# Two formulas for [x.sup.n] being represented by Chebyshev polynomials (1).

Abstract Some very simple formulas to show how [x.sup.n] (n = 1, 2, ...) is represented by Chebyshev polynomials [T.sub.k](x) and [U.sub.k](x)(k = 0; 1, ... and their an application are given in this paper.

Keywords Chebyshev polynomial, represent, orthogonal property, common formula.

[section]1. Introduction

The first type Chebyshev polynomials [T.sub.k](x):

[T.sub.k](x) = 1/2 [[(x + [square root of ([chi square] - 1)].sup.k] + (x - [square root of [([[chi square].sup.2] - 1).sup.k]], k = 0, 1, ...

and the second type Chebyshev polynomials [U.sub.k](x):

[U.sub.k](x) = 1 / 2[[[square root of ([chi square]) - 1]).sup.k+1] - (x - [square root of [([[chi square].sup.2] - 1).sup.k+1]],k = 0, 1, ...

have widely applications in many fields [1-4], contact closely with Fibonacci numbers, Lucas numbers [5-6], and so on. It is a general method that each [x.sup.n] (n = 1; 2; is represented by Chebyshev polynomials [T.sub.k](x) and [U.sub.k](x) (k = 0, 1, ... , n). But so far it is regretted that one can only use some recurrence relations[7], recurrence formulas

[T.sub.k+2] (x) = 2x[T.sub.k+1](x) - [T.sub.k](x); [U.sub.k+2](x) = 2x[U.sub.k+1](x) - [U.sub.k](x); k = 0, 1, ...

where [T.sub.0](x) = 1, [T.sub.1](x) = x, [U.sub.0] = 1, [U.sub.0] = 2x, or some table [2-3]:

1 = [T.sub.0]; x = [T.sub.1]; [x.sup.2] = 1/2 ([T.sub.0] + [T.sub.2]); [x.sup.3] = 1/4 (3[T.sub.1] + [T.sub.3]); [x.sup.4] = 1/8 (3[T.sub.0] + 4[T.sub.2] + [T.sub.4]),

[x.sup.5] = 1/16 (10[T.sub.1] + 5[T.sub.3] + [T.sub.5]); [x.sup.6] = 1/32 (10[T.suub.0] + 15[T.sub.2] + 6[T.sub.4] + [T.sub.6]);

[x.sup.7] = 1/64 (35[T.sub.1] + 21[T.sub.3] + 7[T.sub.5] + [T.sub.7]); [x.sup.8] = 1/128 (35[T.sub.0] + 56[T.sub.2] + 28[T.sub.4] + 8[T.sub.6] + [T.sub.8]),

It is obviously that the bigger n is, the more difficult the problem is. In this paper, we shall give the very simple formulas to solve the problem thoroughly, and give some perfect related results of trigonometric formulas.

Illustration: "!" is the factorial sign and "!!" is the double factorial sign through this paper. For examples,

9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1; 9!! = 9 * 7 * 5 * 3 * 1; 8!! = 8 * 6 * 4 * 2.

[section] 2. Main Result

Theorem 1. Let

[x.sup.n] = 1/2 [a.sub.n]0[T.sub.0](x) + [[infinity].summation over (k = 1)] [a.sub.nk][T.sub.k](x).

Then if k [greater than or equal to] n + 1, [a.sub.nk] = 0; If k [less than or equal to] n, k and n are of opposite parity, then [a.sub.nk] = 0; If k [less than or equal to] n, k and n are of same parity, then

[a.sub.nk] = 2n! / (n - k)!! (n + k)!!.

Proof. Multipy [T.sub.m](x)/[square root of([1 - [chi square]] (m = 0, 1, ...) to the two sides of (1) and definite integral from--1 to 1. Because of the orthogonal property [3,8] of the first type Chebyshev polynomials, we can get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Let x = cos t and use the formula [9]

[b.sub.p] = [[integral].sup.[pi].sub.0] [cos.sup.p] tdt = (p - 1)!! / p!! [pi], when p is even; [b.sub.p] = 0; when p is odd:

We have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So when n + k is odd, or n and k are of opposite parity (whether k > n or k < n), [a.sub.nk] = 0 for [a.sub.n]+k0 = 2/[pi] [b.sub.n+k] = 0; when k > n, n and k are of the same parity, or k = n + 2i (i > 0), [a.sub.nk] = 0 for n - k + 2i = 0; when k < n + 1, n and k are of same parity, or n + k is even, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Theorem 2. Let

[x.sup.n] = [b.sub.n0][U.sub.0](x) + [[infinity].summation over (k = 1)] [b.sub.nk] [U.sub.k] (x).

Then if k [greater than or equal to] n+1, [b.sub.nk] = 0; If k [less than or equal to] n, k and n are of the same parity, then

[b.sub.nk] = 2(k + 1)n! / (n - k)!!(n + k + 2)!!.

Proof. Multiply [U.sub.m](x)[square root of (1 - [chi square])] (m = 0; 1, ...) to the two sides of (2) and definite integral from - 1 to 1. Because of the orthogonal property [3;8] of the second type Chebyshev polynomials, we can get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Let x = cos t and we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So we can get the result of Theorem 2 from Theorem 1. By using theorem 1,2, changing n to 2n or 2n + 1, we can get very simple formulas:

Theorem 3.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

n = 0, 1, 2, ...,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

Therom 3 is very convenient for us and it is suggested to be written in the books. For example, we can immediately get that

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[section]. Some Formulas of Trigonometric Function

Although there are many formulas of trigonometric function, we can still get some new ones.

Theorem 4. For many positive integer n, we have

[cos.sup.2n] x = (2n)! [4.sup.n][(n!).sup.2] + 2(2n)!/[4.sup.n][n.summation over (k = 1)] 1 / (n - k)!(n + k)! cos(2k)x,

[cos.sup.2n + 1] x = (2n + 1)! / [4.sup.n] [n.summation over (k = 0)] 1/(n - k)!(n + k + 1)! Cos (2k + 1) x.

Proof. Making [cos.sup.m] x Fourier cosine expansion, we get

[cos.sup.m] x = 1/2 [a.sub.m]0 + [[infinity].summation over (k = 1)] [a.sub.mk] cos kx, [a.sub.mk] = 2 /[pi] [[integral].sup.[pi].sub.0] [cos.sup.m] x cos kxdx.

Changing m to 2n or 2n + 1, we can complete the proof of Theorem 4 from Theorem 1.

Theorem 5. Let n be an integer,

[G.sub.nk] = [integral] [cos.sup.n] x cos kxdx;

[H.sub.nk] = [integral] [cos.sup.n] x sin kxdx;

[I.sub.nk] = [integral] [sin.sup.n] x cos kxdx;

[J.sub.nk] = [integral] [sin.sup.n] x sin kxdx:

Then,

[G.sub.nk] = 1 / n + 1 [sin.sup.n+1] x sin (k - 1)x + n - k + 2 / n + 1 [G.sub.(n+1)(k-1)],

[H.sub.nk] = -1 / n + 1 [cos.sup.n+1] x sin(k - 1)x + n - k + 2 / n + 1 [H.sub.(n+1)(k-1)],

[I.sub.nk] = 1 / n + 1 [sin.sup.n+1] x cos(k - 1)x - n k + 2 / n + 1 [J.sub.(n+1)(k-1)],

[J.sub.nk] 1 / (n+1) [sin.sup.n+1] x sin(k - 1)x + n - k + 2 / n + 1 [I.sub.(n + 1) (k-1).

Proof. The proof can be completed just using elementary formulas

cos kx = cos x cos(k - 1)x - sin x sin(k - 1)x;

sin kx = sin x cos(k - 1)x + cos x sin(k - 1)x;

and the method of integration by parts.

Theorem 6. For any positive integer n, we have

[sin.sup.2n] x = (2n)! / [[4.sup.n][(n!).sup.2] + 2(2n)! / [4.sup.n] [n.summation over (k = 1)] [(-1).sup.k] / (n-k)!(n+k)! cos(2k)x,

[sin.sup.(2k + 1)]x = (2n + 1)!/[4.sup.n] [n.summation over (k = 0)] [(-1).sup.k] / (n - k)!(n + k + 1)! sin(2k + 1)x.

Proof. Making [sin.sup.m] x Fourier cosine expansion, we get

[sin.sup.m] x = 1/2 [a.sub.m]0 + [[infinity].summation over (k = 1)]([a.sub.mk] kx + [b.sub.mk] sin kx),

[a.sub.mk] = 1/[pi] [[integral].sup.[pi].sub.0][sin.sup.m] x cos kxdx,

[b.sub.mk] = 1[pi] [[integral].sup.[pi].sub.0] [sin.sup.m] x sin kxdx.

Changing mto 2n or 2n + 1, we can complete the proof of Theorem 6 from Theorem 5 and Theorem 1. In formula (5) and formula (7), it is interested that the absolute values of the coefficients of corresponding terms are equal. Those formulas seem wonderful.

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