I know this has been asked various times but I do not understand any of the answers given yet.
I'm working with the function $f(x)=e^{x}$ in the interval [-4,0] and I need to bound $|f(x)-Q_{n}(x)|$ where $Q_{n}(x)$ is an approximation of the function using Lagrange interpolation with equally spaced nodes. I want the error to be less than 0.1 and I need to find an n for when this is true.
I know that if $\sup\limits_{x\in[a,b]}|f^{(n+1)}(x)|\leq M_{n+1}$ then $|f(x)-P_{n}(x)|\leq \frac{M_{n+1}}{(n+1)!}|(x-x_{0})...(x-x_{n})|$. In this case, we can take $M_{n+1}=1$, so we have $|f(x)-P_{n}(x)|\leq \frac{1}{(n+1)!}|(x-x_{0})...(x-x_{n})|$. The problem is that i do not know how to bound the last product. The closest I've get to is assuming each node $x_{i}$ is at a maximum distance from de point x, so we can bound the product by $4^{n+1}$. Using this bound i get that $n=11$ is enough to make the error less than 0.1 but I´ve been hinted that there is a way to bound this more exactly and get a lower n.
Also, I´ve read in the forum that you can bound that last product by $\prod\limits_{i=0}^{n}|x-x_{i}|\leq\frac{h^{n+1}n!}{4}$ but I don´t see how this makes sense since the numerator grows way fasther than the denominator so when n is higher the bound gets worse.
Thank you.
I've found the answer for this: you can assume the product is maximum when your point is at the limit of the interval, -4 in this case. Then you can bound $(x-x_{0})$ and $(x-x_{1})$ by $\frac{4}{n}$, $(x-x_{2})$ by $2\cdot\frac{4}{n}$ and the k factor by $k\frac{4}{n}$.