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I'd like to minimize the integral $$\int_{-1}^{1}\sqrt{1-x^2}\left(f(x)-p_n(x)\right)^2dx$$ given $f(x)\in C^0\left([-1,1]\right)$ and where $p_n(x)$ is allowed to range over all polynomials of degree $\leq n$. The weight is a good indicator that $p_n(x)$ is related to the Chebyshev polynomials, but I'm not sure how I would show that they might be the minimizers of this functional without being able to take derivatives.

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  • $\begingroup$ So $f$ can be any function in $C^0$ you want to fit for and $p_n$ should still be a Chebychev polynomial? $\endgroup$ – mathreadler Mar 2 '17 at 8:01
  • $\begingroup$ With respect to that weight, I'm thinking yes, and it would be of the second kind, but I don't know how to show it to be true. $\endgroup$ – garserdt216 Mar 2 '17 at 8:04
  • $\begingroup$ It's not clear to me. It feels like if anyone is free to pick a function of all in $C^0$ it should be possible to not be a Chebychev that fits best of all the polynomials in the world, but I have no proof. But it seems to be an interesting problem to experiment on in Matlab or similar software. $\endgroup$ – mathreadler Mar 2 '17 at 8:17
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If we consider $L^2(-1,1)$ equipped with the following dot product $$\langle f(x), g(x)\rangle = \int_{-1}^{1} f(x)\,g(x)\,\sqrt{1-x^2}\,dx $$ we have that Chebyshev polynomials of the second kind are a complete base of orthogonal functions: $$ \langle U_m(x), U_n(x)\rangle = \frac{\pi}{2}\,\delta(m,n) $$ By Parseval's theorem, the best $L^2$-approximation of $f(x)$ in terms of polynomials having degree $\leq n$ is given by the projection of $f(x)$ on the subspace generated by $U_0,U_1,\ldots,U_n$:

$$ p_n(x) = \sum_{k=0}^{n} U_k(x)\cdot\frac{2}{\pi}\int_{-1}^{1}f(x)\,U_k(x)\,\sqrt{1-x^2}\,dx.$$

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