User:Egm6322.s12.team1.zheng.zx/R14

Problem 5: Derive the Christoffel-Darboux identity for Legendre polynomials
 Solved without assistance 

Given:
The 3-term recurrence relation for any sequence of orthogonal polynomials;
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\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$
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for n=0,1,2,..., and $$\,\,p_{-1}(x)=0\,\,$$
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\displaystyle A_n\,,B_n\,,C_n \in \mathcal R\,,with\, A_{n-1}A_nC_n > 0 $$
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If the highest coefficient of $$\,\,p_n(x)\,\,$$ is kn, then
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\displaystyle A_n=\frac{k_{n+1}}{k_n} $$
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\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
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Find:

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\displaystyle \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k}=\frac{p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)}{A_nh_n(x-y)} $$
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Solution:
From the three term recurrence relation
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\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$
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multiply both sides by pn(y):
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\displaystyle p_{n+1}(x)p_n(y)=(A_nx+B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x) $$ $$
 * $$\displaystyle (Equation\;5.1 )
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And for the symmetry between the factors
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\displaystyle p_{n+1}(y)p_n(x)=(A_ny+B_n)p_n(x)p_n(y)-C_np_n(x)p_{n-1}(y) $$ $$
 * $$\displaystyle (Equation\;5.2 )
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Substract equation 5.2 from equation 5.1:
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\displaystyle p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)=A_n(x-y)p_n(x)p_n(y)-C_n[p_n(y)p_{n-1}(x)-p_n(x)p_{n-1}(y)] $$
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For
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\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
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Then
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\displaystyle \begin{align} & \frac{p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)}{A_nh_n(x-y)} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_n(x)p_{n-1}(y)-p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)h_{n-1}} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_{n-1}(x)p_{n-1}(y)}{h_{n-1}}+\frac{p_{n-1}(x)p_{n-2}(y)-p_{n-1}(y)p_{n-2}(x)}{A_{n-2}(x-y)h_{n-2}} \\ & \vdots \\ =& \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k} \end{align} $$
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