Chebyshev equation

Chebyshev’s equation is the second order linear differential equation

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+p^{2}y=0$$

where $p$ is a real constant.

There are two independent solutions which are given as series by:

$$y_1(x)=1-\frac{p^2}{2!}{x}^2+\frac{(p-2)p^2(p+2)}{4!}{x}^4-\frac{(p-4)(p-2)p^2(p+2)(p+4)}{6!}{x}^6+\mathrm{\cdots }$$

and

$$y_2(x)=x-\frac{(p-1)(p+1)}{3!}{x}^3+\frac{(p-3)(p-1)(p+1)(p+3)}{5!}{x}^5-\mathrm{\cdots }$$

In each case, the coefficients are given by the recursion

$$a_{n+2}=\frac{(n-p)(n+p)}{(n+1)(n+2)}{a}_n$$

with $y_1$ arising from the choice $a_0=1$, $a_1=0$,and $y_2$ arising from the choice $a_0=0$, $a_1=1$.

The series converge for ; this is easy to see from the ratio test and the recursion formula above.

When $p$ is a non-negative integer, one of these series will terminate,giving a polynomial solution.If $p\ge 0$ is even, then the series for $y_1$ terminates at $x^p$.If $p$ is odd, then the series for $y_2$ terminates at $x^p$.

These polynomials are, up to multiplication by a constant, the Chebyshev polynomials. These are the only polynomial solutions of the Chebyshev equation.

(In fact, polynomial solutions are also obtained when $p$ is a negative integer,but these are not new solutions, since the Chebyshev equation is invariant under the substitution of $p$ by $-p$.)