I want to find a best approximation polynomial $p_1(x)\in P_1$ for $f(x)=x^3$ in $[-1,1]$ w.r.t. $||\cdot||_{\infty}$. I want to use Chebyshev polynomial to do that, but I don't know how to hang on.
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Is $P_1$ the set of all polynomials of degree $\le 1$? – JimmyK4542 Aug 17 '14 at 19:54
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@JimmyK4542 yes. – 89085731 Aug 17 '14 at 19:56
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Let $p(x)$ will be polynomial which you find. First note that $f(x)-p(x)$ is polynomial of degree $3$ with leading coefficient $1$. You know that among the polynomials of degree $3$ with leading coefficient $1$
$$w(x)=\frac{1}{4}T_{3}(x)$$
is the one of which the maximal absolute value on the interval $[−1, 1]$ is minimal, see.
So you know that you must have $f(x)-p(x)=\frac{1}{4}T_3(x)$, so $p(x)=x^3-\frac{1}{4}T_3(x)$.
agha
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