I am trying to understand Chebyshev Interpolation, and having trouble understanding this problem.
How many Chebyshev nodes are necessary to approximate the function $\sin(x\pi)$ to an error of at most $10^{-10}$ within $[-1, 1]$?
I think we can define a $y = x\pi$ so that we can reduce the problem to $\sin(y)$ within $[-\pi, \pi]$. But how can I continue from here, how do I determine how many Chebyshev nodes are needed?
Thank you!
"Necessary" is a tough question. "Sufficient" is easier to answer.
Recall the interpolation error for an $n$-times continuously differentiable $f$ on $[-1,1]$ using Chebyshev nodes $x_k=\cos((2k-1)\pi/(2n))$, $k=1,\dots,n$, is $$ \leq\frac{1}{2^{n-1}n!}\max_{\xi\in[-1,1]}\lvert f^{(n)}(\xi)\rvert. $$ With $f(x)=\sin(\pi x)$, we have $\max_{\xi\in[-1,1]}\lvert f^{(n)}(\xi)\rvert=\pi^n$ and so for sufficiency we just need to guarantee $$ \frac{\pi^n}{2^{n-1}n!}\leq 10^{-10}. $$ so $n\geq 17$ is sufficient.
If you really want the necessary number of nodes, then you go back and test the interpolation for $n=16,15,\dots$ to find out where it fails. It is going to be an ugly computation.