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

where $0<h_3<h_2<h_1<h$.

Of course one can also do this as Taylor series. Consider
Then $g(0)=0$, $g'(0)=2f'(x)$, $g”(0)=0$, $g”'(0)=-4f”'(x)$, thus

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:

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


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

So, we have:

\Delta f(x) &= f(x+h) – f(x)\\
\Delta^2 f(x) &= f(x+2h) – 2 f(x+h) + f(x)

This then yields the approximation:

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