Definition talk:Exponential Function

I'm not sure it's possible to prove $e^x = \exp (x \ln e) = \exp x$ without being circular, as you're using $\exp$ to define itself? Can you elaborate on what you meant there? I know it kind of goes back to what we were saying on Talk:Exp x equals e^x, but please share your thoughts. --GFauxPas 14:00, 19 January 2012 (EST)
 * ... what page is this talk supposed to be in relation to?
 * We have a great pile of definitions of the exponential function, and which are all equivalent. That statement needs to be proved.
 * That is, taking each one as a definition, it should be possible to prove each of the other definitions.
 * Of course it's possible to do it without being circular, it's just that I haven't got my analysis head on at the moment and it's probably tricky. The initial page as it stood defined exp as the inverse of the ln, and from that all the other properties were proved. Then someone came along and decided to state those derived properties as a direct statement of the definition of exp. However, in order to allow those statements actually to be definitions for exp, it is necessary that each one implies the truth of the other definitions - otherwise they can *not* be definitions, merely derived properties.
 * As for that second one, it was stated from the start that it was derived from the definition of the concept of the power, and from the definition of $e$. Its equivalence to the first one follows from the proof of the equivalence of the definitions of $e$. I guess. --prime mover 16:27, 19 January 2012 (EST)
 * Eek, it's supposed to be Definition:Exponential, don't know how this floated out on its own. It's just that defining "power" for irrational numbers without using $\ln$ is apparently difficult, as it involves proving that you can make a series converging to an irrational number with rational numbers only. Or something along those lines. Thanks for your explanation, something to think about I suppose. Sorry if I went against the grain with my edits. --GFauxPas 16:33, 19 January 2012 (EST)
 * I've moved it.
 * As I say, no problem with having multiple definitions for a concept (there's lots of this on proofwiki - IMO it's one of the great reasons why proofwiki rocks) - as long as we make sure it actually does stand up as a definition.
 * Which is why there's a flag on the page to sort it out sometime. Note that this is not an instruction specifically to you to sort out (just because you posted the statement does not mean it's your karma burden to resolve) - it's just "there" for when someone knows where they're going with it. As I have said before: if you're uncertain of your ground in an area, and you haven't got a solid backbone of authoritative textbooks (plural) behind you, then you're probably advised to leave a particular topic alone until you know where you're at. But it's your call - I admire your sense of adventure. --prime mover 16:40, 19 January 2012 (EST)

Beware the initial statement
We can't limit the range of $\exp$ to $\R$. What about $\C$? --prime mover 04:26, 22 January 2012 (EST)
 * well yeah, that's why I added the "the scope of this needs to be expanded to $\C$. The problem is is that the first definition has $\exp x = y \iff \ln y = x$, which doesn't hold in $\C$, right? Because then $e^x$ is periodic. --GFauxPas 07:13, 22 January 2012 (EST)
 * Yes, $e^x$ is periodic in the complex plane, what's the problem? The result still holds. --prime mover 08:49, 22 January 2012 (EST)
 * $\exp(2\pi i) = 1 \iff \ln 1 = 2\pi i$, $\exp 0 = 1 \iff \ln 1 = 0$, $2\pi i = 0$ --GFauxPas 08:58, 22 January 2012 (EST)
 * oh yeah ... whatever ... --prime mover 09:45, 22 January 2012 (EST)
 * I confess that my knowledge of $\C$ is summed up in "$i^2 = -1$ and $e^x$ is periodic". If someone could adjust the page accordingly that would be great, but I can't do so. As I said before, I don't know to what extent I'm being rigorous vs. too nitpicky/annoying/silly. --GFauxPas 10:49, 22 January 2012 (EST)

I hate digging this up a bit, but the problem here is that $\ln$ (in complex setting, $\log$ is *really* preferred) does not have the complex plane as its natural domain. Details are difficult but the main point is that one generally proceeds by a 'branch cut', deleting the negative real axis from $\C$ and defining $\log$ on that set to extend $\ln$ on $\R_{>0}$. The complex part of $\log$ is governed by the 'argument' function, which (by the branch cut) can be defined in a continuous fashion (taking $-\pi < \arg z < \pi$). --Lord_Farin 09:17, 9 July 2012 (UTC)
 * Yes, good call. The whole area of complex analysis needs to be filled out - I started a while back but got bogged down on the topology when establishing the fundamentals and never got back to it. --prime mover 09:59, 9 July 2012 (UTC)