Understanding why Roots of Chebyshev Polynomial Give best Interpolation

numerical methods

So I know that if we choose the nodes of the interpolating polynomial to be the roots of the Chebyshev polynomial then the error is minimized. My textbook doesn't fully prove this, they just give a sketch of the proof.

In the sketch they mention $\int_{-1} ^{1} |(x-x_0) \cdot\cdot\cdot (x-x_n)| dx =2^{-n} $ and give no hint as to how they got here.

I know that we get this by integrating both sides of the usual error statement for interpolating polynomials. I think the rest of the proof mainly makes sense but this step has me stumped.

I tried to use induction to prove this to myself but it wasn't working. Maybe I made a silly mistake, but I can't seem to find my error and I cannot find anything on this on the internet. Can someone please give me a hint on how to prove this?

Thanks!

Best Answer

There is a missing absolute sign and it should be max not integral, i.e. $$ \max_{x\in[-1,1]}\lvert (x-x_0)\dots(x-x_n)\rvert=\max_{x\in[-1,1]}\lvert 2^{-n}T_{n+1}(x)\rvert=2^{-n}. $$

Edit: correcting not integral. The formula for error of polynomial interpolation with nodes $x_i$, $i=0,1,\dots,n$ is $$ f(x)-P_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n (x-x_i) $$ so the maximum error is never more than $\sup\frac1{(n+1)!}\lvert f^{(n+1)}\rvert$ when using Chebyshev nodes.