If we interpolate the function $f(x)=e^{x-1}$ with a polynomial $p$ of degree $12$ using $13$ nodes in $[-1,1]$, what is a good upper bound for $|f(x)-p(x)|$ on $[-1,1]$?
I know that if $f$ is a function of class $C^{n-1}([a,b])$ and if $p_n$ is the polynomial of degree less than or equal to $n$ that interpolates $f$ in $x_0,x_1,...,x_n$ different points in $[a,b]$, then for each $x\in [a,b]$ there is a $\xi_x$ such that $f(x)-p_n(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi_x)\prod_{i=0}^{n}(x-x_i)$
Using the above I get to $f(x)-p(x)=\frac{1}{13!}e^{\xi_x-1}\prod_{i=0}^{12}(x-x_i)$ where $\xi_x,x_i\in (-1,1), i=0,1,...,12$ and so $|e^{\xi_x-1}|<1$, with which $|f(x)-p(x)|=\frac{1}{13!}|e^{\xi_x-1}|\prod_{i=0}^{12}|x-x_i|<\frac{1}{13!}\prod_{i=0}^{12}|x-x_i|$ but I do not know what else to do, could someone help me please? Thank you very much.
Now you need to bound the product. How good a bound you get depends on how much work you are willing to do. An easy one is $|x-x_i| \le 2$ so the product is less than $2^{13}$. If your nodes have six on each side of $0$ plus one at $0$ you have $|x-x_i| \le 1$ for seven of them, so the product is less than $2^6$. Even better is to specify that your $x_i$ are the roots of the $13^{th}$ degree Chebyshev polynomial. It concentrates the nodes near the end of the interval to minimize this product. Better yet is to concentrate your nodes more at the high end of the interval because the derivatives of your function have a factor $e^{x-1}$ which is maximized at $x=1$. Now the $|x-x_i|$ product is small where the $e^{x-1}$ is large. What is good enough in the context of the question is left to the reader.