Problem 1. The central difference approximation for $f''(x)$ is given by, $$ D_0^2(f,x,h) = \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$ where $h \gt 0$. Prove that $$ | D_0^2(f,x,h) - f''(x) | = Ch^2$$ where $C$ depends on $f$, a $\xi \in [x, x+h]$ and a $\zeta \in [x-h,x]$
(Hint : You can start by writing each of $f(x + h)$ and $f(x - h)$ into a third-order Taylor polynomial of $h$ plus a remainder term.)
Problem 2. The two numbers $\xi$ and $\zeta$ in Problem 1 are usually unknown. As a result, $C$ is also unknown. Compute the limit of $C$ as h approaches $0$. (This limit provides an estimate for $C$ when $h$ is small.)
My answer :
I've done problem 1 with the hint but I'm having trouble with the problem 2 because no information about $f$ is given.
But since in the hint it is said to use Taylor polynomial, I assume $f$ is differentiable upto at least order $4$.
Now, $$ | D_0^2(f,x,h) - f''(x) | = Ch^2$$ where $C = \frac{f^{(4)}(\xi) + f^{(4)}(\zeta)}{4!}$ (I've calculated it).
Since $f^{(4)}$ exists then can I say that $$\lim_{h\to 0} C \\ = \frac{f^{(4)}(x) + f^{(4)}(x)}{4!}\\ = \frac{f^{(4)}(x)}{12} \ \ ?$$
I don't think I can since it's not said that $f^{(4)}$ is continuous.
Then what is the limit for $C$.
Someone please help me, it's urgent and I really need it.
Thanks.