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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!

k12345
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1 Answers1

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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.

user10354138
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  • thank you! You are correct I will edit it. Can you please explain the result a bit more? – k12345 Feb 24 '21 at 01:06
  • Thank you! This was very helpful – k12345 Feb 24 '21 at 01:14
  • 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? – k12345 Feb 24 '21 at 01:17
  • Also now that I'm looking at it should the bounds of the integral after the substitution be $\pi$ to $0$? – k12345 Feb 24 '21 at 01:28
  • That makes sense. Is there some way to prove the integral though? That's really what I need to understand the proof I'm going through – k12345 Feb 24 '21 at 01:37
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    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 $n\neq 1$. – user10354138 Feb 24 '21 at 01:43
  • I think I may see the problem. You tried to use the definition of Chebyshev of the 1st kind. Now that I look closer at the proof I believe they are referring to Chebyshev of the 2nd kind. Does this change things? Sorry about that! – k12345 Feb 24 '21 at 01:43
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    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 $2^{n+1}$. – user10354138 Feb 24 '21 at 01:50
  • Thank you so much! I get it now. – k12345 Feb 24 '21 at 02:00