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Given $\mathrm{f}\in C\left[-1,1\right]$ solve:

$$ \min\int_{-1}^{1}\sqrt{\, 1 - x^{2}\,}\,\,\left\vert\,\mathrm{f}\left(x\right)-\mathrm{p}_n\left(x\right)\,\right\vert^{\,2}\,\mathrm{d}x $$ where $\mathrm{p}_n$ is a polynomial of degree $n$ or less.

I'm assuming this problem is related to Chebyshev polynomials of the second kind because of the weight on the integral, but I'm lost after that.

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  • $\begingroup$ Did you really mean to say "interpolation"? Are $f(x)$ and $p_n(x)$ single, continuous functions or are they piecewise-defined? $\endgroup$ – John Barber Dec 10 '17 at 16:04
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You are correct: Chebyshev polynomials of the second kind provide an orthogonal base of $L^2(-1,1)$ with respect to the inner product $$ \langle f,g\rangle = \int_{-1}^{1} f(x)\,g(x)\,\sqrt{1-x^2}\,dx $$ and they fulfill $$ \int_{-1}^{1} U_n(x)\,U_m(x)\sqrt{1-x^2}\,dx =\frac{\pi}{2}\delta(m,n),\qquad \deg U_m(x) = m. $$ Given some $f(x)\in C[-1,1]\subset L^2(-1,1)$ we may assume $$ f(x)\stackrel{L_2}{=}\sum_{m\geq 0}c_m\cdot U_m(x),\qquad c_m=\frac{2}{\pi}\int_{-1}^{1}U_m(x)\,f(x)\,\sqrt{1-x^2}\,dx $$ then, by Parseval's identity, $$\operatorname*{argmin}_{\substack{p\in\mathbb{R}[x]\\ \deg p\leq n}}\int_{-1}^{1}\sqrt{1-x^2}\left[f(x)-p(x)\right]^2\,dx = \sum_{m=0}^{n}c_m U_m(x) $$ where $$\int_{-1}^{1}\sqrt{1-x^2}\left[f(x)-p(x)\right]^2\,dx =\frac{\pi}{2}\sum_{m>n}c_m^2.$$

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  • $\begingroup$ Also: What does $p_n(x)$ end up being then? I assume it's the sum of the first $n$ terms of $f$'s Chebyshev polynomial representation? $\endgroup$ – John Barber Dec 10 '17 at 18:09
  • $\begingroup$ $p_n(x)=\sum_{m=0}^{n}c_m U_m(x)$, i.e. $p_n(x)$ is the projection of $f$ on $\text{Span}(U_0,U_1,\ldots,U_n)$. $\endgroup$ – Jack D'Aurizio Dec 10 '17 at 18:11

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