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I've been trying to understand how the second order derivative "formula" works:

$$\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$

So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after $x+h$ and the before $x-h$" and the $h^2$ is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own.

It's is basically a derivative of a derivative, right? Newtonian notation declares as $f''$ and Leibniz's as $\frac{\partial^2{y}}{\partial{x}^2}$ which dissolves into:

$$(f')'$$ and $$\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$$

So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes.

The original one is rather straightforward:

$$\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$

And can easily be shown that $f'(x) = nx^{n-1} + \dots$ is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the $\lim_{h\to0}$ for simplicity:

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

So, the derivative of the derivative should be:

$$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$

$$f''(x) = \lim_{h\to0} \frac{ \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h} }{h}$$

$$f''(x) = \lim_{h\to0} \frac{ \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h} }{h}$$

$$f''(x) = \lim_{h\to0} \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x) }{h^2}$$

$$f''(x) = \lim_{h\to0} \frac{ f(x+2h) - 2f(x+h) + f(x) }{h^2}$$

What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help.

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up vote 9 down vote accepted

The only problem is that you’re looking at the wrong three points: you’re looking at $x+2h,x+h$, and $x$, and the version that you want to prove is using $x+h,x$, and $x-h$. Start with $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\;,$$ and you’ll be fine.

To see that this really is equivalent to looking at $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x+h)-f\,'(x)}h\;,$$ let $k=-h$; then

$$\begin{align*} f\,''(x)&=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\\ &=\lim_{-k\to0}\frac{f\,'(x)-f\,'(x-(-k))}{-k}\\ &=\lim_{k\to 0}\frac{f\,'(x-(-k))-f(x)}k\\ &=\lim_{k\to 0}\frac{f\,'(x+k)-f(x)}k\;, \end{align*}$$

and renaming the dummy variable back to $h$ completes the demonstration.

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Hah, neat! The answer was lost to me in the very thing I omitted during my work - the limit. It can approach from both ways, how incredibly silly of me and brilliant in expression here. Thank you! – LearningDroid Oct 10 '12 at 5:41
@LearningDroid: You’re very welcome. – Brian M. Scott Oct 10 '12 at 5:44
It seems you've used the following: if $a_n\rightarrow0$ and $b_n\rightarrow0$ then $$\lim_{n\rightarrow\infty}\frac{\lim_{n\rightarrow\infty}a_n/b_n}{b_n} = \lim_{n\rightarrow\infty}\frac{a_n}{b_n^2}$$ This needs to be demonstrated no? – Gregory Grant Mar 19 '15 at 21:09
@Gregory: I've not used it. Possibly the OP was using it, though use of the corresponding non-sequential result seems more likely. That part of the argument, however, was not the point of the question, so there was no reason to address it. – Brian M. Scott Mar 19 '15 at 22:05
Thanks, but I don't see how to use your hint without using this fact about limits. The OP also uses this fact, although not explicitly, he sneaks it past when he replaces $f'(x)$ with $\frac{f(x)+h}{h}$ (no limit). – Gregory Grant Mar 20 '15 at 1:25

Using the Taylor series expansions of $f(x+h)$ and $f(x-h)$,

$$ f(x+h) = f(x) + f'(x)h+f''(x)\frac{h^2}{2} + f'''(x)\frac{h^3}{3!}+\cdots $$

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

Adding the above equations gives

$$ \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = f''(x) + 2\frac{f''''(x)}{4!}h^2+\cdots $$

taking the limit of the above equation as $h$ goes to zero gives the desired result

$$ \Rightarrow f''(x) = \lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} \,.$$

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Thank you very much, this adds much perspective, but Mr. Scott managed to prove it without bringing in Taylor Series, which while correct, complicate the proof process. :D – LearningDroid Oct 10 '12 at 5:43
@LearningDroid: You are welcome. – Mhenni Benghorbal Oct 10 '12 at 6:10
There's one thing I am not completely sure about your prove. You say: $$f''(x)=\lim_{h\to0} \frac{ \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h} }{h}$$ but shouldn't it be $$f''(x) = \lim_{h\to0} \frac{ \lim_{h_1\to0} \frac{ f(x+h) - f(x+h-h_1)}{h} - \lim_{h_2\to0} \frac{ f(x) - f(x-h_2)}{h} }{h}$$ why can you assume the all, $h_1, h_2$ and $h$ are the same. Moreover why can you take them to zero at the same rate? – user120681 Jan 11 '14 at 21:35

Your formula is correct. You can easily check it by using Taylor (or, more formally, if you only have second derivatives, a second order Mean Value Theorem): $$\begin{multline} \frac1{h^2}\left[f(x+2h)-2f(x+h)+f(x)\right]= \\ \frac1{h^2}\left[ f(x)+2hf'(x)+\frac{4h^2}2f''(x)-2(f(x)+hf'(x)+\frac{h^2}2f''(x))+f(x)\right] = \\ \frac1{h^2}\,h^2f''(x) =f''(x). \end{multline}$$ Your deduction is a little shaky, though, as you are unifying two limits into one without justification.

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