# What is the relation between Chebyshev and Taylor polynomials?

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?

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 $n \ne m$.