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# Chebyshev equation

Chebyshev’s equation is the second order linear differential equation

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

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

$y_{1}(x)=1-\tfrac{p^{2}}{2!}x^{2}+\tfrac{(p-2)p^{2}(p+2)}{4!}x^{4}-\tfrac{(p-4% )(p-2)p^{2}(p+2)(p+4)}{6!}x^{6}+\cdots$ |

and

$y_{2}(x)=x-\tfrac{(p-1)(p+1)}{3!}x^{3}+\tfrac{(p-3)(p-1)(p+1)(p+3)}{5!}x^{5}-\cdots$ |

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 $|x|<1$; 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 $p\geq 0$ 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 $-p$.)

## Mathematics Subject Classification

34A30*no label found*

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