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Chebyshev polynomial question

I am trying to prove a property of Chebyshev polynomials.

Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by $$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$$

Show that $T_n(x)= 2^{n−1}(x−x_0)(x−x_1)\cdots(x−x_{n−1})$,, where $x_0,\ldots,x_{n-1}$ are the roots of the polynomial.

I am not even sure how to get started on this and would appreciate any help!

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marked as duplicate by draks ..., Henry T. Horton, Fabian, tomasz, Martin Argerami Dec 26 '12 at 0:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What are $x_0,x_1,...$? Do you want to prove that the roots of $T_n(x)$ are real numbers? In fact, the roots are simply roots and they are in $(-1,1)$. – vesszabo Sep 10 '12 at 15:10

As I understand, you want simply to show that the leading term of $T_n(x)$ is $2^{n-1}x^n$.

By induction, you can see that

  1. this is true for $n=1$, the leading term of $T_1(x)=x$ is indeed $2^0x^1$,
  2. if it is true for a given $n$, i.e. the leading term of $T_n(x)$ is $2^{n-1}x^n$, then by $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$, given that the second term in the sum do not alter the leading term, the leading term is multiplied by $2x$, so it is $2^{n}x^{n+1}.$
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I am not sure that I entirely follow this – Samuel Gregory Sep 10 '12 at 21:34

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