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The symmetric derivative is always equal to the regular derivative when it exists, and still isn't defined for jump discontinuities. From what I can tell the only differences are that a symmetric derivative will give the 'expected slope' for removable discontinuities, and the average slope at cusps. These seem like extremely reasonable quantities to work with (especially the former), so I'm wondering why the 'typical' derivative isn't taken to be this one. What advantage is there to taking $\lim\limits_{h\to0}\frac{f(x+h)-f(x)} h$ as the main quantity of interest instead? Why would we want to use the one that's defined less often?

J. M. ain't a mathematician
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Robert Mastragostino
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2 Answers2

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The symmetric derivative being defined at more places isn't a good thing.

In my mind, the main point of differentiation is to locally approximate a function by a linear function. That is, the heart of saying that the derivative $f'(a)$ exists at a point $a$ is the statement that

$$f(x) = f(a) + f'(a) (x - a) + o(|x - a|)$$

as $x \to a$, and if I were the King of Calculus this is how the derivative would actually be defined. (Among other things, this definition generalizes smoothly to higher dimensions.) Removable discontinuities are a non-issue as they should just be removed, but at a cusp we do not have this property for any possible value of $f'(a)$, so we shouldn't be talking about derivatives at such points at all. (We can talk about left or right derivatives, but this is something different.)

The symmetric derivative at $a$ is not a natural definition. It has the utterly strange property that any weirdness in a neighborhood of $a$ is ignored if it happens to be canceled by equivalent weirdness after reflecting around $a$. Let me give an example. Consider the function $f(x) = 1_{\mathbb{Q}}(x)$ which is equal to $1$ if $x$ is rational and $0$ otherwise. The symmetric derivative of $f$ at any rational point exists and is equal to $0$! Is there any reasonable sense in which $f$ is differentiable at a rational point?

The ordinary derivative, on the other hand, is sensitive to weirdness around $a$ because it compares all of that weirdness to $f(a)$.

Qiaochu Yuan
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  • @Qiochu Yuan This is indeed the correct definition. – ncmathsadist Jul 15 '12 at 01:49
  • @QiochuYuan Linearity makes it exceedingly obvious in retrospect. Always a sign of a fantastic answer. – Robert Mastragostino Jul 15 '12 at 02:10
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    the "King of Calculus"... I'll have to remember that phrase for the next time I teach such a course. – KCd Jul 15 '12 at 05:33
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    Not worth a full answer but: all this hints at the fact that a proper two-sided definition of the derivative of $f$ at $x$ should be the limit of $(f(x+r)-f(x-s))/(r+s)$ when $r\gt0$ and $s\gt0$ both go to $0$. Fortunately (or unfortunately, maybe), this seems to be strictly equivalent to the usual definition (provided removable discontinuities are... well, removed). – Did Jul 15 '12 at 13:01
  • What happens if we remove the restriction that r > 0 and s > 0 (and just ask that they not sum to zero)? What extra properties must a function differentiable at x satisfy for the two-variable difference quotient limit (in r and s) to exist? [E.g., I believe it suffices for the function to be continuously differentiable. How tight a restriction is this?] – Sridhar Ramesh Aug 30 '12 at 04:33
  • @QiaochuYuan: does the sort of derivative you prefer in general have a name? – dfeuer Jun 17 '13 at 02:34
  • Re. "The symmetric derivative at a is not a natural definition." I agree but in real life you can't give me something at 0º K anymore than you can give me a 3ft stick. In both cases, though, you can give me approximations. So, requiring $f$ to be defined at $a$ just to observe "all of that weirdness" for inputs near $a$ does not feel so "natural" to me: $f$ doesn't cease to be smooth near $a$ when $f(a)$ ceases to exist anymore than a sewer pipe with a pinhole you can't even see ceases to be one. So, what would be a concept that would appear reasonable to "just plain folks"? – schremmer Jan 04 '18 at 19:02
  • @Did That looks a lot better to me. Did you ever try it with "just plain folks" by which I mean people not appropriately brainwashed. OOops. I meant prepared. And then, if so, what did the administration, if not the esteemed colleagues, have to say? – schremmer Jan 04 '18 at 19:17
  • @schremmer Not sure what your question is. Please make it clearer. Or not. – Did Jan 04 '18 at 23:49
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    @Did Not sure what you want. :-) Say you are teaching Calculus I to people who don't want to be there but must and that you want to "talk to them". At least in the case of a gap, the lack of continuity does not involve looking beyond $a$. Then they do see that any kind of "smoothness" requires looking at a neighborhood of $a$ but, as I tried to explain, not why they should still look at $a$ which, in their eyes, makes the definition of the ordinary derivative "weird". Hence my interest in the asymmetric derivative you presented and my question as to whether you had presented it to students. – schremmer Jan 06 '18 at 05:32
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    @schremmer I have not. – Did Jan 06 '18 at 14:26
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Following my comment on the Mean Value Theorem. Since MVT fails, anything we prove from MVT is likely to fail as well. For example:

Find the minimum of the (symetrically) differentiable function $f(x) = x+2|x|$ on the interval $[-1,1]$.
Usual solution: find where the derivative is zero. Answer: nowhere! Since $f'(x) = -1$ on $[-1,0)$, $f'(0)=1$, and $f'(x)=3$ on $(0,1]$.

GEdgar
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    An interesting note is that there is a generalization of the MVT for the symmetric difference quotient, which asserts that difference quotients are bounded by the symmetric difference quotient but not necessarily equal to the symmetric difference quotient as with the usual MVT. – Simply Beautiful Art Jun 17 '20 at 22:09
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    I saw today that there is a whole book on this. MVT: A Most Valuable Theorem by C. Smorynski – GEdgar Jan 23 '21 at 12:39