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I'm reading Interpolation and Approximation by Davis, more specifically "Best Approximation" Chapter VII.

Let $n \in \mathbb N$.

Let $C[a,b]$ denote the set of continuous real functions over $[a,b]$ and $f\in C[a,b]$

Let $\cal P_n$ denote the set of polynomials with degree less than $n$.

The author proves that $$\min_{\large P \in \cal P_n} \max_{\large x\in [a,b]} |f(x)-P(x)|$$ is attained for a unique $P_n$ which defines as "the Chebyshev approximation of degree $\leq n$".

Nevertheless he fails to mention any link between $P_n$ and the usual $\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (x^2-1)^k x^{n-2k} $.

  • So my question is: how to prove that $P_n = \sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (X^2-1)^k X^{n-2k}$ ?

The most straightforward way would be to check that the above minimum is achieved for $\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (X^2-1)^k X^{n-2k}$ , but I can't.

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The $P_n$ mentioned by Davis is the best uniform approximation of a given function $f$, so it depends on $f$. The polynomial you mentioned is the plain ordinary Chebyshev polynomial, and is independent of any function $f$. So, they can't be the same thing.

In fact, though, one is a special case of the other. One definition of the Chebyshev polynomial is that it's the polynomial with minimal uniform norm. In other words, it's a polynomial that is a best uniform approximation of the zero function. So, if you go through Davis' argument with $f=0$, then the $P_n$ that you get should be the usual Chebyshev polynomial.

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  • $\begingroup$ Yes you're totally right. I was out of my mind the whole time. $\endgroup$ May 9, 2014 at 15:02
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    $\begingroup$ Wouldnt $P_n = 0?$ be the best appx for f=0? $\endgroup$
    – JustANoob
    Mar 11, 2018 at 1:17

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