3
$\begingroup$

Possible Duplicate:
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!

  • 0
    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 $(-1,1)$.2012-09-10

1 Answers 1

4

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 $2^{n}x^{n+1}.$
  • 0
    I am not sure that I entirely follow this2012-09-10