Use taylor series to arrive at the expression f'(x)=1/h

I’m not really sure how to go about this.. any help is appreciated.

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Apply the extended mean value formula several times to
$$
\frac{-3f(x)+4f(x+h)-f(x+2h)-2hf'(x)}{h^3}
$$


$$
=\frac{4f'(x+h_1)-2f'(x+2h_1)-2f'(x)}{3h_1^2}\\
=\frac{4f”(x+h_2)-4f”(x+2h_2)}{6h_2}\\
=\frac{4f”'(x+h_2+h_3)}{6}\\
$$
where $0<h_3<h_2<h_1<h$.


Of course one can also do this as Taylor series. Consider
$$g(h)=-3f(x)+4f(x+h)-f(x+2h).$$
Then $g(0)=0$, $g'(0)=2f'(x)$, $g”(0)=0$, $g”'(0)=-4f”'(x)$, thus
$$g(h)=2h·f'(x)-\frac23h^3·(2f”'(x+2h_1)-f”'(x+h_1)).$$

For fixed $h$, define the finite difference operator $\Delta$ as:

$$\Delta f(x) = f(x+h) – f(x)$$

We can then write:

$$f(x+h) = \left(1+\Delta\right)f(x)$$

The Taylor series also gives us an expression for $f(x+h)$ in terms of $f(x)$, which can be written in the compact form:

$$f(x+h)= \exp\left(hD\right)f(x)$$

where $D$ is the differential operator. Comparing the expressions for $f(x+h)$ in terms of the finite difference operator and the differential operator yields identity:

$$D = \frac{1}{h}\log\left(1+\Delta\right)$$

Expanding in powers of $\Delta$ till order $\Delta^2$ yields:

$$D \approx \frac{1}{h}\left(\Delta – \frac{\Delta^2}{2}\right)$$

To calculate the action of powers of $\Delta$ on $f(x)$, we write:

$$\Delta = E-1$$

where the operator $E$ is defined as:

$$E f(x) = f(x+h)$$

Therefore:

$$\Delta^n = \left(1+E\right)^n = \sum_{k=0}^n\binom{n}{k}(-1)^{n-k}E^k$$

So, we have:

$$\begin{split}
\Delta f(x) &= f(x+h) – f(x)\\
\Delta^2 f(x) &= f(x+2h) – 2 f(x+h) + f(x)
\end{split}
$$

This then yields the approximation:

$$D f(x)\approx \frac{1}{h}\left( – \frac{3}{2}f(x) + 2 f(x+h)-\frac{1}{2} f(x+2h)\right)$$