# Link between Chebyshev polynomials and best approximants

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.

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.
• Wouldnt $P_n = 0?$ be the best appx for f=0? Commented Mar 11, 2018 at 1:17