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I just read about Chebyshev polynomials and that they are used in approximations.

I don't fully understand them yet.

What is the relation between Chebyshev polynomials and Taylor expansions?

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I'm not entirely sure which relation you have in mind so I will tell you something about expanding functions in terms of these polynomials.

They are orthogonal polynomials. There are two types $T_n(x)$ and $U_n(x)$. If we have a function with a Taylor series $$ f(x) = \sum_{k=0}^\infty a_k x^k $$ we might want to write it in a basis of the polynomials $$ f(x) = \sum_{k=0}^\infty b_k T_k(x) $$ the orthoganality condition is $$ \int_{-1}^1 T_n(x) T_m(x) \frac{dx}{\sqrt{1-x^2}} = \begin{matrix} \pi & n=m=0 \\ \frac{\pi}{2} & n=m\ne 0 \end{matrix} $$ and the integral vanishes if $n\ne m$, so $$ \int_{-1}^1f(x) T_m(x) \frac{dx}{\sqrt{1-x^2}} = \int_{-1}^1\sum_{k=0}^\infty b_k T_k(x) T_m(x) \frac{dx}{\sqrt{1-x^2}} = \begin{matrix} \pi b_k & k=m=0 \\ \frac{\pi b_k}{2} & k=m\ne 0 \end{matrix} $$ we now have a way of getting the expansion in terms of these polynomials $$ f(y) = \frac{T_0(y)}{\pi}\int_{-1}^1f(x) T_0(x) \frac{dx}{\sqrt{1-x^2}}+\sum_{k=1}^\infty \frac{2 T_k(y)}{\pi}\int_{-1}^1f(x) T_k(x) \frac{dx}{\sqrt{1-x^2}} $$ if you can solve these integrals you can write your function in a new basis.

The orthogonality for $U_n(x)$ is $$ \int_{-1}^1 U_n(x)U_m(x) \sqrt{1-x^2}\;dx = \frac{\pi}{2} \;\; n=m $$ and the integral vanishes for $n \ne m$.

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