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Let $n\geq 1$ be an integer and $f(x)=x^{n+2}$ for all $ x \in [−1, 1]$. Find the best uniform approximation of $f$ in $\mathbb{P}_n$.

Attempt: Let's solve this first for $f(x)=x^{n+1}$ instead. Suppose $p \in \mathbb{P}_n$ is the best uniform approximation. Then $g=f-p \in \mathbb{P}_{n+1}$ with leading coefficient $1$. As $g$ has the smallest norm among all polynomials in $\mathbb{P}_{n+1}$ with leading coefficient $1$, $g$ must be the $(n+1)$st Chebyshev Polynomial.

Now, for $f(x)=x^{n+2}$ the same argument does not work. But by Chebyshev Alternation theorem there must exist $n+2$ distinct points $-1 \leq x_1 < x_2 < \cdots < x_{n+2} \leq 1$ such that $g=f-p$ attains its maximum magnitude at those points with alternating signs. As $g' \in \mathbb{P}_{n+1}$, either $x_1=-1$ or $x_{n+2}=1$.

However, I can't make any progress from here. Can I get any hints/insights?

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  • $\begingroup$ Notation: \mathbb{P}_n rather than \mathbb{P_n} unless you want the subscript to appear in the blackboard font as well (unusual). $\endgroup$
    – Sammy Black
    Commented May 16 at 14:56

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The same argument still works because the Chebyshev polynomials $T_k$ only consist of odd or even-degree terms.

The $(n+2)$th Chebyshev $T_{n+2}(x) = 2^{n+2}( x^{n+2} + a_n x^n + a_{n-2} x^{n-2} + \cdots)$, the result will be $2^{-(n+2)}$ following your argument.

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  • $\begingroup$ Thanks for the insight. But I don't quite follow. Why would $x^{n+2}$ and $x^{n+1}$ have the same best uniform approximation in $\mathbb{P_n}$? For $x^{n+1}$ it would be the $\frac{T_{n+1}}{2^n}$ but why would this also be the best uniform approximation for $x^{n+2}$? The Chebyshev polynomial argument only works if we are looking for the best uniform approximation of $x^{m+1}$ in $\mathbb{P_m}$ $\endgroup$
    – miyagi_do
    Commented May 17 at 13:25
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    $\begingroup$ if you take $f\in P_n$ as an approximation to $x^{n+2}$, the difference is $x^{n+2} - f(x)$, which is a polynomial of degree $n+2$ with leading coefficient 1 and $g = 2^{-(n+2)} T_{n+2}$ is the one with smallest magnitude. Therefore $|x^{n+2} - f|_{\infty} \ge |g|_{\infty} = 2^{-(n+2)}$. Due to the specific form of Chebyshev polynomial, you can take $f = 2^{-(n+2)} T_{n+2} - x^{n+2}\in P_n$ as your approximation. @miyagi_do $\endgroup$
    – Yimin
    Commented May 17 at 17:37
  • $\begingroup$ Ah, that's slick. Thanks a lot!! $\endgroup$
    – miyagi_do
    Commented May 17 at 17:55

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