,the coefficient of tn in this product, will be

(57)that is the sun of the products of all combinations, without repeti-

tion of order n-ƒË of the numbers 1, 2, 3, .... , ƒË.

We will show later (74) that

(58)

If we expand the first member of (58) in powers of t and write that in both members the coefficients of tn are identical, we get

(59)

In the sum contained in the second member r1 takes every value of 1,2,3,...., with repetition, but so that rl+r2+....+rm=u. Putting t=1 in (58), we have .

(60)

If we put in the generating function (56) t=-1/z, we obtain the

z expansion of an inuerse factorial into a series of inverse powers

(62)

Multiplied by SkƒË, and summed from ƒË=k to ƒË=•‡, it results that

(63)

This is the expansion of an inverse power into a series of inverse factorials. (This was found out by Stirling himself, v. the paper cited above.) The expansion of the generating function (56) gives the expres-sion of the Stirling number of the second kind obtained before (37). Hence, by Laplace's method we have solved the difference equation (34), of two independent variables.

16. Operation Q. In mathematical analysis an operation often


occurs, which we will denote by O, and define by the relation

(64)

If it is repeated, it follows that

and so on. It is easy to see that

(65)

The numbers Ct above are independent of the function on which the

operation is executed. For their determination we can choose the

most convenient function. Let f(x) =xƒÉ, then we have

Putting these values into (65) and dividing by xƒÉ we get

taking account of (33) we conclude that CƒË=GƒËn and therefore

(66)

This is the expression of the On operation in function of derivations. Let us multiply (66) by Snm and sum from n=1 to n=m+1 it

results that

(67)

This formula gives the m-th derivative in function of the O operation.

If u(x) is the generating-function of f(z), that is, if

them the Q operation gives the power-moments of f(z). In fact we have

This formula is useful in the Calculus of Probability.

17. Operation ƒÕ. In the calculus of finite differences an ana-

logous operation occurs, which we denoto by ƒÕ and define by

270 C.JORDAN:

(68)The operation repeated gives

(69)

The numbers Ci being independent of the function on which

the operation is executed ; to calculate them, we can choose the most

convenient function. Lot f(x)=(x+ƒÉ-1)ƒÉ. Then,

Putting these values into (69) and dividing by (x+ƒÉ-1)ƒÉ we get

Therefore according to (33) we have and finally

(70)

This gives the ƒÕn operation in function of that of the differences.

If we multiply this equation by Snm and sum from n=1 to n=m+1

we get

(71)

which gives ƒ¢m in function of the ƒÕ operations.

18. The operation ƒ¢mD-m. Sometimes a function f(x) must be

successively integrated m times and then the m-th difference of the

result taken. We denote this operation by ƒ¢mD-m, where D-1 stands

for the sign •ç of the indefinite integral.

Let us expand f(x) into a Mac1aurin series

The above mentioned operation carried out on f(x) gives

and if we put x=0, into this expression we have


(71)

Remark 1. The Operation of ƒ¢mD-m is equivalent to that of

ƒ¢ D-1 repeated in times : ƒ¢mD-m=(ƒ¢D-1)m.

Remark 2. We have also, as can be shown :

(72)

an d

(73)

Example. Let f(x)=ext, then it results that

and from (71) it follows that(1)

(74)

This equation has already been used in this paper (58).

The operation Dmƒ¢-m is the inverse operation of the former ;

that is, the m-th derivative of the m-th indefinite sum is to be deter-

mined, ƒ¢-1 stands here for ƒ°.

Let us expand f(x) into a Newton series

The operation Dmƒ¢-m carried out gives

and if we put x=0,

(75)

Example. Let f(x)=(1+t)x, then ƒ¢nf(0) =tn and

These values put into (75) give (1)

(1) The constant which enters into the equation by integration can be neg-

lected; in fact, in D-mf(x) it would only form a polynomial of degree m-1, whose

m-th difference is equal to zero . Same remark in the case of ƒ¢-mf(x).

272 C.JORDAN:

(76)

This important relation has already been used in this paper (10). 19. The Bernoulli polynomial of the first kind of degree

n+1 can be defined by the following two relations

Frotrt these follows the symbolical equation of the polynomial

in the expansion of which, instead of B1 the Bernoulli number B1 must be put. The latter can be calculated by aid of the Stirling numbers

(77)

The Bernoulli polynomia1 of the second kind of degree n+1can be defined by the relation :

From these, the Newton expansion of the polynomial may be

deduced

where the coefficients bm are given by (14)

(78)

The Euler polynomial En(x) of degree n is defined by the relations

The first relation expresses that xn/n! is the mean of the Euler

(1) The symbol M signifies the mean, in the manner written in this equation,


polynomial of degree n. The polynomial may be written in a symbolical manner, like the Bernoulli polynomial of the first kind:

where the numbers en are given by

(79)

20. Different formulae containing Stirling's numbers of the

second kind. Let us start from

But on the other hand, we have also

The coefficients of(x)ƒË in both expressions must be identical, therefore

(80)

Multiplying it by (-1)n(kn) and summing from u=ƒË to n=k+1,

we have, if we write in the result ƒË-1 instead of ƒË and n-1 instead

of n,

(81)

Summing the difference equation (34) from ƒË=1 to ƒË=n+2, we

get

(82)

If we multiply the equation (29') by Gnm and sum from n=1 to n=m+1, we get

(82')

21. Example of application of the deduced formulae . It is

274 C.JORDAN:

known, that the total number of the arrangements with repetition

of order n of n elements is equal to nn.

Let us denote by ƒÕ(n,k) the number of these arrangements in

which only k different elements occur. We want to determine this

number. It is obvious that ƒÕ(n,1)=n. Moreover ƒÕ(n,n) being

identical to the number of permutations of n different elements

we have ƒÕ(n, n)=n!.

To have ƒÕ(n, 2) let us remark that from n elements we ean

choose two elements in different ways. If the first of these ele-

ments occurs in the arrangement n-i times, and the second i times

the corresponding number will be equal to ; and the total

number of the considered arrangements is

From formula (80) it follows that

(80)'

putting into it k=1, it results

To obtain ƒÕ(n, 3) let us remark that we may choose three elements

out of n elements in (n3) different ways. If the first of these ele-

ments occurs in the arrangement (n-i) times, the second (i-ƒË)

times and the third ƒË times, the corresponding number will be

n!/(n-i)!(i-ƒË)!ƒË! and the total number of the corresponding

arrangements is

and putting into (80)' k=2 we get

Continuing in the same manner we obtain the required number


For instance the number of the arrangements of three elements

of the third order, with repetition, in which only two different

elements occur is ƒÕ(3, 2)=3.2 G23=18. In fact 1 and 2 will give

the following 6 arrangements : 112, 121, 211, 122, 212, 221. As 1, 3

and 2, 3 would each give six arrangements again, the total number

will be 18.

Remark 1. If we expand nn in factorials by formula (33) we

have

which gives a decomposition of the number of arrangements with

repetition of order n of n elements into groups containing only one,

two, three and so on n different elements.

Remark 2. Putting Gkn=ƒÕ(n, k)/ (n)k into the difference equation

(31) we obtain a difference equation for the numbers ƒÕ(n, k)

22. Changing the interval. Sometimes in a system of differ-

ences, in which the increment is h, a function f(x) is given by its

differences ƒ¢mnf(0) for m= 0, 1, 2, 3, .... , and it is necessary to express

the function in another system of differences, where the increment

is k, that is, the differences of the function are required for x= O in

this system, i.e. ƒ¢mkf(0), for m=0,1,2)....

It can be shown that these latter quantities may be expressed

by the former in the following way :

where ƒÖ=k/h and

(83')

From this it results that

(84)

Equation (83') permits us to determine a table of the quanti-ties

Example 1. If we divide formula (83) by km and determine the

276 C.JORDAN:

limit for k=0, we get the known expression of the derivatives given

by differences.

Example ,2. Lagrange's polynomial, in the particular case of

xƒË=ƒË/n, may be written in the following way

It is easy to express the differences of this quantity for x=0, in a

system of differences, where the increment is h=1/n. In any other

system it would be complicated. We find

if ƒË•…m•…n ; in any other case it is equal to zero.

By our first formula, we can now express the differences of

LnƒË(x) for x=0 in a system of differences whore the increment is k=1.

For this we put in the formula ƒÖ=n, so that

Now we can write the Newton expansion of LnƒË(x) in the system of increments k=1:

This, multiplied by dx and integrated from x=0 to x=1 gives, as is

well known, the Cotos number CnƒË.

Since we have seen that

where bs is a coefficient in Bernoulli's polynomial of the second kind (No. 8.), we have

For instance, in the particular case n=3 and ƒË=1


23. The Stirling number of the second kind can be obtained by aid of a problem of Probability.

Let an urn be given, which contains the numbers 1, 2, .... ,m. Let us draw successively n numbers, putting back every time the number drawn, before the next drawing. The probability that in

drawings the number one should occur r1 times, the number two r2 times, and so on, finally the numbor m should occur rm times, is, according to the generalised Bernoulli theorem of repeated trials, equal to

Let us now calculate the probability that in n drawings every

number out of 1,2,....,m, should be drawn at least once. This is

given by the theorem of total probabilities :

(85)

The sum above is extended to every value of r1=1, 2, 3, .... , m, with repetition, satisfying the condition that r1+r2+....+rm=n. Moreover according to (59) this sum is equal to m! Gmn, as has been

put in (85). On the other hand this probability can be determined by using

Poincares theorem, which gives the probability that in n trials every event should occur at least once. According to the theorem this probability is obtained by subtracting from one the probability that some one of the events should occur ; this is

then adding the probability that any two of the events should occur

then again subtracting the probability that any three of the events

should occur

and so.Finally we have

278 G.JORDAN:ON STIRLING'S NUMBERS.

Hence, from (85) it follows that

a result already obtained in other ways (36).

On Stirling's Numbers(1), by Charles JORDAN, Budapest. 1 ...

Documents

Transcript of On Stirling's Numbers(1), by Charles JORDAN, Budapest. 1 ...