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)
Monotonicity implies continuity: Use monotonicity to show that \sup\left\{f(q):q\leq x, q\in\mathbb{Q}\right\}\leq f(x)\leq \inf\left\{f(q):q\geq x, q\in\mathbb{Q}\right\}. Then suppose that \sup\left\{f(q):q\leq x, q\in\mathbb{Q}\right\}< \inf\left\{f(q):q\geq x, q\in\mathbb{Q}\right\} and show you can find a rational q0 such that \sup\left\{f(q):q\leq x, q\in\mathbb{Q}\right\}< f(q_0) < \inf\left\{f(q):q\geq x, q\in\mathbb{Q}\right\} which is contradictory. You can then define for x\in\mathbb{R}\backslash\mathbb{Q}, f(x)=f(1)^x:=\sup\left\{f(q):q\leq x, q\in\mathbb{Q}\right\}= \inf\left\{f(q):q\geq x, q\in\mathbb{Q}\right\}.
To conclude, suppose f is not continuous in x_0\in\mathbb{R}. Then \exists\varepsilon_0 > 0,\forall y_1\in f(]-\infty, x_0]), \forall y_2\in f(]x_0,+\infty[), |y_1-y_2|\geq \varepsilon_0. This implies that \inf\left\{f(q):q\geq x_0, q\in\mathbb{Q}\right\}\neq\sup\left\{f(q):q\leq x_0, q\in\mathbb{Q}\right\}, which is not true. Therefore f is continuous everywhere.
—Deimos 28 06:23, 11 September 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

[edit] Equivalence of characterizations 1 and 2

The proof given only works for x \geq 0. In line 6, you are arguing (1-1/n)(1-2/n)\leq 1 implies that (x^3/3!)(1-1/n)(1-2/n)\leq x^3/3! and this requires non-negative x.

Indeed, the statement in Characterization 1 that an = (1 + x / n)n is an increasing sequence may not be valid if x < 0. Fathead99 (talk) 14:24, 30 January 2008 (UTC)

[edit] Taylor series.

From the limit (char-1) follows the differential equation (char-4) by differentiation, and then the taylor series (char-2).

It is unclear whether the article talks about the real or complex exponential function. The reference to positive values of x indicate real exponentials, but the results are important for complex exponentials.

The formulation "in several cases can be extended to any Banach algebra" is confusing; in which cases can it be extended, and in which cases can it not be extended? Is differentiation defined for any Banach algebra? The Banach algebra article didn't tell.

Bo Jacoby 07:29, 23 July 2007 (UTC).

[edit] Real versus complex

For some of the characterizations given here (e.g. power series) one may take the domain to be the whole complex plane, but for others (monotonicity plus the functional equation) one must assume the domain is the real line. (There do exist continuous functions on the complex plane that satisfy the functional equation and the initial condition but are not equal to the natural exponential function; they are nowhere (complex-)differentiable. For example, observe that

g(x + iy) = e^x(\cos(2y) + i\sin(2y))\,

for x and y real, is continuous everywhere in the complex plane and satisfies the functional equation

g(z + w) = g(z)g(w)\,

for z and w complex, and the initial condition

g(1) = e\,

but not the Cauchy-Riemann equations, and is therefore nowhere complex-differentiable.) Michael Hardy 13:16, 26 August 2007 (UTC)

[edit] Motivating the definitions?

I think at least definitions 1., 2., and 5. may seem to a beginning student like rabbits pulled out of a hat. Among all possible limits, why study \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n? Among all possible power series, why study \sum_{n=0}^\infty {x^n \over n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots? Among all measurable functions f satisfing f(x+y)=f(x)f(y), why study the one satisfying f(1)=e? I will try to answer these quations in what follows.

Even definitions 3. and 4. (which are really two sides of the same coin, the relation dy=ydx first when x is considered a function of y and then when y is considered a function of x) may need some motivation. In 3., why start the integral at 1? In 4., why chose y(0)=1? The answer is that otherwise we do not get the properties log(xy)=log(x)+log(y) and e(x + y) = exey.

The differential equation y'=y is sufficiently well-behaved to have a power series solution. From 4. we thus get 2. Of course, 3. also leads to a power series: log(1 + x) = x − x2 / 2 + x3 / 3 − x4 / 4 + .... Perhaps this power series should also be included among the definitions.

Although it is commonplace, I find the choice of the variable name "n" in 1. unfortunate. It suggests an integer variable when any complex number is perfectly permissible. I also prefer to use a variable that approaches 0 to one that approaches infinity, so rather than motivating 1., I will motivate: e^x = \lim_{h \to 0} {(1+xh)}^{1/h}.

If you try to differentiate ax with respect to x in the most straightforward of ways, you get \lim_{h\to0}\frac{a^h-1}{h} a^x. Since we know that the derivative should be log(a)ax, and since a can be any positive real number, we must have come across an expression for the function log. Since the exponential function is just the inverse of log, it is natural to wonder whether you can turn the expression for log into one for the exponential function. You can, and what you get is exactly e^x = \lim_{h \to 0} (1+xh)^{1/h}. (And once we have this expression it is a good idea to use the binomial series to expand it; this constitutes an alternative path to the power series 2.)

As for 5., it is really incomplete as a definition since it presupposes that you already have a definition of e. It seems to me that what we do have is an interesting characterisation of the function f(x) = ax. It is determined once you know: I. that f(x + y) = f(x)f(y), II. that f is measurable at one point, and III. the value of f(x) for any value of x except 0. Mattias Wikstrom (talk) 18:02, 8 February 2008 (UTC)