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 $T_{n+1}(x)=2^{n}(x-x_1)(x-x_2)…(x-x_n)(x-x_{n+1})$ ??
Best Answer
An inductive hypothesis is: $T_n=2^{n-1}x^n+f_n(x)$ where $\deg f\le n-1$.
Addendum: From $T_n=2^{n-1}x^n+f_n(x)$ we have $T_{n+1}=2^nx^{n+1} +2xf_n(x)-2^{n-2}x^{n-1}-f_{n-1}(x)$ and evidently $\deg f_{n+1}(x)\le n$ where $f_{n+1}(x)=2xf_n(x)-2^{n-2}x^{n-1}-f_{n-1}(x)$.
Further if $x_1,\ldots,x_n$ are the roots of $T_n$, we get $T_n=2^{n-1}(x-x_1)\ldots(x-x_n)$.