Question: Let $$f(x)=\sin \left(\frac{\pi}{2}x\right)\qquad -1\le x\le 1$$
Find out if this function is the limit of a sequence of interpolating polynomials $\{p_{4n}(x)\}$ of degree $\le 4n$ using the values of $f$ at the nodes $-2n,-2n+1,\ldots,2n$. Note that $f$ takes on values $\pm1$ and $0$ only at the nodes.
My attempt:
I have used here a theorem that gives the range of error when doing Lagrange interpolation. It says :
I say that since $(4n+1)$ is odd, so then $f^{(4n+1)}(\xi)=\pm\left(\frac{\pi}{2}\right)^{4n+1}\cos\left(\frac{\pi}{2}x\right)$.
But in any case, whether or not it's sine or cosine, as per the definition of $M_{4n+1}$, the value of $M_{4n+1}$ is $\left(\frac{\pi}{2}\right)^{4n+1}$.
Then I argue that the limit of the maximum error as $n\to\infty$ is zero as per the following.
$$\lim_{n\to\infty}\frac{\left(\frac{\pi}{2}\right)^{4n+1}}{(4n+1)!}\max_{x\in [-1,1]}|\pi_{4n+1}(x)|=0$$
And so eventually I infer that as $n\to\infty$, we do have a uniform convergence.
Am I correct in my reasoning? Also is there any significance of the statement in the question "$f$ takes on values $\pm1$ and $0$ only at the nodes" that I didn't take into account?
Best Answer
For $x\in[0,1]$ you get $$ |x(x-1)(x-2)\cdots(x-2n)|\le x(1-x)(2n)! $$ and $$ |(x+1)(x+2)\cdots(x+2n)|\le(2n+1)! $$ The same can be done symmetrically for $x\in[-1,0]$. This gives you as bound for the error of the interpolation polynomial $$ \frac{(\frac\pi2)^{4n+1}(2n)!(2n+1)!}{(4n+1)!}|x|(1-|x|) $$ Now apply some estimates for the factorials like $(\frac{k}3)^k\le k!\le (\frac{k}2)^k$ or the Stirling formulas to get $$ \frac{(2n)!(2n+1)!}{(4n+1)!}\sim\frac{(2n)^{2n}(2n+1)^{2n+1}}{(4n+1)^{4n+1}}\sim 2^{-(4n+1)}. $$ As $\frac\pi4<1$, this gives a geometric bound falling to zero for the error term for $n\to\infty$.