An error bound for a finite difference approximation to the second derivative I want to show that $$|\delta_{h,r}f(x)- f''(x)| \leq \frac{11}{12} h^2 \|f^{(4)}\|_{\infty}$$ where $$\delta_{h,r}f(x)=\frac{1}{h^2} (2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))$$ 
I applied the Taylor expanson at $f(x+h)$, $f(x+2h)$ and $f(x+3h)$ and $f(x_1), f(x_2), f(x_3)$ are the corresponding remainder of each expansion. 
$$f(x+h)=f(x)+h f'(x)+\frac{h^2}{2} f''(x)+\frac{h^3}{6} f'''(x)+\frac{h^4}{24} f^{(4)}(x_1), x_1 \in (x,x+h)$$
$$f(x+2h)=f(x)+2hf'(x)+2h^2 f''(x)+\frac{4}{3} h^3 f'''(x)+\frac{16}{24}f^{(4)}(x_2), x_2 \in (x,x+2h)$$
$$f(x+3h)=f(x)+3hf'(x)+\frac{9}{2}h^2 f''(x)+\frac{27}{6}h^3 f'''(x)+\frac{81}{24} h^4 f^{(4)}(x_3), x_3 \in (x,x+3h)$$
Substituting at $$\delta_{h,r}f(x)=\frac{1}{h^2} (2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))$$ we get 
$$\delta_{h,r}=f''(x)+\frac{h^2}{24}(64 f^{(4)}(x_2)-5f^{(4)}(x_1)-81 f^{(4)}(x_3))$$
Since $$|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)| \\ \leq |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2|\max_x |f^{(4)}(x)| \\ = |-\frac{22}{24}h^2| \|f^{(4)}\|_\infty = \frac{22}{24}h^2 \|f^{(4)}\|_{\infty}$$ 
we have
$$|\delta_{h,r}f(x)- f''(x)| \leq \frac{11}{12} h^2 \|f^{(4)}\|_\infty$$
($h>0$) 
Is this correct? 

Using the extended mean value theorem several times one can reduce the number of evaluation points for the 4th derivative
\begin{align}
&\frac{2f(x)−5f(x+h)+4f(x+2h)−f(x+3h)-h^2f''(x)}{h^4}
\\ \\
&=\frac{−5f'(x+h_1)+8f'(x+2h_1)−3f'(x+3h_1)-2h_1f''(x)}{4h_1^3}
\\ \\
&=\frac{-5f''(x+h_2)+16f''(x+2h_2)−9f''(x+3h_2))-2f''(x)}{12h_2^2}
\\ \\
&=2·\frac{-f''(x)+2f''(x+h_2)-f''(x+2h_2)}{12h_2^2}
\\[0.4em]
&\qquad+9·\frac{-f''(x+h_2)+2f''(x+2h_2)−f''(x+3h_2))}{12h_2^2}
\end{align}
where $0\le h_2\le h_1\le h$
Now as a separate calculation we confirm the standard result
$$
\frac{g(y-s)-2g(y)+g(y+s)}{s^2}=\frac{-g'(y-s_1)+g'(y+s_1)}{2s_1}=g''(y-s_1+s_2)
$$
where $0\le s_1\le s$ and $0\le s_2\le 2s_1$ so that $y_2=y-s_1+s_2\in[y-s,y+s]$. 
In combination of both calculations one gets (setting $g=f''$, $y=x+h_2$ resp. $y=x+2h_2$ and $s=h_2$)
\begin{align}
&-2·\frac{f''(x)-2f''(x+h_2)+f''(x+2h_2)}{12h_2^2}
\\[0.4em]
&\qquad-9·\frac{f''(x+h_2)-2f''(x+2h_2)+f''(x+3h_2))}{12h_2^2}
\\ \\
&=-\frac{2}{12}·f^{(4)}(x_{41})-\frac{9}{12}·f^{(4)}(x_{42})
\end{align}
with $x_{41}\in[x, x+2h_2]$ and $x_{42} \in [x+h_2, x+3h_2]$
Now we can conclude
$$
|δ_{h,r}f(x)−f′′(x)|≤\frac{2}{12}h^2·|f^{(4)}(x_{41})|+\frac{9}{12}h^2·|f^{(4)}(x_{42})|\le\frac{11}{12}h^2·\|f^{(4)}\|_∞
$$
