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Given that $$\frac{d}{dx}T^l_k(x)=\frac{1}{l}\sum_{m=0 \\(k+m)odd}^{k-1}\frac{4k} {\xi_m}T^l_m(x)$$ AND $$\phi_j(x)=x(l-x)T^l_j(x)$$ Where $$ \xi_m = \left\{ \begin{array}{ll} 2 & m = 0 \\ 1 & otherwise \end{array} \right. $$ Prove that $$\frac{d}{dx}\phi_j(x)=\frac{4}{l}\sum_{m=0 \\ (j+m) odd}^{j-1}\frac{3j-2m}{\xi_m}\phi_m(x)+\eta^l_j(x)$$ where $$\eta^l_j(x) = \left\{ \begin{array}{ll} l-2x &j \ \ \ even \\ -l & j \ \ \ odd \end{array} \right.$$ What i have got so far is $$\frac{d}{dx}\phi_j(x)=(l-2x)T^l_k(x)+\frac{4}{l}\sum_{m=0 \\(j+m) odd}^{j-1}\frac{j}{\xi_m}\phi_m(x)$$ Edited $$T^l_k(x) =\cos(k \cos^{-1}(2x/l -1 ))$$ How to get the required result?

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  • $\begingroup$ $T^l_k$ is shifted chebyshev on $x \in [0,l]$ $\endgroup$ – TopSpin Aug 16 at 9:56
  • $\begingroup$ Sorry it's supposed to be $k$ not $i$.I have edited my post $\endgroup$ – TopSpin yesterday
  • $\begingroup$ In the equation after, "what I have got so far", shouldn't $T_k^l$ be $T_j^l$? $k$ appears nowhere else in the equation. $\endgroup$ – saulspatz yesterday
  • $\begingroup$ that's right it's $T^l_j$ $\endgroup$ – TopSpin yesterday

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