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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!

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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.

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  • $\begingroup$ thank you! You are correct I will edit it. Can you please explain the result a bit more? $\endgroup$
    – k12345
    Commented Feb 24, 2021 at 1:06
  • $\begingroup$ Thank you! This was very helpful $\endgroup$
    – k12345
    Commented Feb 24, 2021 at 1:14
  • $\begingroup$ Very quick follow up question. I get that you took a change of variable and that's how the bounds of the integral changed to pi's but should it have been $|cos((n+1) \theta)|$ inside the integral? $\endgroup$
    – k12345
    Commented Feb 24, 2021 at 1:17
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    $\begingroup$ Sorry the integral doesn't work: $\int_{-1}^1 \lvert T_2\rvert=\frac23(2\sqrt2-1)$ and without the absolute sign we have $\int_{-1}^1 T_n=\frac{1+(-1)^n}{1-n^2}$ if $n\neq 1$. $\endgroup$ Commented Feb 24, 2021 at 1:43
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    $\begingroup$ That does. So you want $U$ instead of $T$ which works because $\int_{-1}^1\lvert U_n(x)\rvert\,\mathrm{d}x=2$ from $\int_0^\pi \lvert U_n(\cos(\theta))\rvert\sin\theta\,\mathrm{d}\theta$. The integrand is $\lvert\sin((n+1)\theta)\rvert$ which we know the integral on every $\pi/(2(n+1))$ intervals starting at $0$, and the leading coefficient of $U_{n+1}$ is $2^{n+1}$. $\endgroup$ Commented Feb 24, 2021 at 1:50

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