Talk:Characterizations of the exponential function
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Give me a sec, I think the "intuitive" proof actually links defs 1 and 3, not 1 and 2, and might actually be able to replace the more clunky Havil proof I have here. But, I'm pooped and will return later. Revolver 01:21, 24 Jul 2004 (UTC)
Okay, I'm going to revamp this article a bit. In particular, I think that 2 of the arguments (the "technical" and "intuitive" ones) can be used to show the equivalent ways of defining the exponential function, not just the number e. Then, the equivalent definitions of e follow trivially by taking x = 1. Hopefully, this works! So, I'm going to change the title of the article to "definitions of the exponential function". Revolver 07:12, 24 Jul 2004 (UTC)
What would people think of moving this to characterizations of the exponential function? Michael Hardy 20:08, 24 Jul 2004 (UTC)
PS: see characterization (mathematics).
- I don't really care, to be honest. Call it what you want. Revolver 22:42, 24 Jul 2004 (UTC)
Howcome self-derivative isn't included? It's always been the characteristic that jerked my knee... Kwantus 19:44, 4 Sep 2004 (UTC)
- The main problem that pops into my mind is that it's not immediately obvious why such a function exists, i.e. why there is y s.t. y' = y and y(0) = 1. Without falling back on one of the others defs, I mean. If anyone knows of a nice way to prove this existence independent of the other 3 defs, I'd be happy to include it. Revolver 08:35, 20 Sep 2004 (UTC)
This is really a beautiful article. I like it very much. :) MathKnight 01:20, 1 Oct 2004 (UTC)
Characterization 4
Sorry, we have this problem on the german Exponentialfunktion. Maybe someone knows the exavt proof. I have only a german account. So Greetings Roomsixhu --83.176.135.233 02:50, 31 May 2005 (UTC)
Got it! How do I proove this? roomsixhu --83.176.135.227 19:02, 31 May 2005 (UTC)
Take characterization 4, which is:
- <math>y'=y,\quad y(0)=1.<math>
Divide both sides by y and integrate. Get characterization 3, which is:
- <math>\int_{1}^{y} \frac{dt}{t} = x.<math>