What is the Solution to the Chebyshev Polynomial Problem?

[Muh. Fauzi M.]

This is something Chebyshev polynomial problems. I need to show that:

$\sum_{r=0}^{n}T_{2r}(x)=\frac{1}{2}\big ( 1+\frac{U_{2n+1}(x)}{\sqrt{1-x^2}}\big )$

by using two type of solution :
$T_n(x)=\cos(n \cos^{-1}x)$ and $U_n(x)=\sin(n \cos^{-1}x)$ with $x=\cos\theta$,

I have form the complex superposition:

$T_n(x)+iU_n(x)=(x+i\sqrt{1-x^2})^n$

and expand it by binomial theorem to get :

$T_n(x)=x^n-\dbinom{n}{2}x^{n-2}(1-x^2)+\dbinom{n}{4}x^{n-4}(1-x^2)^2-...$

and

$U_n(x)=\sqrt{1-x^2}\big[ \dbinom{n}{1}x^{n-1}-\dbinom{n}{3}(1-x^2)+... \big]$

I try to change $T_n(x)$ to $T_{2r}(x)$ and $U_n(x)$ to $U_{2n+1}(x)$, but still stuck and can't solve the problem.

Any one can help solve this?

 

[Ray Vickson]

This is something Chebyshev polynomial problems. I need to show that:

$\sum_{r=0}^{n}T_{2r}(x)=\frac{1}{2}\big ( 1+\frac{U_{2n+1}(x)}{\sqrt{1-x^2}}\big )$

by using two type of solution :
$T_n(x)=\cos(n \cos^{-1}x)$ and $U_n(x)=\sin(n \cos^{-1}x)$ with $x=\cos\theta$,

I have form the complex superposition:

$T_n(x)+iU_n(x)=(x+i\sqrt{1-x^2})^n$

and expand it by binomial theorem to get :

$T_n(x)=x^n-\dbinom{n}{2}x^{n-2}(1-x^2)+\dbinom{n}{4}x^{n-4}(1-x^2)^2-...$

and

$U_n(x)=\sqrt{1-x^2}\big[ \dbinom{n}{1}x^{n-1}-\dbinom{n}{3}(1-x^2)+... \big]$

I try to change $T_n(x)$ to $T_{2r}(x)$ and $U_n(x)$ to $U_{2n+1}(x)$, but still stuck and can't solve the problem.

Any one can help solve this?

Note that $\sum_{r=0}^n T_{2r}(x) = \sum_{r=0}^n \cos(2 r \theta)$, where $x = \cos(\theta)$. Can you evaluate that last summation?

 

[Muh. Fauzi M.]

Note that $\sum_{r=0}^n T_{2r}(x) = \sum_{r=0}^n \cos(2 r \theta)$, where $x = \cos(\theta)$. Can you evaluate that last summation?

Thanks for your respond. I've made it by choosing an arbitrary $n$ and then evaluate both $\sum_{r=0}^n T_{2r}(x)$ and $U_{2n+1}(x)$, for example $n=1$, and, voila...

 

[Ray Vickson]

Thanks for your respond. I've made it by choosing an arbitrary $n$ and then evaluate both $\sum_{r=0}^n T_{2r}(x)$ and $U_{2n+1}(x)$, for example $n=1$, and, voila...

How did you evaluate $\sum_{r=0}^n T_{2r}(x)$? For example, are you able to evaluate this for $n = 10,000$ or $n = 5,000,000$? The problem requires that you do it for all possible finite values of $n$.

 

[Muh. Fauzi M.]

How did you evaluate $\sum_{r=0}^n T_{2r}(x)$? For example, are you able to evaluate this for $n = 10,000$ or $n = 5,000,000$? The problem requires that you do it for all possible finite values of $n$.

I see mr. that's my problem actually. But for accomplishing an assignment in the short of time, I fall to just using a deductive reasoning.
Well, let use your advice, so
$\sum_{r=0}^n T_{2r}(x)=\sum_{r=0}^n\cos(2r\theta)=1+\cos(2\theta)+\cos(4\theta)+...+\cos(2n\theta)$
Then... Can't see the pattern.

 

[geoffrey159]

Try to work with the complex exponential (http://en.wikipedia.org/wiki/Euler's_formula)

 

[Ray Vickson]

I see mr. that's my problem actually. But for accomplishing an assignment in the short of time, I fall to just using a deductive reasoning.
Well, let use your advice, so
$\sum_{r=0}^n T_{2r}(x)=\sum_{r=0}^n\cos(2r\theta)=1+\cos(2\theta)+\cos(4\theta)+...+\cos(2n\theta)$
Then... Can't see the pattern.

See. eg., http://mathworld.wolfram.com/Cosine.html