Triangle Construction for Coefficients of Polynomials Represent

the Multiple of (x+j)Faez Hassan Ali Al_Azawi.

M.Sc. in Mathematic /College of Science/Al_MustansiriyahUniversity

Email: faezhasan@yahoo.com

AbstractIn this paper a new triangle of the polynomials

coefficients produced from the multiple of the term (x+j)will introduced. Depending on this triangle we will suggesta fast method to solve interpolation and numericaldifferentiation problems which are based on Newton-forwardformula.

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ن6 وت�� ي� . ن�� ة� م�ام�ي� Nالا1. Introduction

In mathematics, Pascal's Triangle is a geometricarrangement of the binomial coefficients in a triangle.Pascal's Triangle is named after Blaise Pascal in much ofthe western world, although other mathematicians studied itcenturies before him in India, Persia, China, and Italy[1].

Pascal's Trait'e du Triangle Arithmetique (in English translationin [2]) makes a systematic study of the numbers in hisTriangle. They have simultaneous roles in mathematics asfigurate numbers, combination numbers, and binomialcoefficients, and he elaborates on all these. Given theirmultifaceted nature, it is no wonder that these ubiquitousnumbers had already been in use for over 500 years, inplaces ranging from China to the Islamic world [3].

This work describes the construction of new triangle ofthe coefficients of polynomial represents the (x+j). Theproposed triangle will contributes in finding an easy and

1

fast method to solve any mathematical field involve(x+j), like interpolation, numerical differentiation(first, second,… derivative),…etc.

2. Pascal’s Triangle ConstructionThe rows of Pascal's triangle are conventionally

enumerated starting with row zero, and the numbers in eachrow are usually staggered relative to the numbers in theadjacent rows (see figure (1)) [4].

i=0 C00

i=1 C10 C11

i=2 C20 C21 C22

i=3 C30 C31 C32 C33

i=4 C40 C41 C42 C43 C44

Figure (1) The general form of Pascal’s Triangle

A simple construction of the triangle proceeds in thefollowing manner. On the zeroth row, write only the number1. Then, to construct the elements of following rows, addthe number directly above and to the left with the numberdirectly above and to the right to find the new value. Ifeither the number to the right or left is not present,substitute a zero in its place. For example, the firstnumber in the first row is 0 + 1 = 1, whereas the numbers 1and 3 in the third row are added to produce the number 4 inthe fourth row, and so on as in figure (2) which isdescribes the binomial (x+y)n [4].

i=0 1i=1 1 1i=2 1 2 1i=3 1 3 3 1i=4 1 4 6 4 1

Figure (2) The coefficient values of (x+y)n

Pascal’s Triangle

While The coefficient values of (x-y)n Pascal’s Triangleis depicted in figure (3).

i=0 1

2

i=1 1 -1i=2 1 -2 1i=3 1 -3 3 -1i=4 1 -4 6 -4 1

Figure (3) The coefficient values of (x-y)n

Pascal’s Triangle

This construction is related to the binomial coefficientsby Pascal's rule, which states that if:

(x + y)n =

Then:

for any nonnegative integer n and any integer k between 0and n.

Pascal's triangle has higher dimensional generalizations.The three-dimensional version is called Pascal's pyramid orPascal's tetrahedron, while the general versions are calledPascal's simplices [1].

3. Multiple of (x+bj)The multiple of (x+bj) meant the form:

=(x+b1)(x+b2)(x+bi) …(1)

Where x is real variable, i and j are non-negative integernumbers, bj any known real number, 1ji. e.g. for:

i=2, =(x+b1)(x+b2) = x2+(b1+b2)x+b1b2.

i=3, = (x+b1)(x+b2)(x+b3) = x3+(b1+b2+b3)x2+

(b1b2+b1b3+b2b3)x+b1b2b3.To formulate the general form (ith form) of equation (1),

let’s make the following definitions.Let i be non-empty set consists of i elements, i =

{b1,b2,,bi}. Let’s define:Rr

k(i) be the multiple of k of the numbers mj with index r,1jk, distinct m1,m2,,mki.This mean:

3

Rrk(i) = , Where r=1,2,Ci

k. Cik is the combination of

k from i.We define Rt

0(i)= R10(i) = 1.

Example(1): for:i=1, 1={b1}, C1

0 = 1, C11 = 1, t=1, then R1

0(1) = 1, R11(1) =

b1.i=2, 2={b1,b2}, C2

0 = 1, C21 = 2, C2

2 = 1, t=1,2, then: k=1, R1

1(2) = b1, R21(1) = b2.

k=2, R12(2) = b1b2.

i=3, 3={b1,b2,b3}, C30 = 1, C3

1 = 3, C32 = 3, C3

3 = 1, t=1,2,3then: k=1, R1

1(3) = b1, R21(3) = b2, R3

1(3) = b3. k=2, R1

2(3) = b1b2, R22(3) = b1b3, R3

2(3) = b2b3. k=3, R1

3(3) = b1b2b3.And so on.

Now let’s define Dij which is the coefficient of xi asfollows:

Dij = …(2)

For the mentioned coefficients Dij, a triangle can beconstructed we call it Dij Triangle with Positive bj (DTPbj)(see figure (4)).

i=0 D00

i=1 D10 D11

i=2 D20 D21 D22

i=3 D30 D31 D32 D33

i=4 D40 D41 D42 D43 D44

Figure (4) The general form of DTPbj.

So the general form of equation (1) using equation (2)for level i is:

=Di0xi + Di1xi-1 + + Di,i-1x + Dii =

…(3)4. Multiple of (x+j)

4

As special case of equation (3) when j=bj, then i =

{1,2,,i}.

Example (2): recall example (1), for:i=1, 1={1}, R1

1(1) =1, then D10=1 and D11=1.i=2, 2={1,2}: k=1, R1

1(2) =1, R21(1) =2, then D21 = 1+2 = 3.

k=2, R12(2) = 12 = 2, then , D22 = 2.

i=3, 3={1,2,3}: k=1, R1

1(3) =1, R21(2) =2, R3

1(3) =3, then D31=1+2+3 = 6. k=2, R1

2(3) =12 = 2, R22(3) = 13 = 3, R3

2(3) =23 = 6then D32=2+3+6 = 11. k=3, R1

3(3) =123 = 6, then D33 = 6.And so on.

Notice that Di0=1, i.From equation (3) and as in figure (4) another DT can be

constructed, we call it DTPj as shown in figure (5).

i=0 1i=1 1 1i=2 1 3 2i=3 1 6 11 6i=4 1 10 35 50 24

Figure (5) The coefficient values of DTPj

For multiple (x-j) we take i = {-1,-2,,-i}, so anotherDP can be suggested negative integer, we call it DTNj asshown in figure (6).

i=0 1i=1 1 -1i=2 1 -3 2i=3 1 -6 11 -6i=4 1 -10 35 -50 24

Figure (6) The coefficient values of DTNj

Notice that from figures (5) and (6) the row i can bededuced from row i-1, e.g. row i=2 can obtained from rowi=1 as follows:

5

= (x-2) = (D10x + D11)(x-2) = (x–1)(x-2),

D10x2

+ D11x = x2

- x

+ - 2D10x - 2D11 = - 2x

+ 2

D10x2

+ (-2D10+D11)x

+ (-2D11+D12)

= x2

- 3x

+ 2 …(4)

Considering Di,i+1=0, i.Equation (4) will be:D20x2 + D21x + D22 = x2 -3x + 2,s.t. D21 = -2D10+D11 = (-2)(1)+(-1) = -3, and D22 = -2D11+D12 =-2(-1) + 0 = 2.

And that is the same result in the right side.This result can be proved in the next theorem.

Theorem (1):

, where Di,j = -iDi-1,j-1 + Di-1,j, for i2,

1ji.Proof:

Let the proposition P(i): .

1. T.P. P(1) is true.

Left Side: = x-1.

Right side: = D10x + D11, D10 = 1, D11 = -1 (from

DTNj table).Since the two sides are equal, then P(1) is True.

2. Assume P(i) is true,T.P. P(i+1) is true, this mean we have T.P.

.

= Di0xi+ + Di1xi + + Di,ix

6

1

+ - (i+1)Di0xi - - (i+1)Di,i-1x - (i+1)Dii

Di0xi+

1+ (-

(i+1)Di0+Di1

)xi

+ + (-(i+1)Di,i-

1+Dii)x- (i+1)

Dii

= Di+1,0xi+1 + Di+1,1xi ++ Di+1,i-1x + Di+1,i = ,

Where Di+1,0= Di0, and,Di+1,1= -(i+1)Di0+Di1,, Di+1,i-1= -(i+1)Di,i-1+Dii, Di+1,i= -(i+1)Dii. P(i+1) is true. P(i) is true i.

Remark (1):

The polynomial x3-3x2+2x represents the multiple ,

where i=3 (start from j=0), so in general when applyingTheorem (1):

…(5)

Where Dij = -(i-1)Di-1,j-1 + Di-1,j.

Remark (2): from Remark (1) we have:1. Di+1,j = -iDi,j-1 + Di,j.2. Di+2,j = -(i+1)Di+1,j-1 + Di+1,j.

So the derived DT (we denoted by DTNj0) from DTNj which isshown in figure (6) using Remarks (1) and (2), will beshown in figure (7).

i=1 1 D10

i=2 1 -1 D20 D21

i=3 1 -3 2 D30 D31 D32

i=4

1 -6 11 -6 D40 D41 D42 D43

Figure (7) The new coefficient values of DTNj0

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Notice that the coefficients in figure (7) are as the

coefficients in figure (6) but it’s shifted up by one row,while i starts from 1.

5. Newton-Forward Interpolation ProblemsAs known before that Newton Forward Formula (NFF) which

is used in interpolation field is [5]:

Pn(xt) = y0+ y0+ 2y0++ ny0 = y0 + iy0

…(6)

Where iy0 represent the ith difference of y0 values, is

combination i from t Cti, and:

t= , xt is the interpolated value, and h is constant

argument between any two successive points.

=

…(7)Where 1in.

As an example of equation (7), when:

i=1, then = t,

i=2, then = ,

i=3, then = ,

i=4, then = , and so

on, if:

i=k, then = ,

this means we can depend on equation (5) to find thegeneral form when i=k (considering t=x as a new variable).

It may difficult to calculate the coefficients Dij of ti-j

of equation (5), so we suggest using the coefficient valuesof DTNj0 which are shown in figure (7).

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To understand how DTNj0 can be used in interpolation

problem, let’s see the following example.

Example (3):Let’s have the following interpolation table:

x 0 1 2 3 4y 1 5 7 10 15

Using equation (6):

P4(xt) = y0+ y0+ 2y0+ 3y0+ 4y0

The difference table will be:x y y 2y 3y 4y0 1

41 5 -2

2 32 7 1 -2

3 13 10 2

54 15

By using equation (5):

…(8)Now we will show the benefit of DTNj0 which be shown in thenext theorem.

Theorem (2): Pn(xt) = y0 +

Proof:From equation (6) obtain:

Pn(xt) = y0 + iy0 = y0 + iy0

=y0+ =y0+ …(9)

9

=y0+

=y0+

= y0 +

In more compact form, the above equation will be:

Pn(xt) = y0 +

…(10)If the triangle in figure (7) (DTNj0) rotated by 90

counter clock wise, we call it RDTNj0, so it will takes theshape shown in figure (8) (for simplicity use n=4).

y0/1! 2y0/2! 3y0/3! 4y0/4!D10 D21 D32 D43 t

D20 D31 D42 t2

D30 D41 t3

D40 t4

Figure (8) the coefficients values of RDTNj0.From equation (10) and figure (8) the following relation

can be concluded:

P4(xt) = y0 + t +

t2 +

t3 +

t4 …

(11)Equation (11) represents the polynomial depending on powersof t, s.t. t=(xt-x0)/h.

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Example (4):

Recall, example (3) and RDTNj0 information mentioned inequation (11):

4/1!

(-2)/2!

3/3!

(-2)/4!

( 1 + -1 + 2 + -6 ) t( 1 + -3 + 11 ) t

2

( 1 + -6 ) t3

( 1 ) t4

P4(xt)=1+

.

After taking the accuracy more than 2 digits, P4(xt) willbe:

P4(xt) = 1+6.5t – 3.417t2 + t3 - 0.083t4 …(12)

e.g. if xt = 2.5, this implies t = 2.5 (using x0 = 0), thenP4(2.5) = 8.2765625.

6. Newton-Forward Numerical Differentiation Problems6.1 First Derivative of Numerical Differentiation Problems

In first derivative problems, the following equation isderived from NFF in interpolation [5]:

11

In more compact form:

…(13)In figure (9), two DT will be shown, (c) is DTNj0 shown

in figure (7), figures (a) and (b) are the derived DT weshall denoted by D1TNj0 since its represents the firstderivative.

i=0 D00 1 1

i=1 D10 D11 2 -1 1 -1

i=2 D20 D21 D22 3 -6 2 1 -3 2

i=3 D30 D31 D32 D33 4 -

18 22 -6 1 -6 11 -6

(a)

(b)

(c)

Figure (9) the coefficients valuesof Dij of D1TNj0.

In next lemma, the fact Dij = (i+1-j) Di+1,j can be proven.

Lemma (1):The derived coefficients Dij = (i+1-j) Di+1,j, for

j=0,1,,i-1.Proof:From equation (9):

Pn(xt) = y0+

= …(14)

Notice that the degree of the polynomial P(xt) reducedfrom n to n-1, so the coefficients (i-j)Dij will be inlevel (row) i-1 of D1TNj0 so by comparing equation (13) and(14), we obtain:

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Di-1,j = (i-j) Di,j in the level i-1,

Remark (3):From lemma (1), 1. Dij = (i+1-j) Di+1,j.2. Di+1,j = (i+2-j) Di+2,j.

Example (5): for:i = 1, D10 = 2D20, D11 = D21,i = 2, D20 = 3D30, D21 = 2D21, D22 = D32,i = 3, D30 = 4D40, D31 = 3D31, D32 = 2D42, D33 = D43, and soon.

Lemma (2):

Proof:Let us recall equation (10) and using equation (13), then:

…

(15)Where Di,i-k = kDi,i-k , 1kn.

In solving numerical differentiation problems, we stilluse figure (8) to use the rotating DP.

Example (6):Recall, example (4) and D1TNj0 information mentioned inequation (15):

4/1!

(-2)/2!

3/3!

(-2)/4!

( 1 + -1 + 2 + -6 ) t0

( 2 + -6 + 22 ) t1

( 3 + -18 ) t

13

2

( 4 ) t3

P3(xt)=

.

After taking the accuracy more than 2 digits, P3(xt) willbe:

P3(xt) = 6.5 – 6.834t + 3t2 - 0.332t3

…(16)e.g. if xt = 2.5, this implies t = 2.5 (using x0 = 0), thenP3(2.5) =2.9775.

This approach is long if we need to pass throughinterpolation, therefore, equation (16) can be obtained bydifferentiating equation (11) directly (or specially byequation (12)).

In next lemma the relation between Dij and Di-1,j isfound.

Lemma (3):

Dij = -iDi-1,j-1+ Di-1,j, for 1j<i.

Proof: from lemma (1):Dij = (i+1-j)Di+1,j …(a)AndDi-1,j = (i-j)Dij , then

Dij = Di-1,j, (for j< i)

…(b)And

Di,j-1 = Di-1,j-1 = Di-1,j-1, (for j1)

…(c)From remark (1), since:Dij = -(i-1)Di-1,j-1 + Di-1,j, then:Di+1,j = -iDi,j-1 + Dij …(d)Substitute (b) and (c) in (d) obtain:

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Di+1,j = Di-1,j-1 + Di-1,j

…(e)Substitute (e) in (a) obtain:

Dij = -iDi-1,j-1+ Di-1,j, for 1j<i

…(17) Remark (4):

1. for j = 0, equation (17) will be:

Di0 = Di-1,0 …

(18)2. for j = i, equation (17) will be:Dii = -iDi-1,i-1 …(19)

Example (6):

1. D30 = D20 = (3)=4 (by equation (18)).

2. D31=-3D20+ D21=-3(3)+ (-6)=-18

D32=-3D21+ D22=-3(-6)+ (2)=22 (by equation (17)).

3. D33= -3D22 = -3(2) = -6 (by equation (19)).Remark (5): From lemma (3), its obvious that Di+1,j = -

(i+1)Di,j-1+ Dij.

6.2 Second Derivative of Numerical Differentiation ProblemsIn the second derivative problems, we recall equation

(13) which is represents the first derivative of NFF [5]:

…

(20)In more compact form:

…

(21)In next lemma the fact Dij = (i+1-j)(i+2-j) Di+2,j is proven.

15

In figure (10), we will show two DT, (c) is DTNj0 shown

in figure (9-c), figures (a) and (b) are the 2nd derivativeof DT, we shall denoted by D2TNj0 since its represents the2nd derivative.

i=0 D00 1 1

i=1 D10 D11 6 -6 1 -

3i=2 D20 D21 D22 12 -

36 22 1 -6 11

i=3 D30 D31 D32 D33 20 -

120210

-100 1 -

10 35 -50

(a)

(b)

(c)

Figure (10) the coefficients valuesof Dij of D2TNj0.

Lemma (4):The derived coefficients Dij = (i+1-j)(i+2-j) Di+2,j, for

j=0,1,,i-2.Proof:From equation (14):

…(22)Notice that the degree of the polynomial P(xt) reduced

from n to n-2, so the coefficients (i-j)(i-j-1)Dij will bein level (row) i-2 of D2TNj0 so by comparingequation (21) and (22), we obtain: Di-2,j = (i-j)(i-j-1)Dij.

Remark (6): from lemma (4): Di-1,j = (i-j)(i+1-j)Di+1,j andDij = (i+1-j)(i+2-j)Di+2,j.

Remark (7): Its obvious by using lemma (1) that:1. Dij = (i+1-j) Di+1,j.2. Di-1,j = (i-j) Dij.3. Di-2,j = (i-j-1) Di-1,j.

Example (7): for:

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i = 1, D10 = 2(3)D30, D11 = 1(2)D31,i = 2, D20 = 3(4)D40, D21 = 2(3)D41, D22 = 1(2)D42,i = 3, D30 = 4(5)D50, D31 = 3(4)D51, D32 = 2(3)D52, D33 =1(2)D53, and so on.Or:i = 1, D10 = 2D20, D11 = D21,i = 2, D20 = 3D30, D21 = 2D31, D22 = D32,i = 3, D30 = 4D40, D31 = 3D41, D32 = 2D42, D33 = D43, and soon.

Lemma (5):

Proof:Let us recall equation (15), then

…(23)Where Di,i-k = (k-1)Di,i-k=k(k-1)Di,i-k, 1kn.In solving numerical differentiation problems, figure (8)still using the rotating DT.

Example (8):Recall, example (4) and D1TNj0 information mentioned in

equation (15):

(-2)/2!

3/3!

(-2)/4!

( 2 + -6 + 22 ) t0

( 6 + -36 ) t1

( 12 ) t2

P2(xt)=

17

After taking the accuracy more than 2 digits, P2(xt) willbe:P2(xt) = – 6.834 + 6t – t2 …(24)

e.g. if xt = 2.5, this implies t = 2.5 (using x0 = 0), thenP2(2.5) =1.916.

This approach is long if we need to pass throughinterpolation, therefore, equation (24) can be obtained bydifferentiating equation (16) directly.

In next lemma the relation between Dij and Di-1,j isshown.

Lemma (5): Dij = -(i+1)Di-1,j-1+ Di-1,j, for 1j<i.

Proof: from remark (7):Dij = (i+1-j)Di+1,j, and …(a)Di-1,j = (i-j)Dij , then

Dij = Di-1,j, (for j< i), and

…(b)

Di,j-1 = Di-1,j-1 = Di-1,j-1, (for j1)

…(c)From remark (4):

Di+1,j = -(i+1)Di,j-1 + Dij …(d)

Substitute (b) and (c) in (d) obtain:

Di+1,j = Di-1,j-1 + Di-1,j…(e)

Substitute (e) in (a) obtain:

Dij = -(i+1)Di-1,j-1+ Di-1,j, for 1j<i

…(25)Remark (7):1. for j = 0 equation (25) will be:

Di0 = Di-1,0 …

(26)2. for j = i notice that

Dii = -(i+1)Di+1,I …(27)

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Example (9):

1. D20 = D10 = (6)=12 (by equation (26)).

2. D21=-3D10+ D11=-3(6)+ (-6)=-36 (by equation

(25)).3. D22= D32 = 22 (by equation (27)).Remark (5): From lemma (5), its obvious that Di-1,j = -iDi-

2,j-1+ Di-2,j.

7. The Derivation of Coefficients Relations Diagram

The idea of this work can be summarized in a blockdiagram, which is depends on (Theorem (1), Lemmas (1), (4),(5) and Remarks (2), (3), (5), (6), (7) and (8)) todescribe the relation between the coefficients Dij, Dij andDij in three levels i, i+1 and i+2 (or i-1 and i-2) in twodimension. In horizontal derivative (extension), therelation between one derivative and another and(k=0,1), and vertical derivative, the relation betweenand (k=0,1,2) can be found. This block diagram can beshown if figure (11).

19

20

DijDi-1,jDi-2,j

-iDi,j-1+Dij-iDi-1,j-1+Di-1,j-iDi-2,j-1+Di-2,j

(i-j)(i-j-1)Dij

(i-j-

1)Dij

(i-j)Dij

i

Di+1,jDijDi-1,j

-(i+1)Di+1,j-1+Di+2,j-(i+1)Di,j-1+Dij-(i+1)Di-1,j-1+Di-1,j

(i-j)(i+1-j)Di+1,j

(i-j)Dij

i+1

i-1

i-2

ii-1

Di+2,jDi+1,jDij

(i+1-j)(i+2-j)Di+2,ji i

+1

i+2

Di-1,j

-(i-1)Di-1,j-1+Di-

1,j

i-1

Figure (11) block Diagram of Derivation of coefficients relations.

(i+1-j)Di+1,j

(i+2-j)Di+2,j

(i+1-j)Di+1,j

8. Conclusions and Recommendations1. The block diagram mentioned in figure (11) describesthe relation between the coefficients Dij, Dij, and Dij sowe can conclude first, that the other levels (rows) forthe same coefficients ( , ,…) in verticalextension. Secondly, we can find the other columns forthe derivation of the coefficients ( , ,…).

2. We can construct proposed triangle for the othermultiple or double relations for coefficients producedanother form of polynomials.

3. We can construct a proposed 3-dimensions pyramid forthe sample (a+b+c)n.

4. as future work we can find the nth form for theserelation horizontally and vertically, as an example, wecan conclude the general vertical extension formula of Dij

like the formula: Di+k,j = -(i+k-1)Di+k-1,j-1 + Di+k,j, kZ.

References[1]. Wikimedia Foundation, Inc. , “Pascal's Triangle”,

Wikipedia, the free encyclopedia.htm, 17 May 2009.[2]. Mortimer Adler, “Great Books of the Western World”, (ed.),

Encyclopedia Britannica, Inc., Chicago, 1991.[3]. David Pengelley, “Pascal's Treatise on the Arithmetical

Triangle: Mathematical Induction, Combinations, the BinomialTheorem and Fermat's Theorem A Historical Project in Three Parts”,Version 1.8, Nov. 23, 2005.

[4]. V.J. Katz, “A History of Mathematics: An Introduction”,(second ed.), Addison-Wesley, New York, 1998.

[5]. Steven T. Karris, “Numerical Analysis Using MATLAB andSpreadsheets”, Second Edition Orchard Publicationswww.orchardpublications.com, 2004.

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