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)

Contents

[edit] 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:

y'=y,\quad y(0)=1.

Divide both sides by y and integrate. Get characterization 3, which is:

\int_{1}^{y} \frac{dt}{t} = x.

is an increasing sequence which is bounded above. Since every bounded, increasing sequence of real numbers converges to a unique real number, this characterization makes sense.

Do you call that a proof? It's not much more than a tautology. It should be explained why it is increasing and why it is bounded.--Army1987 20:31, 19 July 2005 (UTC)

[edit] What did I miss?

The proof of the equivelance of charactaristics 1 and 3 isn't actually a proof, it's a circular argument and I've no idea why it's there. It claims that:

\lim_{n\to\infty}\ln\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}n\ln \left(1+\frac{x}{n}\right).

which should only hold surely, if we were dealing with a logarithm, but that's exactly what we're trying to proove, that ln x is the same thing as loge(x)

  • That's easy to prove; integral of dt/t from 1 to x^n, by the substitution u^n=t, becomes the integral of du n u^(n-1)/u^n = n u/du for u = 1 to x. Any good calculus book should have this proof. User:Ben Standeven as 70.249.214.16 03:05, 14 October 2005 (UTC)

[edit] Continuity required for characterization 5?

Is there an example of a discontinuous function f(x) (\mathbb{R} \to \mathbb{R}) satisfying f(x + y) = f(x)f(y) and f(0) = 1, but for which f(x)\neq e^{cx} for all c?

Of course f(x) = ecx if f(x) is continuous, as proved in the article (I believe this is a homework assignment in Rudin), but I'm curious to see a counterexample showing that continuity is necessary. (And if it's not a necessary assumption, it would be good to note this in the article and to give a reference.)

—Steven G. Johnson 04:57, 22 November 2006 (UTC)

I answered my own question, finally: continuity is required, as one can prove the existence of a discontinuous counter-example. I'm not aware of any constructive counter-example; the existence proof seems to require the axiom of choice.
I came up with the proof of a discontinuous f(x) satisfying f(x + y) = f(x)f(y) on my own after quite a bit of puzzling. Then a friend of mine pointed out that this is also shown in Hewitt and Stromberg, Real and Abstract Analysis (exercise 18.46)...it turns out to be easy to do once you have proved the existence of a Hamel basis for R/Q.
—Steven G. Johnson 00:30, 23 February 2007 (UTC)

[edit] Equivalence of 1 and 3

Hi,

The point raised on the discussion page about the equivalence of 1 and 3 is worthy of being addressed in the main article, I believe, as a perceptive reader will see that one is using a result whose prior establishment is not clear. (Indeed, I thought there was a problem and, having satisfied myself that there wasn't, inserted an edit to clear it up, before I had read the point on the discussion page. [Sorry for not checking first]) I hope the edit is still deemed worthy of retention.

Hugh McManus 14:56, 26 February 2007 (UTC)Hugh McManus

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