Using Taylor series expansions, derive the error term for the formula
\begin{equation}
f''(x)\approx \frac{1}{h^{2}}\left [ f(x)-2f(x+h)+f(x+2h) \right ].
\end{equation}
I've tried it on my own way. We see that
\begin{align*}
f(x+h)&=\sum_{k=0}^{3}\frac{h^{k}}{k!}f^{(k)}(x)+E_{n}(h)\\
&=f(x)+hf'(x)+\frac{h^{2}}{2}f''(x)+\frac{h^{3}}{6}f'''(x)+E_{3}(h)
\end{align*}
\begin{align*}
f(x+2h)&=\sum_{k=0}^{3}\frac{(2h)^{k}}{k!}f^{(k)}(x)+E_{n}(2h)\\
&=f(x)+2hf'(x)+2h^{2}f''(x)+\frac{4h^{3}}{3}f'''(x)+E_{3}(2h)
\end{align*}
and
\begin{equation}
f(x+2h)-2f(x+h)=-f(x)+h^{2}f''(x)+h^{3}f'''(x)+E_{3}(2h)-E_{3}(h)
\end{equation}
then by isolating $f''(x)$ we get
\begin{equation}
f''(x)=\frac{1}{h^{2}}\left [ f(x+2h)-2f(x+h)+f(x) \right ]-hf'''(x)-\frac{1}{h^{2}}\left [E_{3}(2h)-E_{3}(h) \right ]
\end{equation}
which isn't the right way to do since the term of "$-hf'''(x)$" is added (that's not mentioned in this problem). The rest is the error term. How do I answer it correctly?
Best Answer
With the purpose of not leaving this question unanswered, this answer contains what OP already knows: stopping at $E_2$ is the cleanest way to go, since we already know how to express the error term of a Taylor series in convenient ways (Lagrange form, integral form). Some algebra quickly yields the error $h^{-2}(E_2(2h)-2E_2(h))$.