Chebyshev equationChebyshevEquation2002-11-21 19:59:452002-11-21 20:18:39Definition% this is the default PlanetMath preamble. as your knowledge
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% define commands hereChebyshev's equation is the second order linear differential equation
$$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2y = 0$$
where $p$ is a real constant.
There are two independent solutions which are given as series by:
$$
y_1(x) = 1 - \tfrac{p^2}{2!}x^2 + \tfrac{(p-2)p^2(p+2)}{4!}x^4
- \tfrac{(p-4)(p-2)p^2(p+2)(p+4)}{6!}x^6 + \dotsb
$$
and
$$
y_2(x) = x - \tfrac{(p-1)(p+1)}{3!}x^3 + \tfrac{(p-3)(p-1)(p+1)(p+3)}{5!}x^5 - \dotsb
$$
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 $|x| < 1$; 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 $-p$.)