CONSTRUCTION OF CHEBYSHEV POLYNOMIALS

by

JOLIANE LEA STORM, B.S.Ed.

A THESIS

IN

MATHEMATICS

Submitted to the Graduate Faculty of Texas Tech University in

Partial Fulfillment of the Requirements for

the Degree of

MASTER OF SCIENCE

Approved

Accepted

May, 1994

0 0

0 0

g:~

Av fltJ5

T3 ,qqif #D· 7

<!bf'~

~uJ !JEll 7377

d~ Q).fH../q1

Copyright, 1994, Joliane Lea Storm, Dr. Clyde Martin

ACKNOWLEDGMENTS

I am deeply grateful to Dr. Clyde Martin for his

1mmense help and guidance on this thesis. He has helped

make it possible for me to achieve this very important

milestone in my life, and he had the confidence in me that I

lacked. I am also grateful to Dr. John White for his help

and direction with difference equations, which I found are

really fun. I also thank Dr. Dalton Tarwater for being a

member of my committee, along with Dr. Martin and Dr. White.

My parents deserve a great big thanks for their love and

support of me all of my life, and I wish to dedicate this

thesis to my husband, Greg Storm, who married me while I was

working on and writing this.

11

TABLE OF CONTENTS

ACKNOWLEDGMENTS . . . . . . . . . . . . . . • . . . . . . 11

I. INTRODUCTION . . . . . . . . . . . II. THE DIFFERENCE EQUATION APPROACH .

III. THE OPERATOR APPROACH . . . . IV. CONCLUSIONS . . . . .

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . APPENDICES

A. CHEBYSHEV POLYNOMIALS . . . . . . . . . . . . . B. DERIVATIVES OF THE CHEBYSHEV

RECURRENCE RELATION . . . . . . . . . .

c. THE CHEBYSHEV POLYNOMIALS EVALUATED AT ZERO . . D. THE FIRST DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . . . . E. THE SECOND DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . . . . F. THE THIRD DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . . . G. THE FOURTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . . . . H. THE FIFTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . . . . I. THE SIXTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO . . . . . . • . . J. THE SEVENTH DERIVATIVES OF THE CHEBYSHEV

K.

L.

M.

POLYNOMIALS EVALUATED AT ZERO . . . . . THE EIGHT DERIVATIVES OF THE CHEBYSHEV POLYNOMIALS EVALUATED AT ZERO . . . . .

THE NINTH DERIVATIVES OF THE CHEBYSHEV POLYNOMIALS EVALUATED AT ZERO . . . . .

THE TENTH DERIVATIVES OF THE CHEBYSHEV POLYNOMIALS EVALUATED AT ZERO . . . . .

~~~

. . . .

. . . .

. . . .

. . . .

. 1

. 4

. 20

. 33

35

36

37

38

39

40

. 41

42

43

44

45

46

47

48

N. THE CASSORATI . . . . . . . 0. INDEFINITE SUMS . . . . P. DETAILS OF THE GENERALIZATION

FROM CHAPTER II . . . . . . .

. l.V

. . . . . . . .

. . . . . . . . .

49

50

51

CHAPTER I

INTRODUCTION

There are several ways to develop the Chebyshev

polynomials but there seems to be nothing in the

literature which constructs the Chebyshev polynomials

of the first kind by explicitly constructing their

coefficients. Two methods for explicitly constructing

the coefficients of the Chebyshev polynomial equations of

the first kind are presented here. Both of these methods

begin with the recurrence relation

~+ 1 (x) = 2xTn (x)- Tn_ 1 (x),

and with initial conditions

To ( x) = 1 and ~ ( x) = x ,

( 1.1)

{ 1. 2)

where ~(x) denotes the Chebyshev polynomials of the first

kind [1]. Both also compute each ~(0) for use in the

Taylor series expansion,

~ (X) = Tn ( 0) + (X - 0) T,/!OJ + (X - 0 l Tn:~OJ + • • . + (X - 0 l 'l!kt~J 1

(1.3)

where ~(0) denotes the kth derivative of the nth Chebyshev

polynomial evaluated at zero.

The first method is presented in Chapter II. It is the

method by which the Chebyshev recurrence relation, (equation

1.1) hereafter referred to as the CRR, is evaluated at

x = 0 and rearranged into a difference equation (see

1

equation 2.3) [2]. The particular solution to this

difference equation, ~(0), (see equation 2.7) is obtained

and used in the next phase of development. This next phase

involves finding the first derivative of the CRR, evaluating

it at x = 0, and rearranging it into another difference

equation (see equation 2.9). The particular solution of ,

that difference equation, ~ (0), (see equation 2.25) is

obtained and used in the next difference equation, which 1s

obtained by rearranging the second derivative of the CRR and

evaluating it at x = 0 (see equation 2.27). The process

continues in this manner. The result is that the Chebyshev

polynomials, or Tn ( x) • s, are found by using the ~ ( 0) 's in

the Taylor series expansion of ~(x) about x = 0 (equation

1.3). Each ~(0), when divided by k! in the Taylor series

expansion, is the coefficient of the xkth term. In

Chapter II, all the ~(O)'s up to k = 5 are found and used

to find ~ (x) up to n = 5. A generalization of ~ (0) is

obtained (see equation 2.70 and Appendix P), but it does not

lead to a computation of ~(0) nor does it lead to a closed

form for ~(0).

The second method for obtaining the Chebyshev

polynomials is presented in Chapter III. This method, like

the first method, begins with the CRR (equation 1.1), finds

the ~(O)'s, and then uses them in the Taylor series n

expansion (equation 1. 3) to find each Tn ( x). The ~ ( 0) 's,

however, are found by using a series of operators motivated

2

by the operators presented in [3]. This process begins by

finding all of the derivatives of equation 1.1 (see equation

3.5) and then by applying an •averaging• operator to them

k- 1 times (see equation 3.15). Thus, ~(0) is solved for

in ter.ms of a power of the inverse of the "averaging"

operator (see equation 3.18). This inverse, on its own,

does not provide any helpful information, so it is broken

down in terms of a forward differencing operator (see

equation 3.20). The inverse of this new expression may be

found more easily (see equation 3.24) with the aid of one

more operator, a differential operator (see equation 3.29).

The application (see equation 3.32) of this inverse operator

results in an equation for ~(0) which is then simplified.

A theorem is stated toward the end of Chapter III clarifying

this, and the results are shown.

Also in Chapter III, a link between the Chebyshev

polynomials and the Bernoulli numbers is found [1].

However, this link does not serve to simplify the

computation of ~(0) but only to complicate it. This

connection, nevertheless, could aid in other research.

3

CHAPTER II

THE DIFFERENCE EQUATION APPROACH

The first method of development of the Chebyshev

polynomials of the first kind, symbolized by ~(x), is the

method which uses difference equations. The preparation for

this process begins by computing all the initial values

needed for calculating particular solutions to each

difference equation. The initial values are listed in

Appendices C through M. The process for computing each

~(0) is the same, however, each successive calculation

becomes more and more involved. Recall that ~(0) is the

kth derivative of the nth Chebyshev polynomial evaluated at

x = 0 which will be substituted into the Taylor series

expans1on. The process begins by evaluating the Chebyshev

recurrence relation, CRR, (equation 1.1) at zero, then

rearranging it (see equation 2.3), and then using difference

equation methods [2] to solve it. That solution, ~(0),

(see equation 2.7) is then used in the first derivative of

the CRR which is evaluated at zero, rearranged, (see ,

equation 2.9) and then solved. In turn, that answer, ~ (0),

(see equation 2.25) is used in the second derivative of the

CRR evaluated at zero, rearranged, (see equation 2.27)

solved, and so on. Each successive ~(0), when divided by

k! in the Taylor series expansion, equations 1.3 or 2.1,

below, is the coefficient of the xkth term in the nth

4

Chebyshev polynomial, ~ ( x). Note that ~ ( x) ~s a

polynomial of degree n. I •

7;, (X ) = Tn ( 0 ) + (X - 0) '1;, 1 t!o J + (X - 0 l T, 2 r! o J + . . . + (X - 0 /' -r:kr ~ J •

( 2 . 1)

Here is the process in detail. Begin with the

CRR

Tn+ 1 (x) = 2xTn (x)- Tn_ 1 (x), ( 2. 2)

[1], and evaluate it at x = 0. Then rearrange it to obtain

Tn + 1 ( 0 ) + Tn _ 1 ( 0 ) = 0 . ( 2 . 3 )

For typographical ease, let y(n + 1) = Tn+l (0) and let

Y ( n - 1) = 7;,_ 1 ( 0) obtaining from equation 2 . 3

y(n + 1) + y(n- 1) = 0. ( 2. 4)

It is difficult, however, to work with n + 1 and n - 1, so

replace n with n + 1 so that the homogeneous difference

equation, equation 2.4, ~s now

y(n + 2) + y(n) = 0. ( 2. 5)

To solve equation 2.5 let m2 = y(n + 2) and 1 = y(n)

obtaining m2 + 1 = 0. Thus, m2 = -1 and m = + i, and

consequently, the general solution to the homogeneous

equation is

y(n) .!!l1 . nil - c 1 cos 2 + c 2 s~n 2 ( 2. 6)

from [2] .

To find the particular solution, use the initial data

from Appendix C which is To ( 0) = 1 and ~ ( 0) = 0 so that

To ( 0) = y( 0) = 1 = c 1 cos 0 + c 2 sin 0 ~ c 1 = 1

and

5

~ (0) - y(l) - 0 - cos I] + c 2 sin I] - c2

• 1 => c2

- 0

so that

y ( n) = cos ~ = ~ ( 0) • ( 2. 7)

This is the coefficient of the x 0 term in each of the

Chebyshev polynomials, and it is the particular solution of

the homogeneous equation, equation 2.5.

In order to find the coefficient for the x 1 terms,

begin from equation 2.2 and take the derivative, obtaining

~+ 1 (X) = 2 X~ (X) + 2 ~ (X) - ~- 1 (X) •

Now evaluate this at x = 0 and rearrange it to obtain

~+ 1 ( 0} + ~- 1 { 0} = 2 Tn { 0}.

( 2. 8)

{ 2. 9)

As in the previous process, choose a variable to represent ,

the T • s. Let g ( n + 1) = ~+ 1 ( 0), and let g ( n - 1) = Tn_ 1 ( 0).

Recall that y( n) = Tn ( 0), so that equation 2. 9 1s

g(n + 1) + g(n- 1) = 2y(n). (2.10)

Now replace n with n + 1, so that from equations 2.10, 2.7,

and the trigonometric identity for cos(a +b), equation 2.10

becomes

g( n + 2) + g( n) = 2y( n + 1) = 2 cos (n+;Jn = -2 sin nr. (2.11)

The solution to the homogeneous part of this equation is

equation 2.6 so a general solution and a particular solution

for the non homogeneous case must be found.

The process for finding the ~(D)'s will be the same

for each successive problem. Realize that g(n) and y(n)

are used alternately in each successive iteration for

6

clarity. The general solution for each of the difference

equations, including equation 2.11, is

g ( n ) = c 1Y 1 ( n ) + c 2y 2 ( n ) + u 1 ( n ) y 1 ( n ) + u 2 ( n ) y 2 ( n ) , (2.12)

[2] where, in every case (from equation 2.6)

y1 (n) = cos nr and y2 (n) = sin~. (2.13)

Therefore g(n) (or y(n)) in each case looks like

g(n) = C1 COS nr + C2 sin nr + U1 (n) COS~ + U2 (n) sin nr 1

which is formally used as

g ( n ) = cos nr ( c 1 + u 1 ( n ) ) + sin nr ( c 2 + u 2 ( n ) ) • (2.14)

Furthermore, the values of u 1 (n) and u 2 (n) are found in the

following manner:

and

u 1

( n) - - Li-1 tfn)y2 fn+1) Cfn+ 1)

( ) _ A-1 t(n)y1 (n+ 1)

U2 n - Ll C(n+1) I

[ 2] 1 where Y1 ( n + 1) = COS (n+i)fl = - sin nr 1 and

(2.15)

(2.16)

y 2 (n+1) = sinrn+;Jn = cosnf. Computing U 1 (n) and u 2 (n) will

require knowing the Cassorati, C(n + 1), and the indefinite

A-1 sum, Ll , of several different functions. The Cassorati is

the same for each iteration, and is calculated and shown ~n

Appendix N. The indefinite sum is different for each

iteration and is shown in Appendix 0. Conveniently,

C(n + 1) equals 1, so that now U1 (n) and u 2 (n) in equations

2.15 and 2.16 become

(2.17)

and

(2.18)

7

The f ( n) 's are equal to the previously found ~ ( 0), but

figured as y(n + 1) (or g(n + 1)). See equation 2 .11. Also,

in every case in which sin~ and cos~ are multiplied

together, they equal sinnll, which is zero. In other

words, in every case, either u 1

( n) = constant or

U2 (n) =constant, since for either u1(n) or u

2(n),

Li-1 sin ~ cos~ = Li-1 0 = constant. (2.19)

Another common occurrence in u1 (n) or u2 (n) is that either

or

cos ~ . cos ~ = 1 ( 1 + ( -1 t )

sin~·sin~ = ~(1-(-1f).

In the case described in equation 2.11,

g(n + 2) + g(n) = -2 sin~.

The f(n) in this case is -2 sin~. The u 1 (n) 1n this

case is u 1 ( n) = - Li-1 ( -2 sin nf) cos ~ = Li-1 0 = constant,

and u 2 (n) is

(2.20)

(2.21)

(2.22)

u2 (n) = -Li-1 (-2 sin~) sin nf, (2 .23)

which from equation 2.21 is u 2 (n) = Li-1 (1- (-1t ). This in

turn is equal to u 2 ( n) = Li-1 1 - Li-1 ( -1 t, which is, from

Appendix 0, u2

(n) = n + ~ (-1)n. Now the particular solution

is, from equation 2.14,

g(n) = cos nf (c1

+ constant) + sin .nfl (C 2 + n + 1 (-1)n ). { 2. 24}

To find c 1 and c 2 , use the initial values in Appendix , ,

D, To (0) = g(O) = 0 and ~ (0) = g(1) = 1. Substitute in

these values, and obtain c1 = -constant and C2 1.

- 2. This

implies that the particular solution, g(n), is

8

, g(n) = (~+n+1f-1rJsintf- Tn (0) (2 .25)

In fact, this answer checks.

To derive the coefficients of x 2 for the Chebyshev

polynomials, take the derivative of equation 2.8 to obtain

~~ 1 (X) = 2 X~' (X) - ~'_ 1 (X) + 4 ~ (X) •

Evaluate this at x = 0 and rearrange it to obtain

~~ 1 ( 0) + ~~ 1 ( 0) = 4 ~ ( 0) .

(2 .26)

(2 .27)

As in the previous process, let y ( n + 1) = ~~ 1

( 0) and let

y(n- 1) = ~'- 1 (0). Recall from equation 2 .25, that

g ( n J = ~ ( 0 J , so that

y(n + 1) + y(n- 1) = 4g(n). (2 .28)

Now, replace n with n + 1 so that from equations 2.28,

2.25, and the trigonometric identity for sin(a +b),

y(n + 2) + y(n) - 4g(n + 1) = 4(~ + n + 1 + ~ (-1f+ 1) sin rn+gJn

- ( 6 + 4 n - 2 ( -1 r ) cos nr . (2 .29)

The solution to the homogeneous equation 1s equation 2.6, so

the particular solution for the non homogeneous case must be

found. Equation 2.14 is the particular solution, and now

u 1 (n) and u 2 (n) must be found. The f(n) in this case is

from equation 2.29,

f ( n )= ( 6 + 4 n - 2 ( -1 r ) cos nr . (2.30)

This results in u1 (n) equaling

u1 ( n) = - Li-1 ( 6 + 4 n - 2 ( -1 r ) cos nr cos nr. (2.31)

Substituting equation 2.20 into equation 2.31, and then

multiplying and simplifying produces

u 1 ( n ) = - Li-1 ( 2 + 2 n + 2 ( -1 )n + 2 n ( -1 r ) . (2.32)

9

Taking ~-1 of each term in equation 2.32 g1ves

U 1 (n) = -n 2- n + (-1t (n + ~). (2.33)

In order to find u 2 (n), use equation 2.18 and obtain

u 2 ( n) = - ~-l ( 6 + 4 n - 2 ( -1 t ) cos nr sin nJI . ( 2 • 3 4)

From equation 2.19,

u 2 (n) =constant. (2.35)

Now the solution, from equation 2.14, is

y(n) = cos~ (c1 + -n 2 - n + (-1t (n + 1)) +sin~ (c 2 + constant).

(2.36)

To find the particular solution of equation 2.36, use

the initial values in Appendix E, 'Yo'(O) = y(O) = 0 and

'Ji'(O) = y(1) = 0. Substitute these into equation 2.36, and

obtain c 1 = - ~ and c 2 = -constant. This implies that the

particular solution, y(n), is

y ( n) = (- ~ - n 2 - n + ( -1 t ( n + ~ ) ) cos ~ = ~, ( 0) . ( 2 • 3 7 )

In fact, this answer checks when substituted into equation

2.28.

Now, to derive the coefficients of x 3 for the Chebyshev

polynomials, first take the derivative of equation 2.26

obtaining

~: 1 (x) = 2x~"(x)- ~~~ 1 (x) + 6~'rxJ• (2.38)

Next, evaluate this at x = 0 and rearrange it to obtain

~: 1 ( 0 ) + ~~ 1 ( 0 ) = 6 ~, ( 0 ) . ( 2 • 3 9 )

Also, let g(n + 1) = ~: 1 (0) and g(n- 1) = ~~ 1 (0). Remember

that from equation 2.37, y(n) = ~'(0), so that

g(n + 1) + g(n- 1)- 6y(n).

10

(2.40)

Now replace n with n + 1 so that from equations 2.40, 2.37,

and the trigonometric identity for cos(a +b)

g(n + 2) + g(n) = 6y(n + 1)

= 6 (- ~ - ( n + 1 i - ( n + 1) + ( -1 t ( n + 1 + -t) cos rn;lJ •

Simplify this and obtain

g(n + 2) + g(n) = 6(~ + 3n + n 2 + (-1t (n + ~))sin tf. (2 .41)

As noted previously, the solution to the homogeneous

equation is equation 2.6, thus, the particular solution for

the non homogeneous case must be found. Equation 2.14 is

the particular solution, so now u 1 (n) and u 2 (n) must be

found. The f(n) in this case is from equation 2.41,

f ( n) = 6 ( ~ + 3 n + n 2 + ( -1 t ( n + ~))sin n§l,

so that u 1 (n) from equation 2.17 is

(2.42)

ul (n) = -L1-l 6(~ + 3n + n 2 + (-1t (n + ~))sin nr cos~. (2 .43)

Therefore, from equation 2.19,

u1 (n)=-constant. ( 2 . 44)

In order to find u 2 (n), use equations 2.18 and 2.42 to

obtain

u2 (n)=-Li-1 6(~ + 3n + n 2 + (-1t (n + ~))sin ~sin~. (2.45)

Replace sin~· sin~ with its equivalent, -t (1- (-1t ), from

equation 2.21, and multiply, obtaining

u2

{n)= -3L1-1 (1 + 2n + n 2 - (-1t - 2n(-1t - n 2 (-1t ). Take L1-1

of each of these terms, simplify, and obtain

u 2 ( n ) = -3 ( n; + 1 n 2 + ~ n - ~ + 1 ( -1 r ( n 2 + n ) ) . (2.46)

Now the particular solution is, from equation 2.14,

11

g (n) =cos~ ( c 1 + constant} +

sin nr ( c 2-3 ( n; + 1 n 2 + ~ n - ~ + 1 ( -1 t ( n 2 + n } } . (2.47)

To find the variables, c 1 and c 2 , use the initial values ~n

Appendix F that 'Po"(O} = g(O} = 0 and ~, = g(1} = 0.

Substitute these into equation 2.47, and obtain

C1 = -constant and c 2 = 0. This implies that the particular

solution, g(n}, is

g ( n } = (- n 3 - ~ n 2

- ~ n + ( -1 t (- ~ n 2 - ~ n } } sin nr = ~"( 0} .

(2.48)

This answer also checks when substituted into equation 2.41.

Now to derive the coefficients of x 4 in the Chebyshev

polynomials take the derivative of equation 2.38 obtaining

~:1 (x} = 2x~v (x}- ~~1 (x} + B~"(x}. (2 .49)

Evaluate this at x = 0 and rearrange it to obtain

~:1 ( 0} + ~~1 ( 0} = 8 ~"( 0}. ( 2. 50)

Next let y(n + 1} = ~:1 (0} and y(n- 1} = ~~1 (0}. From

equation 2. 48, g( n} = ~"( 0}, so that

y(n + 1} + y(n- 1} = Bg(n}. (2.51)

Replace n with n + 1, so that from equations 2.51, 2.48, and

the trigonometric identity for sin(a +b),

y(n + 2} + y(n} = By(n + 1} = (-(n + 1)3 - j (n + 1l - -j- (n + 1) + (-1/+ 1 (-1 (n + 1)2

- -i (n + 1)) sin fn+}m.

Simplify this and obtain

y(n + 2) + y(n) = 4(-2n 3 - 9n 2

- 13n- 6 + 3(-1f (n2 + 3n + 2)) cos .a.¥.

(2.52)

12

The solution to the homogeneous equation 1s equation 2.6 so

use equation 2.14 to find the particular solution, and then

find u 1 {n) and u 2 (n). The f{n) in this case is from

equation 2.52,

f{n) = 4(-2n3

- 9n 2 - 13n- 6 + 3(-1t (n 2 + 3n + 2)) cos tf 1

so that u 1 {n) from equation 2.17 1s

u 1 ( n) =

( 2 • 53)

-L1-1 4(-2n

3 - 9n 2

- 13n- 6 + 3(-1t (n 2 + 3n + 2)) cos~ cos~ 1

(2.54)

and from 2.20 is

u 1 ( n) = - .d-1 4 ( -2 n 3 - 9 n 2

- 13 n - 6 + 3 ( -1 r ( n 2 + 3 n + 2)) ~ ( 1 + ( -1 r ) .

After simplifying, this is

u 1 ( n ) = L1-1 ( 4 n 3 + 12 n 2 + 8 n + 4 n 3

( -1 )n + 12 n 2 ( -1 t + 8 n ( -1 )n ) •

(2.55)

Taking L1-1 of each term in equation 2.55 gives

u 1 {n)= n 4 + 2n 3 - n 2

- 2n + (-1t (-2n 3 - 3n 2 + 2n + ~)). (2 .56)

In order to find u 2 (n), use equation 2.18 and obtain

u2 (n) = -.d-1 4(-2n3 - 9n 2

- 13n- 6 + 3(-1f (n 2 + 3n + 2)) cos~ sin~,

and from equation 2.19,

u 2 (n) =constant. ( 2 . 57)

Now the solution is, from equation 2.14,

y(n) = cos~ (c1 + n 4 + 2n 3 - n 2

- 2n + (-1)n (-2n 3 - 3n 2 + 2n + ~)) +

sin ~ ( c 2 + constant). (2.58)

To find c1 and c 2 , use the initial values in Appendix G,

~v (0) = y(O) = 0 and rr;v (0) = 0. Substitute these into

13

equation 2.58, and obtain C1

= and C 2 -constant.

This implies that the solution, y(n), is

y(n) = (n 4 + 2n3

- n2

- 2n- -j- + (-1f (-2n 3 - 3n 2 + 2n + -j-)) cos 'f

- ~v ( 0) ( 2 . 59 )

This answer checks when substituted into equation 2.52.

To derive the coefficients of x 5 in the Chebyshev

polynomials, first take the derivative of equation 2.49

obtaining

Evaluate this at x = 0 and rearrange it to obtain

T:+ 1 ( 0) + T:_ 1 ( 0) = 1 0 ~v ( 0). ( 2 . 61)

Now let g(n + 1) = ~~1 (0) and g(n- 1) = ~v (0). From

equation 2. 59 y(n) = ~v (0), so that

g(n + 1) + g(n- 1) = 10y(n). (2.62)

Upon each iteration replace n with n + 1 so that from

equations 2.62, 2.59, and the trigonometric substitution for

cos (a + b), g ( n + 2) + g ( n ) = 1 0 y ( n + 1 )

- 1 0 ( n + 1/ + 2 ( n + 1/ - ( n + 1 )2 - 2 ( n + 1 ) - ~ +

(-1t+ 1 (-2(n + 1/ - 3(n + 1/ + 2(n + 1) + 1JJ cos~, which

simplifies to

g(n + 2) + g(n) -

-10(- ~ + n 4 + 6n 3 + 11n 2 + 6n + (-1t (2n3 + 9n

2 + 10n + 1JJ sin~.

{ 2 . 63)

The solution to the homogeneous equation is equation 2.6 so

the particular solution for the non homogeneous case must be

found. Find the particular solution (equation 2.14) by

14

first finding U1 (n) and u 2 (n). The f(n) 1n this case 1s

from equation 2.63,

f(n) = -10(- -23 + n

4 + 6n 3 + 11n 2 + 6n + (-1j (2n 3 + 9n 2 + 10 + 3 )) 'n .IJl1 n 2 s~ 2 ,

(2.64}

so that U1 (n) from equation 2.17 is

From equation 2.19 this means that

u1 (n) =constant. (2.65)

In order to find u 2 (n), use equations 2.18 and 2.64

and obtain

-Li-1 (-10(--f + n 4 + 6n3 + lln 2 + 6n + (-lt (2n3 + 9n 2 + lOn + 1JJJ sin .llf sin lf.

( 2. 66)

Replace sin nf ·sin~ with its equivalent, 1 (1- (-lt ), from

equation 2.21, and multiply, obtaining

u 2 fnJ=Li-1 5(n 4 + 4n 3 + 2n 2 - 4n- 3 + (-l)n (-n 4

- 4n 3 - 2n 2 + 4n + 3)).

Take ~-1 of each of these terms, simplify, and obtain

u 2 (n) = n 5 + 1n 4- 5n 3 -10n 2

- ~n + (-lt(jn 4 + 5n 3 -10n 2-

2] n).

Now the particular solution from equation 2.14 1s

g(n) = cos nf (cl + constant)+

(2.67}

(2.68}

To find c 1 and c 2 , use the initial values 1n Appendix H that

Tav ( 0) = g ( 0) = 0 and ~v ( 0) g(l) = 0.

15

Substitute these into equation 2.68 and obtain

C 1 = -constant and C 2 = 0. This implies that the solution,

g(n), 1s

(2.69)

This answer checks when substituted into equation 2.63.

The process of evaluating the derivatives of the CRR at

x = 0, rearranging them, and solving them using difference

equation methods [2] continues indefinitely. A compact

formula for this process could be obtained if a formula

could be found for each A-1 in Appendix 0.

A generalization has been obtained which helps in

understanding the process, nevertheless, it does not help 1n

the computations. The generalization writes ~(0) as

(2.70)

where g0 [n] - 1 and g 0 (n) = 0. Before A-1 is taken, the

polynomials which are represented by the g's are the same.

However, after A-1 is taken, these polynomials are

different. For more details, consult Appendix P. This is

the reason that the key to a compact formula lies with A-1•

Results

Recall that ~(x) is a polynomial of degree n. Each

~(0) is substituted into the Taylor series expansion of

~ ( x) about x = 0, equation 2 . 1,

~ ( 0) 2 "I; ( 0) ( 0 )Jc ~ ( 0) ~(x) = Tn(O) + (x- 0)1! + (x- 0) --rr-+ ... + x- Jc! •

16

• To find T0 (x), use Tn (0) = cos~ (equation 2. 7) with

n - 0 obtaining To ( 0) = cos 0 = 1. Substitute this into

equation 2.1, and obtain the Chebyshev polynomial,

'I;;(x) = 1. ,

• To find ~ (x), use ~ (0) = (1 + n + 1 (-l)n) sin nf

(equation 2.25) and equation 2.7 evaluated at n = 1 to

obtain:

~ ( 0) = COS ~ = 0 1 (from equation 2.7) ,

~ ( 0) = ( ~ + 1 - ~ ) sin if = 1 . (from equation 2.25)

Substitute these into equation 2.1 1 and obtain the Chebyshev

polynomial, ~ (x) = x.

• To find ~ (x), use

~, ( 0) = (- ~ - n 2 - n + ( -1 ;n ( n + 1 ) ) cos n§I ( equation 2 . 3 7 ) and

equations 2.25 and 2.7 evaluated at n = 2 to obtain:

~ ( 0) = cos n = -1' (from equation 2. 7) ,

~ (0) = 3 sin n = 01 (from equation 2.25)

" T2 (0) = -4 cos ll = 4. (from equation 2 .37)

Substitute these into equation 2.1, and obtain the Chebyshev

polynomial, T2 ( x) = 2 x 2 - 1 .

• To find ~ (x), use

T~"(O) = (-n 3- ~n2 -1n + (-1tf-1n 2

- ~n))sin ~

(equation 2.48) and equations 2.37,2.25, and 2.7 evaluated

at n = 3 to obtain:

~ ( 0) = COS 3f = 01 (from equation 2.7)

I

~ (0) = 3 sin 3f = -3 1 (from equation 2 .25)

" ~ (0) = -16 cos 3f = 0, (from equation 2 .37)

17

T;, ( 0 J = -24 sin 3f = 24. (from equation 2.48)

Substitute these into equation 2.1, and obtain the Chebyshev

polynomial, 1; ( x) = 4 x 3 - 3 x.

• To find ~ ( x), use

~v (0) = (n4 + 2n 3

- n 2 - 2n- ~ + (-lr (-2n 3

- 3n 2 + 2n + -%JJ cos nf

(equation 2.59) and equations 2.48, 2.37, 2.25, and 2.7

evaluated at n = 4 to obtain:

T4 ( 0) = cos 2 ll = 1, (from equation 2. 7) ,

T4 (0) = 5 sin 2ll = 0, (from equation 2 .25)

" T4 (0) - -16 cos 2ll = -16, (from equation 2 .37)

T~, ( 0) - -120 sin 2 ll = 0, (from equation 2. 48)

~v (0) - 192 cos 2ll = 192. (from equation 2 .59)

Substitute these into equation 2.1, and obtain the Chebyshev

polynomial, T4 ( x) = Bx4 - Bx2 + 1.

• To find Ts ( x), use

(equation 2.69) and equations 2.59, 2.48, 2.37, 2.25, and

2.7 evaluated at n- 5 to obtain:

T5

( 0) = cos 5§1 = 0, (from equation 2. 7) ,

T5

(0) = 5 sin 5§1 = 5, (from equation 2 .25) ,

T5 ( 0) - -3 6 cos 5§1 = 0 , (from equation 2.37)

T~" ( 0) - -120 sin 5f - -120 I (from equation 2. 48)

~v (0) - 1152 cos 5f - 0, (from equation 2. 59)

Tsv ( 0) = 1920 sin 5§1 = 1920. (from equation 2. 69)

Substitute these into equation 2.1, and obtain the Chebyshev

polynomial, T5 (x) = 16x5- 20x3 + Sx.

18

Each of these, ~(x) through ~(x), are indeed the

first five Chebyshev Polynomials. They may be checked

with the Chebyshev Polynomials from [1] in Appendix A.

The difference equation method for generating the ~(O)'s,

which takes each derivative of the CRR evaluated at x = 0

and solves it, is an effective but lengthy method since a

closed formula was not obtained.

19

CHAPTER III

THE OPERATOR APPROACH

The second method for constructing the Chebyshev

polynomials, like the first method, involves writing each

Chebychev polynomial as a finite Taylor series expansion

about x = 0 (equation 1.3, repeated below). In other

words, it involves writing each Tn (x) (with the

restrictions that k > 0 and n > 1} as

Tn (X J = Tn ( 0) + (X - 0) -r:,/~ 1 + (X - 0 l ~(~) + ... + (X - 0 l ~(~) 1

with initial conditions 1 ~ ( 0) = 1 1 'If ( 0) = 0 I and

'Ya ( 0) = 0. All the ~ ( 0) 's can be generated and are listed

in Appendix C. The ~(D)'s, however, are generated by using

a series of operators motivated by [3] which will be

detailed in this chapter.

To begin the process of finding each ~ ( 0), k > 0 I

n > 1, a formula is found for the kth derivative of the CRR

and evaluated at x = 0 (see equation 3.7). After

rearranging the equation, an averaging operator, ~, 1s

applied to the equation k - 1 times, creating ~ ( 0) in the

form of a finite series (see equation 3.16). Thus, ~(0) is

solved for in terms of a power of ~-1 (see equation 3.19).

However, this provides no helpful information, so ~ is

written in terms of a forward differencing operator, El

which causes ~(0) to be written as an infinite series (see n

equation 3.22). Now E can be applied directly to the

20

~(O)'s, eliminating the presence of operators in the

equation and turning the infinite series into a finite

series (see equation 3.23). At one point in working with

E, an infinite series must be raised to a power, and thus,

a differential operator, D, along with the Taylor power

ser1es expansion of ( 11xJx is used (see equations 3.29 and

3.30). Once this is obtained, the resulting equation is

simplified. This simplified equation (see equation 3.48) 1s

the form for writing each ~ (0).

Here is the process. Beginning with the CRR, (also

equation 1.1)

Tn+ 1 (X) - 2 xTn (X) - Tn-1 (X) I ( 3 . 1)

and initial conditions,

~(x) = 1, Tj(x) = x, ~ (x) = 0, ( 3. 2)

(with the ~ ( 0) • s calculated by Tn ( 0) = cos ~) calculate the

kth derivative with respect to x, and obtain

~+ 1 (X) = 2 X~ (X) - ~- 1 (X) + 2 Tn (X).

Take the second derivative and obtain

~~ 1

(X) = 2 X~' (X) - ~'_ 1 (X) + 4 ~ (X) •

( 3 • 3)

( 3. 4)

Notice that the following relationship holds when k > 0 and

n > 1:

T!+1 (x) = 2x~ (x)- rr;_ 1 (x) + 2k~- 1

(x). ( 3. 5)

This will be proven by showing that the (k + 1)st derivative

. 1S

~:~ (x) = 2x~+1 (x)- ~:; (x) + 2(k + 1)~ (x). ( 3 . 6)

PROOF: Beginning with equation 3.5, take the derivative of

21

each term and obtain

{ ~+ 1 (X))' = { 2 X~ (X) - ~- 1 (X) + 2 k~- 1 (X) J . Differentiate

individual parts obtaining

[~+1 (x)]' [2x~ (x)]' - rrr;_1 (x)J + [2k~-1 {x)J' which gives

~:; (x) = 2x~+1 {x) + 2~ (x)- rr;~; {x) + 2krr:-1+1 (x), and add the

~(x) terms obtaining

~:; (x) = 2xrr;+1 {x)- rr;+1 (x) + 2(k + 1)~ (x), which is equation

3.6. QED.

Next, s1nce the objective is to expand ~(x) about

x- 0 using the Taylor series, take equation 3.5 and let

x - 0, obtaining ~+ 1 (0) = 2 · 0~ (0)- ~- 1 (0) + 2k~- 1 (0).

Rearrange this and obtain

~+1 (0) + rr;_1 (0) = 2krr:-1

(0) ( 3. 7)

(only for k > 1) with initial conditions from equation 3.2.

Fork= 0, use equation 3.1 with x = 0. For typographical

convenience, write rr:+ 1 (0) as x~+ 1 , so that equation 3.7

becomes k k 2kxk-1

Xn+1 + Xn-1 = n I

where the initial conditions are, from equation 3.2,

x~ = 1 , x~ = 0 I and x~ = 0 1

and from Appendix C,

X~ - -1 , X~ = 0, X~ = 1 1 X~ - 0 1 X~ = -1 1 • • • •

( 3. 8)

( 3 • 9}

The object is to find The next step involved in

the process is to define a linear "averaging• operator on a

sequence, Jl, as

Jl(Yn)- Yn+1 + Yn-1' (3.10)

22

so that J.I.X~ - and from equation 3o8,

(3011)

for all ko

Since the goal is to find a general formula for each

x~, begin applying~ over and over again to equation 3o11

obtaining:

~ ( j.JX~ ) = 2 kJ.JX!- 1

~ 2 x! = 2 k 2 ( k - 1 ) x~- 2

~(~2x~) = 2k2(k- 1)/J.X~- 2

~ 3x~ = 2k2(k- 1)2(k- 2)x~- 3

( 3 o12)

{3013)

(3o14)

(3015)

~k- 1 X~ - 2 k 2 ( k - 1) 2 ( k - 2) o o o 2 ( k - k + 2) X~-k+ 1 - 2k- 1 k ! X~

(3o16)

2k2(k- 1)2(k- 2) o o o 2(k- k + 2)2(k- k + 1)xk-k _ n

(3.17)

so that one sees that

"kxk = 2k k ' xo ,... n • n• (3o18)

A problem ar1ses when solving equation 3.18 since the

initial data for equations 3012 to 3.16 for k>1 is x~ - 0

and x: = 0, while equation 3.18 has initial data x~ = 1 and

0 - 0 x1 - . The problem is with x~ = 1. To eliminate the

problem simply use equation 3.16 where ~ is taken to the

k- 1 power instead of solving equation 3.18. This involves

solving

23

xk = 2k-l k 1 11 -k+l 1 n • ,.... Xn I

which involves finding Jl-k+ 1 •

Upon the continuation of the process to find x~,

define another operator, E, such that ~Yn = Yn+k and

(3.19)

Yn-k· Now from equation 3.10, J1 defined in terms of

E is Jl = E + E""" 1 = E (I + E""" 2 ) , so that

Jl-1 = r I + g-2 rl p;l = ( l+~2 ) p;l • (3 .20)

Note: Although _1_ . not defined, simply apply the Taylor 1+~2

1S

ser1es expansion of ( 1~x ) where ao

( l~x) = L (-1;nxn, lxl < 1, (3.21) n=O

and substitute E"""2 for x. Formally, obtain from equations 3.20 and 3.21, Jl-

1 =( 1+~2 )E'""" 1 =( i,(-l)m(E"""2 )m)g-1 so that m=O

ao

Jl-1 = { L ( -1 )m p;2m } p;l • ( 3 • 2 2 ) m=O

By applying Jl-1 to ao ao

J.L-lx~ = ( L ( -1 )m e-2m ) p;1 X~ = ( L ( -1 )m e-2m ) x~-1 m=O m=O

[(-1 )o e-2·o + (-1 / e-2·1 + (-1 )2 g-2·2 + (-1 l p;2·3 + .. ·]x~_ 1 [I - g-2 + g-4 - e-6 + - ... ]x~-1

I 0 ~2 0 + ~4 0 - ~6 xo + Xn-1 - L Xn-1 L Xn-1 L n-1

0 0 0 0 xn-1 - xn-3 + xn-5 - xn-7 + - .... (3.23)

Notice, however, that this series is finite since all terms

after + x~ (which is equal to 1) or + x~ (which is equal to

0) do not exist. If n is odd, each n in x~ is even, and the

last term in the series is x~=1. However, if n is even,

the last term in the series is x~=O.

Next, find ( J.L-1 l- 1 • It would seem logical to begin

24

with equation 3.22 and ra1se each side to the power of

k - 1, obtaining

(J.L-1 l-1 = [( 1+~2 )~1 f-1 = { { i (-1)m ~2m} ~1} k-1 which lS

m=O CIO

Jl-k+1 = ( l+~2 l-1 ~k+1 = { L (- 1 r ~2m) 1-1 g-k+1. ( 3 • 24 ) m=O

This, however, raises an infinite ser1es to a power, and an

easier method for obtaining Jl-k+1 comes from applying the

Taylor Series expansion of ( 1 ~x)k to ( 1+~2)k. In order to do

this, define an operator, D, such that py = y'. Take each

derivative of (~)1 , beginning with the first:

D2 (-1_ )k 1+x

D( 11xl = (-1)k(1 + xr'k+11 = (-1)k( 1 ~xl+ 1

-k(-1)(k + 1)(1 + xr(k+ 2) = (-1l k(k + 1)( 1~xl+ 2

D] (_1_ )k 1+x (-1/ k{k + 1}{k + 2}( 11xl+]

thus, obtaining the mth derivative which is

Dm ( 1 1 X l = ( -1 )m k ( k + 1 ) ( k + 2 ) • • • ( k + m - 1 ) ( 1 ~X l + m •

Next, evaluate Dm( 1 ~x)k at x 0 obtaining

Dm ( 1 ~ x } I x= 0 -( k + m -1 J ! ( -1 }m

(k-1)! .

(3 .25)

(3.26)

(3.27)

(3 .28)

(3 .29)

The Taylor Series expansion of equation 3.29 about x = 0 CIO .

1S: CIO ~ ( m + Jc -1 J ! ( -1 }m L .t.J (k-1)! m!

m=O

~ (m+k-1)! (-1 )m xm .t.J (k-1)! m!

m=O

CIO

= ~ m=O

Substitute ~2 for x, and obtain

25

ClD

( 1+~-2 l = :!: m=O

( m+k-1) k - 1 ( -1 }m ( e-2 }m • To the k - 1 power

this is

ClD

( _.L_ )k-1 = """ 1+~2 Ll

m=O

From equations 3.24 and 3.30,

(3 .30)

k+1 co ( m + k - 2 ) Jl- :!: k - 2 ( -1 r e-2m e-k+ 1

1 SO that from equation m=O

3.19,

ClD

X~ = 2k-l k ! }2 (3 .31) m=O

Simplify this further by applying e-k+l to x~, and obtain:

ClD

X~ =2k-l k ! }2 m=O

( m+k-2) k - 2 ( -1 )m e-2mx~-k+l' and apply e-2m to

x!_k+1 , to obtain:

ClD

2k-1 k ! :!: m=O

( m ~ k 2 2

) (-lr x~_ •• 1_ 2m· (3.32)

But what is x~-k+l- 2m? From equation 2. 2 5,

X~ ( ~ + n + -t (-1 t) sin nr. Notice that if n l.S odd,

X~ - n sin nr I and if n is even, X~ = 0. Therefore I

substituting n - k + 1 - 2m for n in equation 2. 25, results

l.n

x 1 = ( n - k + 1 - 2m) sin (n-k+~- 2m)U I n-k+l-2m (3.33)

where n k + 1- 2m is odd. Furthermore, from the subscript

of x in equation 3 .32, n- k +1-2m > 0, implying that

m < n-~+ 1 • This is interesting Sl.nce the series is finite.

This implies that the range of the sum is from m = 0 to

[n-~+ 1 ]. Combining equation 3.32 with the range of m,

26

equation 3.32 becomes

[o-;+1) 2k-1 k! I:

m=O

{ m ~ k 2 2} (-l)'"(n- k + 1- 2m)sin ln-k+~->mm.

(3.34)

This needs to be simplified further, so break down • (n-k+l-2m)D · ( (n k+l)U } s~n 2 1.nto sin - 2 - mll I which, from the

trigonometric identity for sin(a- b) . l.S

S .;n ( fn-k+lJU mfl) · rn-k+l)U mfl rn-k+lJll · mfl ~ 2 - =s~n 2 cos - cos 2 s~n . Since

rn is either 0 or a positive integer, cos mll = (-1 t and

sin mi1 = 0, yielding

sin ( (n-k2+l)ll - mfl) =( -1 t sin rn-k2+1Jfl • (3.35)

Substitute equation 3.35 into equation 3.34 and obtain:

[~] ( ) x~ = 2k-l k 1 ± m ~ k 2

2 * m=O

(-1)m (n - k + 1 - 2m)(-1)m sin (n-k;l)ll. (3 .36)

Now notice that in equation 3.36, (-1)m(-1)m = 1, and that

sin rn-k; 1'n does not depend upon rn so equation 3. 36 becomes

[n-~+1) ( ) x~ 2k-l k! sin rn-k;lJU I: rn ~ k

2

2 ( n - k + 1 - 2m).

m=O

(3.37)

Notice in equation 3. 37 that if ~~-~+ 1 is even sin rn-k;1'n =0 1

and therefore x~ = 0. This means that both n and k must

either be both even or both odd, which implies the sum's

limit of [ w-~+1] . simply equal to 11-lc so that equation upper l.S 2 I

3.37 looks like o-1<

(m+k-2) xk 2k-1 k ! sin rn-k;lJD

-:l (3 .38) - :E k-2 (n - k + 1- 2m).

n m=O

27

Now separate (n- k + 1- 2m) into (n- k + 1) + (-2m) so that

equation 3.38 becomes

m=O

( m+k-2) k _ 2 ( n - k + 1 )+

(3.39)

To simplify equation 3.39 break it down into 2 parts and

first look at n-lc -2

I: m=O

( m+k-2) k-2 (n-k+1). (3.40)

Since (n - k + 1) does not depend upon m it can be factored

out leaving

Now consider

r>-lc -2

(n-k+1)L m=O

( m+k-2) k-2 .

r>-lc -2

~ ( m+k-2) k-2 m=O

r

From [4] p 62, problem 7, ~ v=O

Apply this to equation 3.42 with

m = V 1 k : k + 1 1 and r : n2k 1 and obtain

(3 .41)

(3.42)

n-lc -2

I: m=O

{ m ~ k 2

2 ) = { ;+~- 2

1 ) , which simplifies equation 3. 41

to n-lc -2

(n- k + 1) L m=O

( m + k - 2 ) -( n - k + 1) ( n+~-2 ) k - 2 - k - 1

so that equation 3.39 now becomes

x! = (21- 1 k!sin (II-A:;nn)

[ ( n+k-2 )

(n - k + 1) k: 1

D-Ie -2

+I: m=O

28

(3.43)

(3.44)

To continue the process of simplification, consider the

sum n-1< -2

1: m=O

which can be written as n-1<

( m + k- 2) (-2 m) k-2

2 -:z

r k= 21 , L ( m + k - 2) ( m + k - 3 ) · · · ( m + 2) m • m=O

(3 .45)

(3.46)

(Note: Sums such as equation 3.46 can be evaluated us1ng

the Euler-Maclaurin summation formula, [5]

g ( 0

) + g ( 1 ) + . . · + g ( N - 1) + g ( N) , where N = ,_/ . This 2 2

development leads to a link with the Bernoulli numbers, but

consequently does not lead to any simplifications.)

Gathering each change made to equation 3.39, it now

becomes

x: 2k-1 k! sin rn-k;lJfl [ (n - k + 1) { ;·t~) -

,k!21! 'f ( k + m - 2) · · · ( m + 1 )m ] . m=D

Replace x~ in equation 3. 4 7 with rr: ( 0) and obtain

-r: ( 0 )= 2•- 1 k ! sin rn-k;lln [ ( n - k + 1) { ;•t2

1 ) -

2 ~I< ( k + m - 2) · · · ( m + 1 ) m ] (k-2)! f-J •

m=O

This leads to the following theorem.

(3.47)

(3 .48)

Theorem: For k and n integers, with k > 0 and n > 1, and

with k and n either both even or both odd, the ~ (O)'s

where

29

T:n ( 0 )= 2k-1 k ! sin fn-k;1JD [ ( ( n+k-2 ) n - k + 1) k :._ 1

-

n-t ] 2 -2

(k-2)! r ( k + m - 2) ... ( m + 1) m m=O

present in the Taylor polynomial expansion of

r:r: (X) = Tn ( 0) + (X - 0 i Tn:f!DJ + (X - 0 l 'I;.:f!DJ + ... + (X - 0 /' -r:l<r~J

construct the Chebyshev polynomial equations of the first

kind.

Results

Each ~(0) is substituted into the Taylor ser1es

expansion of ~(x) about x = 0, equation 1.3. Since each

Chebyshev polynomial, ~(x), 1s a polynomial of degree n,

each ~ ( 0) with k > n is equal to 0. Also, since this

process has the restrictions of k > 0 and n > 1, the

results must begin with n = 1 and k - 1.

• To find ~ (x), use equation 3.48 with n - 1 and k - 1

and obtain: .

'If ( 0) = sin ~ [ ( g ) ] =1 . From Appendix C, 'If ( 0) = 0.

Apply the Taylor expansion and obtain the Chebyshev

po lynomi a 1 , ~ ( x) = 0 + 1 x = x .

• To find ~(x) use equation 3.48 with n- 2 to obtain

~(OJ= 2k-1 k!sin f3-:Jn[(3-k) (k ~ 1)-

n-lr -2

(k!2)! r (k + m- 2) .•. (m + 1)m]. m=O

First 1 from Appendix C 1 1; ( 0 J = -1. Next 1 let k = 2 to 1 0

obtain Tff ( 0) = 4 sin ~ [ ( 1 ) -2 L ] =4. Apply the Taylor m=O

30

expansion, and obtain the Chebyshev polynomial,

T2 (x) = 2x2 - 1.

• To find ~ {x), use equation 3.48 with n - 3 to obtain

~ {x) = 2k-l k! sin r4-:Jn [ (4-k) ( l~k } -k-1

J-lc 2 -2

tk-2J! I: (k + rn- 2) · · · (m + 1)m]. m=O

First, from Appendix C, ~(0) = 0. Next, let k = 1 to , 1

obtain ~ (0) = sin 3f [3 ( 0 )] = -3, and let k = 3 to obtain

~ ( 0) = 24 sin ~ [1-0] = 24. Apply the Taylor expansion, and

obtain the Chebyshev polynomial, ~ ( x) = 4 x 3 - 3 x.

• To find T4 ( x), use equation 3. 48 with n = 4 to obtain

~ {0) = 2k-1 k! sin rs-:Jn [ (5-k) ( k2~k 1} -

4-11: 2 -2

tk- 2J! I: (k + rn- 2) · · · (m + 1)m]. m=O

Next, from Appendix C, obtain T4 ( 0) = 1. Then let k = 2 to n 2 1

obtain T4 ( 0) = 4 sin 3f [ 3 ( 1 ) -2 I: rn] = -16, and let k = 4 to m=O

obtain ~ ( 0) = 192 sin IJ [ 1 - 0] =192. Again, apply the Taylor

expans1on, and obtain the Chebyshev polynomial,

'1:, ( x) = 1 - x 2 ~ ~ + x 4 1it = 8 x4

- 8 x2 + 1 .

• To find T5 (x), use equation 3.48 with n- 5 to obtain

rr: (0) = 2k-1 k! sin t6-:Jn [ (6-k) ( kJ~k 1}-5-lc -2

rk!21 ! I: (k + rn- 2) · · · (m + 1)m]. m=O

First, from Appendix C, obtain T5 ( 0) = 1. Then let k = 1 to

, ( 2 ) . obtain T5

( 0) = sin 5f [ ( 5 0 ] =5, and let k = 3 to obta1n

rr; ( 0) = 2 4 sin 3 f [ 3 (~ ) -2 ± ( rn + 1 ) rn] = -12 0 . F ina 11 y, 1 e t k = 5 m=O

to obtain 'If ( 0) = 1920 sin ~ = 1920.

31

Apply the Taylor expansion, and obtain the Chebyshev

polynomial, T5 ( x) = 16x5 - 20x3 + Sx.

Each of these polynomials may be checked with the Chebyshev

polynomials in Appendix A.

The process may continue in this manner indefinitely.

This method, which uses operators to generate the

coefficients of the Chebyshev polynomials, began with the

"averaging" operator, J.L, then the forward differencing

operator, E, and used the differential operator, D. It was

effective 1n generating the ~(O)'s for use in generating

the Chebyshev polynomials after T0 ( x).

32

CHAPTER IV

CONCLUSIONS

There are advantages and disadvantages for both

methods of development of the Chebyshev polynomials. The

disadvantage to the difference equation method presented

in Chapter II is that no closed form for the ~(O)'s was

obtained. In several places it looks as if one might be

possible, and in fact, a generalization was formed

where

~ ( 0) = [ gk [n] + ( -1 r g( n )] sin nr * * ( cos nf. if k is even) where g 0 [ n 1 = 1 and g 0 ( n ) = 0 .

The term, gk[n], k>O, stands for a polynomial not

multiplied by (-1)k and gk(n), k>O represents a

( 4 .1)

polynomial which is multiplied by (-1)k. Before ~-1 of

these polynomials is taken, they are the same. However,

once ~-1 is taken, they differ. That is why both

polynomials are represented with a •g• and one has

brackets and the other contains parenthesis. The key to

a closed form 1 ies within a closed formula for ..1-1

( -1 r nk.

In research thus far, none is apparent. Therefore, as

far as the generalization, 4.1, is concerned, it is

neither practical nor convenient. Another disadvantage

is that the algebra for each computation of ~(0) becomes

progressively more and more tedious and time-consuming.

An advantage, however, is that the method works.

33

Difference equations can be used in conjunction with the

Taylor series expansion to find the Chebyshev polynomial

equations.

The second method, presented in Chapter III,

contains more advantages than disadvantages since a

closed form, equation 3.46, was obtained. It is also

relatively simple to use. Another advantage is that a

link has been found between the Bernoulli numbers and the

Chebyshev polynomials. This link, when applied, served

only to complicate the computation of equation 3.46, and

was consequently left out. However, it could be useful

for other research. The disadvantages seem to be that

this method of computation of the Chebyshev polynomials

1s no easier that current methods, and the restrictions

on n and k prohibit this method from being effective for

finding T0 ( x).

•

34

REFERENCES

[1] P. Davis: Interpolation and Approximation, Dover Publications, Inc., New York, 1975.

[2] J. White: Notes on Difference Equations, unpublished manuscript, 1965.

[3] F.B. Hildebrand: Introduction to Numerical Analysis, Second Ed., Dover Publications, Inc., New York, 1974.

[4] W. Feller: An Introduction to Probability Theo~ and Its Applications, Vol I, John Wiley and Sons, Inc., New York, 1972.

[5] J. Stoer and R. Bulirsch: Introduction to Numerical Analysis, Springer-Verlag, New York, 1980.

35

APPENDIX A

CHEBYSHEV POLYNOMIALS

The following list of polynomdals was taken from [1).

T0

(X) = 1

~ (x) = X

'11 (x) = 2x2 - 1

~ (x) = 4x3 - 3x

~(x) = 8x4 - Bx2 + 1

Ts (x) = 16x5 - 20x3 + Sx

~(X)= 32x6 - 48x4 + 1Bx2 - 1

T1 (X) = 64x7 - 112x5 + 56x3

- 7x

Tg (x) = 128x8 - 256x6 + 160x4

- 32x2 + 1

~ (xJ = 256r - 576x7 + 432x5 - 120x

3 + 9x

~0 (x) = 512~0 - 1280x8 + 1120x6

- 400x4 + 50x

2 - 1

36

Oth:

1st:

2nd:

3 rd:

4th:

5th:

6th:

7th:

8th:

9th:

lOth:

Tn+1 (X) -

T"' (X) n+l -

IV Tn+1 (X) -

APPENDIX B

DERIVATIVES OF THE CHEBYSHEV

RECURRENCE RELATION

2 XT0

(X) - T0

_1 (X)

2 XT~ (X) - T~_ 1 (X) + 2 T0

(X)

2 xT~' ( X ) - T~~ 1 ( X ) + 4 T~ ( X )

2 XT~" ( X ) - T~'~ 1 ( X ) + 6 T~' ( X )

2 xT~v ( X ) - T~~ 1 ( X ) + 8 T~" ( X )

IX 2 xTniX (X) - Tni-X1 (X) + 18 TnVIII (X) Tn+1 (X)

T x (X) - 2 XT0x (X) - T

0x_ 1 (X) + 2 0 Tnix (X)

n+1

37

T0

(x)

~(x)

T2 ( x)

T3 (X)

T4 (X)

T5 (X)

T6 (X)

T7 ( x)

T8 (X)

- 1

APPENDIX C

THE CHEBYSHEV POLYNOMIALS EVALUATED

AT ZERO

=X

- 2x2 - 1

- 4x3 - 3x

- 8x4 - 8x2 + 1

- 16x5 - 20x3 + Sx

- 32x6 - 48x4 + 18x2 - 1

- 64x7 - 112x5 + 56x3 - 7x

- 128x8 - 256x6 + 160x4 - 32x2 + 1

T9(x)- 256x9

- 576x7 + 432x5- 120x

3 + 9x

~0 (x) = 512JC 0 - 1280x8 + 1120x

6 - 400x

4 +

50x2 - 1

38

'rc,(O) = 1

T1

( 0) - 0

T2

( 0) - -1

T3 ( 0) - 0

T4 ( 0) - 1

T5 ( 0) = 0

T6

( 0) ·= -1

T7 (0) = 0

T8 (0) = 1

Tg(O) = 0

TlO ( 0) = -1

APPENDIX D

THE FIRST DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

T; (X) - 0

T; (X) - 1

T; (X) - 4x

T; (X) - 12 x 2 - 3

T; ( X ) - 3 2 X 3

- 16 X

T; ( x) - 80 x 4 - 60 x 2 + 5

T:(x)- 192x5- 192x3 + 36x

T; ( x) - 448x6 - 560x4 + 168x

2 - 7

T; ( x) - 1024x7 - 1536x5 + 640x

3 - 64x

T9

(x) - 2304x8 - 4032x6 + 2160x4

- 360x2 + 9

Tio (x) = 5120JC - 10240x7 + 6720x5

- 1600x3

+

lOOx

39

T;(O)- 0

T; ( 0) - 1

T; { 0) - 0

T; { 0) - -3

T; ( 0) - 0

T; { 0) - 5

T; ( 0) - 0

T; ( 0) - -7

T: ( 0) = 0

T; ( 0) = 9

T;0 { 0) = 0

APPENDIX E

THE SECOND DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

'J.1"( X) - 0

T;'(x) - 0

T;'( X) - 24x

T;'( X ) - 9 6 x 2 - 16

T;C x > - 320x3 - 120x

T;'( x) - 960x4 - 576x2 + 36

T;'( x) - 2688x5 - 2240x3 + 336 x

T8"(x) - 7168x6

- 7680x4 + 1920x2

- 64

'I';(x) = 18432x7 - 24192x5 + 8640x

3 -

720x

~~ (x) = 46080x8 - 71680x

6 + 33600x4

-

4800x2 + 100

40

~(0) = 0

T;'( 0) - 0

T~( 0) - 0

T1~ ( 0) - 100

T;'t X) -

T;'t X) -

T;'t X) -

T;'t X) -

T;'t X) -

T~'t X) -

T:'{ X) -

T;'t X) -

T"{ X) -8

T;'t X) -

APPENDIX F

THE THIRD DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 T;'t 0)

0 T"t 0) 1

0 T;'t 0)

24 T;'t 0)

192x T;'t 0)

960x 2 - 120 T~'t 0)

3840x3 - 1152x T:'t 0)

13440 x 4 - 6720x2 + 336 T;'t 0)

43008x5 - 30720x3 + 3840x T;'t 0)

129024x6 - 120960x4 + 25920x

2 - 720 T;'t 0)

3

- 0

- 0

- 0 -

- 24

- 0

- -120

- 0

- 336

- 0 -

- -720

'Ji~ (X) = 368640x7 - 430080x5 + 134400x - T;~ ( 0) = 0

9600x

41

T~v (X) -

Tlrv (X) -

T;v (X) -

T;v (X) -

T:v (X) -

T:v (X) -

T:v (X) -

T~v (X) -

T:v (X) -

T:v (X) -

rr:: (X) -

APPENDIX G

THE FOURTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 T~v ( 0)

0 T:v ( 0)

0 T;v ( 0)

0 T)IV ( 0)

192 T;v ( 0)

1920x T:v ( 0)

11520x2 - 1152 T[' ( 0)

53760x3 - 13440x T;v ( 0)

215040x4 - 92160x2 + 3840 T;' ( 0)

774144x5 - 483840x3 + 51840x T9IV ( 0)

- 0

- 0

- 0

- 0

- 192

- 0

- -1152

- 0

- 3840

- 0

2580480x6 - 2150400x

4 + T11~ ( 0) = -9600

403200x2 -9600

42

Tov (X) -

Tlv (X) -

T: (X) -

T3v (X) -

T: (X) -

Tsv (X) -

T: (X) -

T; (X) -

Tav (X) -

T9v (X) -

~~ (x} =

APPENDIX H

THE FIFTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 T;;v ( 0) -

0 ~v (0} -

0 ~v (0) -

0 T3v(0) -

0 ~v (0) -

1920 Tsv (0) =

23040x T: ( 0) =

161280x2 - 13440 T~ ( 0) =

860160x3 - 184320x ~v(O)=

3870720x4 - 1451520x

2 + 51840 T: ( 0) =

0

0

0

0

0

1920

0

-13440

0

51840

15482880x5 - 8601600x

3 + ~~ ( 0) = 0

806400x

43

T;1 (x) -

~VI (X) -

T;1 (x) -

1;VI (X) --

~VI (X) -

T5VI (X) -

T:1

(X) --

T:1 (x) -

T;1 (x} -

T:1 (x) -

T,VI (X) -10

APPENDIX I

THE SIXTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 1'oVl ( 0) -

0 ~VI ( 0) --

0 T;1 ( 0) -

0 T]Vl ( 0) =

0 T.tVI ( 0) =

0 T5VI ( 0) -

23040 T6Vl ( 0) -

322560x T:1

( 0) -

2580480x2 - 184320 TaVI ( 0) -

15482880x3 - 2903040x T9VI ( 0) =

77414400x4 - 25804800x2 + rr:; ( 0) = 806400

44

0

0

0

0

0

0

23040

0

-184320

0

806400

T:II (X) -

~VII (X) -

~VII (X) -

~VII (X) -

T4VII (X) -

T;n (x) -

T6VII (X) -

T;II (x) -

TaVII (X) --

~VII (X) -

~~II (X) -

APPENDIX J

THE SEVENTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 ToVII ( 0) =

0 ~VII ( 0) =

0 T2VI1 ( 0) -

0 ~VII ( 0) -

0 ~VII ( 0) -

0 T:II ( 0) -

0 T:II ( 0) -

322560 T:II ( 0) =

5160960x TaVli ( 0) =

46448640x2 - 2903040 ~VII ( 0) =

309657600x3 - 51609600x ~v:I ( 0) =

45

0

0

0

0

0

0

0

322560

0

-2903040

0

T;ni (x) -

T,_VIII (X) -

'l'JIII (X) --~VIII (X)

'J',Vlii (X) -

T;zn (x) -

T:III (X) -

T:III (X) -

'1':111 (X) -

'I'gVIII (X) -

T,.V:II (X) -

APPENDIX K

THE EIGHTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 'roVIII ( 0) : 0

0 'r,_VIII ( 0 ) : 0

0 TJIII (0) = 0

0 ~VIII (0) : 0

0 T:III (0) = 0

0 T;III (0) = 0

0 T:III (0) = 0

0 T:III (0) = 0

5160960 T:III (0) = 5160960

92897280x 'I'gVIII ( 0) : 0

928972800x2 - 51609600 T,.V:II ( 0 ) : -51609600

46

Tox (X) =

'Ifx (x) =

~x (x) =

'Itx (X) =

'!'Ix (x) =

Tsx (X) =

Tfx (x) =

T;x (x) =

r,x (x) =

'l'Jx (X) =

'If: (x) =

APPENDIX L

THE NINTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

0 Tax (0) = 0

0 'l'Jx (0) = 0

0 ~X (0) = 0

0 'Itx (0) = 0

0 ~}l (0) = 0

0 ~X (0) = 0

0 Tix (0) = 0

0 T,X ( 0) = 0

0 rz;x ( 0) = 0

92897280 ~X ( 0) = 92897280

1857945600x 'If: (0) = 0

47

'Jt (X} = 0

'If (X} = 0

'zt (X} = 0

'If (x} = 0

rr: (X} = 0

~(X}= 0

'It (X) = 0

'r: (x} = 0

'If (X} = 0

'If (x} = 0

APPENDIX M

THE TENTH DERIVATIVES OF THE CHEBYSHEV

POLYNOMIALS EVALUATED AT ZERO

rr: (0} = 0

'If (0) = 0

':rJ (0} = 0

'r: (0) = 0

,; (0) = 0

~ (0} = 0

'It (0} = 0

rr: (0) = 0

rr: (0} = 0

'Iff (0} = 0

2fo (x} = 1857945600 '1fo ( 0) = 1857945600

48

APPENDIX N

THE CASSORATI

Information concerning the Cassorati comes from [2].

The Cassorati, C(n + 1) =

y 1 (n + 1) y 2 (n + 1) y 1 (n + 2) y 2 (n + 2)

COS tn•;m - COS tn•;m

sin (n+ lJfl 2

sin tn•;m --sin nr - cos .LJfl

cos 'f -sin nr - 1,

where y 1 (n) and y 2 (n) are linearly independent solutions

of y ( n + 2) + y ( n ) = 0 , equation 2 . 5 .

49

APPENDIX 0

INDEFINITE SUMS

The following information comes from [2] .

.d-1 ( 1) = n

Li-1n - 1. n2 2 -l.n 2

.d-1 n2 - t nJ - .1 n2 2 +l.n 6

.d-1 nJ - .; n4 - l. nJ 2 + 1 n2

.d-1 n4 ~ ns - l. n4 2 + 1. nJ 3

-..Ln 30

.d-1 (-1f n = -t (-1f (-2n + 1)

.d-1 (-1)n n 2 - - ~ n(-1f (n- 1)

.d-1 (-1f n 3 -1- (-1f (4n3 - 6n2 + 1)

50

APPENDIX P

DETAILS OF THE GENERALIZATION FROM CHAPTER II

Generalization: ~ (0) = [g 0 [n] + (-1t g0

(n)]cos11fl,

where g 0 [ n J = 1 and g 0 ( n ) = 0 •

• 1st derivative: , ,

'!;,+ 1 ( 0) + Tn_ 1 ( 0) = 2 ~ ( 0) •

Replace n with n + 1, so , ,

Tn+2 ( 0) + Tn ( 0) = 2 Tn+ 1 ( 0).

• General solution: ,

~ ( 0) = ( c 2 + u 2 ) sin tf

U 2 (n) = 2L1-1 [g0[n + 11 + (-1f+ 1 g

0(n + 1)1(cos rn+;Jn l

= L1-1 [g0 [n + 1]- (-1t g 0 (n + 1)1(1- (-1t).

This is the point at which the two polynomials are the same:

= L1-1 { g 0 [ n + 1 ] + g 0 ( n + 1 ) - ( -1 t [ g 0 [ n + 1 1 + g 0 ( n + 1 ) 1} •

Once L1-1 is taken, however, these differ:

L1-1 [ g

0 [ n + 1] + g 0 ( n + 1) ~ L1-1

( -1 t ( g 0 [ n + 1 1 + g 0 ( n + 1)). ,

Tn ( 0) = [ c 2 + L1-1 [ g 0 [ n + 11 + g 0 ( n + 1)-

L1-1 ( -1 r ( g 0 [ n + 1 1 + g 0 ( n + 1 ) 1 sin ~ .

• Now replace n with n + 1, and replace

c2

+ L1-1 {g0 [n + 11 + g 0 (n + 1) with g 1 [n1, and

- L\-1 ( -1 t ( g

0 [ n + 1 1 + g 0 ( n + 1)) with g 1 ( n). This

generalization can continue in this manner, with any

constants being incorporated into the g's.

51