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Suppose that $f$ has continuous second derivatives. How do I show that

$$\frac{f(x+h) + f(x-h) - 2f(x)}{h^2}$$


$$2\frac{f(x+h) - f(x) - f'(x)h}{h^2}$$

both tend to $f''(x)$ as $h \rightarrow 0$?

For the first expression, I can rewrite it as

$$\lim_{h \rightarrow 0} \frac{1}{h}(\frac{f(x+h) - f(x)}{h} - \frac{f(x) - f(x-h)}{h})$$

which I can sort of see should tend to $f''(x)$, but I can't seem to show it rigorously. For the second expression, I can rewrite it as

$$\lim_{h\rightarrow 0} \frac{(f(x+h)-f(x))/h - f'(x)}{h}$$

I'm not sure where the factor of 2 comes in, but I guess it should have to do with the fact that we're trying to take limits "simultaneously" for $f'$ and $f''$. Can anyone help? Thanks.

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Hint: if $f$ has continuous second derivatives then you can use Taylor's theorem. – Chris Taylor Mar 22 '12 at 8:27
up vote 3 down vote accepted

Apply L'Hospital's rule to differentiate numerator and denominator with respect to $h$: $$ \begin{align*} \lim_{h \to 0} \frac{f(x+h) + f(x-h) - 2f(x)}{h^2} &= \lim_{h \to 0} \frac{f'(x+h) - f'(x-h)}{2h} = f''(x). \end{align*} $$ (Add and subtract $f'(x)$ in the numerator to see the second equality.) Similarly, $$ \begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x) - f'(x)h}{h^2}\, & = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{2h} = \frac{1}{2} f''(x). \end{align*} $$

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By the way, this proof shows that you do not need continuity of the second derivative, you only need its existence at the point $x$. – Nick Strehlke Mar 22 '12 at 9:36
As always, very good answer Nick! – Daniel Montealegre Mar 22 '12 at 19:39

A further proof of this finite difference approximation is as follows. This avoids the use of L'Hopital's rule, but relies upon continuity of second derivatives. The trick is to use Taylor's theorem.

$$f(x+h) = f(x) + f'(x)h + f''(\xi) h^2/2$$ $$f(x-h) = f(x) - f'(x)h + f''(\xi) h^2/2$$

for some $\xi$ between $(x, x+h)$. Now take the difference and let $h$ tend to zero. By continuity, we will obtain the second derivative.

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