Approximating the second derivative with Taylor’s Series

The Taylor’s Series tells us:

f(x+h)=\frac{h^0 f(x)}{0!}+\frac{h^1 f'(x)}{1!}+\frac{h^2 f''(x)}{2!}+\frac{h^3 f'''(x)}{3!}+...

f(x-h)=\frac{(-h)^0 f(x)}{0!}+\frac{(-h)^1 f'(x)}{1!}+\frac{(-h)^2 f''(x)}{2!}+\frac{(-h)^3 f'''(x)}{3!}+...

adding the two together yields:

f(x+h)+f(x-h) = \frac{h^0 f(x)}{0!}+\frac{(-h)^0 f(x)}{0!}+\frac{h^1 f'(x)}{1!}+\frac{(-h)^1 f'(x)}{1!}

+ \frac{h^2 f''(x)}{2!} + \frac{(-h)^2 f''(x)}{2!} +\frac{h^3 f'''(x)}{3!} +\frac{(-h)^3 f'''(x)}{3!} + ...

Notice how odd terms cancel each other which leaves us with:

f(x+h)+f(x-h) = \frac{h^0 f(x)}{0!}+\frac{(-h)^0 f(x)}{0!} + \frac{h^2 f''(x)}{2!} + \frac{(-h)^2 f''(x)}{2!} + ...

Combining like terms and simplifying yields:

f(x+h)+f(x-h) = 2f(x) + h^2 f''(x) + ...

Now solve for the second derivative:


Truncating the series gives an error of O(h^2) .

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