Seeing Finite Difference Approximations of the Derivatives |

Consider function $f$ that has first and second derivatives at each point.

There are two finite difference approximations to its derivatives at a particular point $x$ :

[1] $\frac{d}{dx}f(x) \approx \frac{ f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}$

Fig. 1

Intuitively (see Fig. 1),

$\frac{d}{dx} f(x) \approx \frac{ f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\quad\quad\quad(1)$

[2] $\frac{d^2}{dx}f(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$

Let

$g(x) = \frac{d}{dx}f(x) \overset{(1)}{\approx} \frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}\quad\quad\quad(2)$

so that

$\frac{d^2}{dx}f(x) = \frac{d}{dx}(\frac{d}{dx}f(x))=\frac{d}{dx}(g(x)) \overset{(1)}{\approx} \frac{g(x+\frac{h}{2})-g(x-\frac{h}{2})}{h}.\quad\quad\quad(3)$

Substituting

$g(x+\frac{h}{2}) \overset{(2)}{=} \frac{f((x+\frac{h}{2})+\frac{h}{2})-f((x+\frac{h}{2})-\frac{h}{2})}{h}=\frac{f(x+h)-f(x)}{h}$

and

$g(x-\frac{h}{2}) \overset{(2)}{=} \frac{f((x-\frac{h}{2})+\frac{h}{2})-f((x-\frac{h}{2})-\frac{h}{2})}{h}=\frac{f(x)-f(x-h)}{h}$

into (3) gives

$\frac{d^2}{dx}f(x) \approx \frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}=\frac{f(x+h)-f(x)-f(x)+f(x-h)}{h^2}=\frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$

i.e.,

$\frac{d^2}{dx}f(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$

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