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Is the symmetric definition of the derivative (below) equivalent to the usual one?

\begin{equation} \lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h} \end{equation}

I've seen it used before in my computational physics class. I assumed it was equivalent but it seems like it wouldn't matter if there were a hole at $x=h$ in the symmetric derivative, whereas with the usual one it wouldn't be defined. Which is kinda interesting...

If they're not equivalent - is there a good reason as to why we should use the common one? Or is the symmetric one actually more useful in some sense because it "doesn't care" about holes?

Eric Wofsey
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doublefelix
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2 Answers2

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As I noted in a comment to the other answer, Milly's computation is incorrect. I am posting this answer to rectify the situation. The symmetric derivative is defined to be \begin{align} \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}. \end{align} If $f$ happens to be differentiable, then the symmetric derivative reduces to the usual derivative: \begin{align} \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} &= \lim_{h\to 0} \frac{f(x+h)-f(x)+f(x)-f(x-h)}{2h} && (\text{add zero}) \\ &= \frac{1}{2} \left( \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + \lim_{h\to 0} \frac{f(x) - f(x-h)}{h}\right) \\ &= \frac{1}{2} \left( \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - \lim_{-h\to 0} \frac{f(x+h) - f(x)}{-h}\right) && (\ast)\\ &= \frac{1}{2} \left( f'(x) + f'(x) \right) && (\text{since $f'(x)$ exists}) \\ &= f'(x). \end{align} I'm begin a little trixy at $(\ast)$. Notice that we can think about the difference quotient as being the slope of a secant line through the points $(x,f(x))$ and $(x+h,f(x+h))$. This slope is given by \begin{equation*} \frac{f(x+h)-f(x)}{(x+h)-x}. \end{equation*} Multiplying through by $-1$, this becomes \begin{equation*} -\frac{f(x+h)-f(x)}{(x+h)-x} = \frac{f(x+h)-f(x)}{x-(x+h)} = \frac{f(x+h)-f(x)}{-h} \end{equation*} Taking $h$ to zero (which is the same as taking $-h$ to zero) on the left gives $-f'(x)$, justifying $(\ast)$. Again, this proves the key statement:

Proposition: If $f$ is differentiable at $x$ (in the usual sense), then $f$ is symmetric differentiable at $f$, and the symmetric derivative agrees with the usual derivative.

The converse does not hold. The usual example is the absolute value function which is not differentiable at zero, but which is symmetric differentiable at zero (with derivative zero): \begin{equation*} \lim_{h\to 0} \frac{|0+h|-|0-h|}{2h} = \lim_{h\to 0} \frac{|h|-|h|}{2h} = \lim_{h\to 0} \frac{0}{2h} = 0. \end{equation*} In particular, this demonstrates that the symmetric derivative is a generalization of the usual derivative. It cannot be equivalent, because it can meaningfully define the derivative of a larger class of functions.

As to the why? of the symmetric derivative (and why we don't use it instead of the usual derivative), I think that adequate answers can be found attached to this question and in comments elsewhere on MSE.

Xander Henderson
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  • FYI, I posted a lot about the symmetric derivative in several comments here, and one day I intend to write up a rather thorough survey of this topic (how non-differentiable can a symmetrically differentiable function be, where Dini derivates and other variations are also involved). I didn't do this back then because I was working on something else. – Dave L. Renfro Jul 13 '18 at 15:47
  • @DaveL.Renfro Nifty! I thought about mentioning Dini derivatives and Radon-Nikodym derivatives here, but it was going to take more time that I was willing to spend on this. Thanks for the links. – Xander Henderson Jul 13 '18 at 16:04
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$$\lim_h \frac{f(x-h)-f(x+h)}{2h}= \frac12 \lim_h \frac{f(x-h)-f(x)+f(x)-f(x+h)}{h}$$ $$=\frac12(\lim_h \frac{f(x-h)-f(x)}h+\lim_h \frac{f(x)-f(x+h)}h)=\frac{1}{2}(-f'(x)-f'(x))=-f'(x)$$ However this definition can be "strange" sometimes (in the sense it does not catch the idea of "derivative as slope"): let $f(x):=1/x^2$, and $$\lim_h \frac{f(-h)-f(h)}{2h}=0.$$ Even without asymptotes, you can have something like $g(x):=|x|$, and $$\lim_h \frac{g(-h)-g(h)}{2h}=0.$$

Milly
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    How do you know $\lim_h \frac{f(x-h)-f(x)}h+\lim_h \frac{f(x)-f(x+h)}h)$ exists in the first place? – Lemon Dec 04 '14 at 06:23
  • I just assumed it, to show (if it exists) it is opposite to the usual derivative (for the "equivalency" part in OP's question). – Milly Dec 04 '14 at 06:26
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    You've just shown, in fact, that OP's correct that the two definitions of derivative aren't equivalent; the symmetric definition suggests that $f(x)=|x|$ is differentiable at $x=0$, whereas the traditional definition does not. – Steven Stadnicki Dec 04 '14 at 06:41
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    There is an error in the computation here. The symmetric derivative is defined to be $$ \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}, $$ which is the opposite of what is given in the first line of the computation. When this is corrected, we discover that the symmetric derivative is equal to the usual derivative when a function is differentiable. – Xander Henderson Jul 13 '18 at 04:15