# Prove $T_n(x)$ of Chebyshev Polynomial given the recurrence relation [closed]

Using the recursion formula for Chebyshev polynomials, show that $T_n(x)$ can be written as $$T_n(x)=2^{n-1}(x-x_1)(x-x_2)...(x-x_n)$$ where $x_i$ are the $n$ roots of $T_n$

The recurrence relation: $T_0(x)=1$,$T_1(x)=x$, and $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$

I have an intuition that I need to use induction,but what is my inductive hypothesis?

Here is my work:

Assume $T_n(x)=2^{n-1}(x-x_1)(x-x_2)...(x-x_n)$

Prove $T_{n+1}(x)=2^{n}(x-x_1)(x-x_2)...(x-x_n)(x-x_{n+1})$

Use the definition of Chebyshev polynomial, $T_n(x) = cos(n\theta),x =cos(\theta)$

$T_0(x)=1$ and $T_1(x)=x$

And then use the recursion formula $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$, I get

$T_{n+1}(x) = 2x[2^{n-1}(x-x_1)(x-x_2)...(x-x_n)]-[2^{n-2}(x-x_1)(x-x_2)...(x-x_{n-1})]$

Then

$T_{n+1}(x) = 2x(2^{-1})(x-x_n)[2^n(x-x_1)(x-x_2)...(x-x_{n-1})]-2^{-2}[2^n(x-x_1)(x-x_2)...(x-x_{n-1})]$

Rearrange I get

$T_{n+1}(x) = [2x(2^{-1})(x-x_n)-2^{-2}][2^n(x-x_1)(x-x_2)...(x-x_{n-1})]$

What should I do next(assume what I did above is correct) so that I can get $T_{n+1}(x)=2^{n}(x-x_1)(x-x_2)...(x-x_n)(x-x_{n+1})$ ??

• This question is better suited for math.stackexchange.com . And, please, use the homework flag if this is your homework.
– Jan
Oct 23 '13 at 6:16
• This question appears to be off-topic because it is about a homework problem in mathematics, and is better suited for math.stackexchange.com. Oct 23 '13 at 6:56
• Note that you can write any polynomial as $C\prod_k(x-x_k)$ where the $x_k$ are its roots. Note also that the roots, i.e. the $x_k$, are not the same for $T_n(x)$ and $T_{n+1}(x)$. Oct 23 '13 at 7:25
• The question was cross posted - math.stackexchange.com/questions/536085/…
– Jan
Oct 23 '13 at 7:33
• Given the use of Chebyshev bases in numerical analysis, this question would still be on topic here, although it's not a common question. (For instance, there have been a few questions that pertain to optimization, but are more on the theoretical side, and are on-topic here.) Oct 23 '13 at 23:03