# Chebyshev Diff EQ

Find a power series solution about $x_0=0$ for the Chebyshev differential equation $$(1-x^2)y''-xy'+n^2 y=0,$$ as a function of of the integer $n$. Show that the solutions form a terminating expansion for each value of $n$. What is the orthogonality relationship for these polynomials?

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Welcome to math.stackexchange! In order to make the answers helpful for you, you should tell us what you tried and where you are stuck. – Davide Giraudo Sep 27 '12 at 20:00
Here is a detailed solution. – Mhenni Benghorbal Sep 27 '12 at 20:14
@DavideGiraudo I think I solved for the power series expansion first and got this as my recurrence relation. 0=a_(n+2) (n+2)(n+1)+(h^2-n^2 ) a_n Not sure where to go from there as all my coefficients turned out really wild. – Charles Sep 27 '12 at 20:25
What is $h$ here? – Davide Giraudo Sep 27 '12 at 20:27
sorry to avoid confusing myself, I changed the n in the primary equation to an h – Charles Sep 27 '12 at 20:29

The power series around zero is $$y(x) = \sum_{k=0}^\infty a_k \,x^k.$$ Therefore $$y' = \sum_{k=0}^\infty k \,a_{k} \,x^{k-1} =\sum_{k=1}^\infty k \,a_{k} \,x^{k-1}=\sum_{k=0}^\infty \,(k+1) \,a_{k+1} \,x^{k},$$ and $$y'' = \sum_{k=0}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=1}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=0}^\infty \, (k+1)\,(k+2) \, a_{k+2} \,x^{k}.$$ Substituting these series into differential equation, we get
\begin{aligned} 0 & = \left(1-x^2\right)y''-xy'+n^2 y= \\ &= \left(1-x^2\right)\sum_{k=0}^\infty \left( (k+1)\,(k+2) \, a_{k+2} \,x^{k}\right) -x\sum_{k=0}^\infty \left((k+1) \,a_{k+1} \,x^{k}\right)+n^2 \sum_{k=0}^\infty a_k \,x^k = \\ &= \sum_{k=0}^\infty \Big( (k+1)\,(k+2) \, a_{k+2} \left(1-x^2\right)x^{k} (k+1) -\,a_{k+1} \,x^{k+1} +n^2 a_{k} \,x^k\Big) = \\ & = \big(2 a_2 + n^2 a_0 \big) + \Big(\big(n^2 -1\big)a_1 + 6a_3 \Big)\, x + \\ & \phantom{=\big(} \sum_{k=2}^\infty \Big( (k+1)\,(k+2) \, a_{k+2} + \left(n^2 - k^2\right)a_k \Big) \,x^k =0 \end{aligned} Thus, \begin{aligned} 2 a_2 + n^2 a_0 =0 , \\ \big(n^2 -1\big)a_1 + 6a_3 = 0. \end{aligned}\label{*}\tag{*} By induction, for integer $k \geq 2$ $$a_{k+2} = \frac{(k-n)\,(k+n)}{(k+1)\,(k+2)} a_{n}$$ Determining initial coefficients for odd $k$ and for even $k$ from the system $\eqref{*}$, you will be able to get explicit formula for both even and odd part of the series.
Note that the series terminates at $k=n$.