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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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    sorry to avoid confusing myself, I changed the n in the primary equation to an h2012-09-27

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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 $k=n$.

Orthogonality for a proper weighted inner product is discussed here.