Taylor polynom, residual for symmetric values

Question

When creating the taylor polynom for a $C^3$-function around a certain point i get the formula $f(z+h)=f(h)+hf'(z)+\frac{h^2}{2}f''(z) + \frac{h^3}{6}f'''(z) + R$

Now lets say I create the polynom around $f(z-h)$. The essential question for me is now: Can $R$ be assumed to be the some or do I have to consider $R_1$ and $R_2$? The reason beeing im trying to calculate the error $|f'(z) - \frac{f(z+h) - f(z-h)}{2h}|$ and estimating it by $|f'(z) - \frac{f(z+h) - f(z-h)}{2h}| \leq L\cdot h^2$. If I assume $R_1 = R_2$ I can say that $L = \frac{|f'''(z)|}{6}$.

Answer 1

I believe $R_1\neq R_2$ since $R_1 = \frac{h^4}{4!}f^{''''}(\xi_1)$ and $R_1 = \frac{h^4}{4!}f^{''''}(\xi_2)$ for $\xi_1\in(z,z+h)$, $\xi_2\in(z-h,z)$.

But there is a silver lining because:

$|\frac{R_1-R_2}{2h}|=\frac{h^3}{2\times 4!}|(f^{''''}(\xi_1)-f^{''''}(\xi_2))|\leq \frac{2h^3}{2\times4!}\sup_{\xi \in (z-h,z+h)}|f^{''''}(\xi)|$

If we take the lower order, then:

$R_1 = \frac{h^3}{3!}f^{''''}(\xi_1)$, but $R_2 = -\frac{h^3}{3!}f^{''''}(\xi_2)$

so then $|\frac{R_1-R_2}{2h}|=\frac{2h^3}{12}|f^{'''}(\xi_1)+f^{'''}(\xi_2)|\leq\frac{h^3}{3}\sup_{\xi \in (z-h,z+h)}|f^{'''}(\xi)|=Lh^3$

To conclude:

$f(z+h)=f(z)+hf'(z)+h^2f''(z)+R_1$, $f(z-h)=f(z)-hf'(z)+h^2f''(z)-R_1$,

Thus: $|f'(z)-\frac{f(z+h)-f(z-h)}{2h}|=|f'(z)-\frac{(f(z)-f(z))+2hf'(z)+h^2(f''(z)-f''(z))+R_1+R_2}{2h}|$

$=|f'(z)-f'(z)+\frac{R_1+R_2}{2h}|=|\frac{R_1+R_2}{2h}|\leq\frac{h^3}{3}\sup_{\xi \in (z-h,z+h)}|f^{'''}(\xi)|=Lh^3$.

Which is your desired bound.

Answer 2

Consider the fraction $$ \frac{f(x+h)-f(x-h)-2hf'(x)}{h^3} $$ and apply the extended mean value theorem thrice.

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With a succession of intermediate points $h>h_1>h_2>h_3>0$ you get that this fraction has the same value as $$ \frac{f'(x+h_1)+f'(x-h_1)-2f'(x)}{3h_1^{\,2}} $$ and then the same value as $$ \frac{f''(x+h_2)-f''(x-h_2)}{6h_2} $$ which gives via the same strategy the value $$ \frac{f'''(x+h_3)+f'''(x-h_3)}6 $$ or via the standard mean value theorem $$ \frac{f'''(x_3)}3 $$ where $x-h_2<x_3<x+h_3$.