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Mathematics
Calculus
Chebyshev Differentiation Matrix
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[QUOTE="Leonardo Machado, post: 6378001, member: 599520"] [B]TL;DR Summary:[/B] It is a question about the method to obtain the Chebyshev coefficients for differential operators. Hi everyone. I am studying Chebyshev Polynomials to solve some differential equations. I found in the literature that if you have a function being expanded in Chebyshev polynomials such as $$ u(x)=\sum_n a_n T_n(x), $$ then you can also expand its derivatives as $$ \frac{d^q u}{dx^q}=\sum_n a^{(q)}_n T_n(x), $$ with the following relation $$ a^{(q)}_{k-1}= \frac{1}{c_{k-1}} ( 2 k a^{(q-1)}_k+ a^{(q)}_{k+1}), $$ being $c_k=2$ for k=0 and 1 if k>0. It all together defines the Chebyshev differentiation matrix, which is $D$ in $$ a^{(1)}_i=D_{ij} a^{(0)}_j. $$ Now I would like to know if there is any way of doing $$ x^l \frac{du}{dx}=\sum_n a^{(x)}_n T_n(x), $$ I am looking for it everywhere in the literature but I can't find a way of dealing with this kind of operator that appears in the Laplacian. I can't describe every linear operator without it [/QUOTE]
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Chebyshev Differentiation Matrix
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