Timeline for Uses of the symmetric derivative $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2011 at 19:41 | comment | added | Dave L. Renfro | @kahen: Any additive function, that is a function $f$ such that $f(x+y)=f(x)+f(y)$ for all real numbers $x$ and $y$, has a symmetric derivative equal to $0$ at each point, and discontinuous additive functions are quite pathological (their graphs are dense in the plane, they are non-Lebesgue measurable in each open interval, etc.). For more about the relation between the symmetric derivative and the ordinary derivative, see groups.google.com/group/sci.math/msg/d58ce3669a91243a and mathforum.org/kb/message.jspa?messageID=5056119 | |
| Sep 19, 2011 at 7:49 | comment | added | anon | @Didier: Oh wow, that fact went right over my head. I didn't process that for some inexplicable reason. | |
| Sep 19, 2011 at 7:45 | comment | added | Did | @anon, your last comment made me wonder whether you realized that for this function $f$ and for any rational $x$, $f(x+h)-f(x-h)=0$ for every $h$ hence the pseudo-derivative exists, but $f(x+h)-f(x)=-1$ for some $h$ with $|h|$ as small as one wants hence the (true) derivative does not exist. So, no, the idea of this example is not just that the derivative of a constant function is zero. (And apparently you did not get the point of my answer either...) | |
| Sep 18, 2011 at 22:29 | comment | added | anon | @Didier: To clarify: in my head at least I was thinking of "usual derivatives" as defined by the limit definition, and the domain of the arguments as a potentially customizable feature in that definition. Here this is just the idea that a constant function is differentiable and its derivative is zero, which I don't think has any direct implications specifically for the side-mixed limit in the original question. | |
| Sep 18, 2011 at 20:46 | comment | added | Did | @anon, I do not get your point. Usual derivatives, one-sided or not, are not restricted to the rationals, are they? That a function as highly irregular as the indicator function of the rationals HAS a pseudo-derivative in this sense is an excellent hint that, in fine, the notion has little to do with actual derivation. | |
| Sep 18, 2011 at 20:32 | comment | added | Shay Ben-Moshe | I see, well it gives us information actually. From the symmetry of $1_\mathbb{Q}$, it doesn't matter which value does the function get around $x=0$ the change in the function is zero, just like the limit says. | |
| Sep 18, 2011 at 20:28 | comment | added | kahen | It's the usual notation for the indicator function of the rationals | |
| Sep 18, 2011 at 20:27 | comment | added | Shay Ben-Moshe | I don't understand what you wrote, what does $1_\mathbb{Q}$ mean? | |
| Sep 18, 2011 at 20:24 | history | answered | kahen | CC BY-SA 3.0 |