# Is there a (deep) relationship between these various applications of the exponential function?

Here is a list of some applications of the exponential function.

1) The exponential mapping in Lie theory.

I put this first because my intuition tells me that this must be the most fundamental, or deep, way of thinking about the exponential function. I have often been misled by my intuition however, and the main reason I feel strongly about this is because of how fundamental I consider Lie theory to be.

2) Fourier Series

3) Roots of Unity

4) Gaussian Distribution

5) Boltzmann Distribution

There are certainly other applications, but it always kind of bothered me that I couldn't use symmetry methods to see how (all of) these applications are related. Is it possible that there is no way to do this, i.e. that it's just a happy accident that the exponential function has these applications and it is unrelated to any continuous symmetries?

• (4) and (5) are related to combinatorial limits and the Central Limit Theorem, whereas (1) - (3) are mutually connected in obvious ways. So the challenge comes down to finding a deep link between any of {1,2,3} and {4,5}. One possible avenue of investigation is via the "characteristic function" (Fourier transform) of the Gaussian, which is a key tool in most proofs of the CLT. Another promising route is via the Heat equation, diffusions, and random walks. Sep 4, 2010 at 18:49
• Since this got bumped: Fourier Series and Integrals by Dym and McKean may be of interest. Nov 7, 2015 at 16:27
• I don't think that the exponential map in the Lie-group context should be first, but this is clearly just a matter of taste. Probably the most fundamental fact about it is that it is the only measurable function for which $f(x+y)=f(x)f(y)$ for all $x,y$. From here you get that it is essentially the only function coinciding with it's derivative, and these two together encompass a whole world of algebra and analysis. Nov 10, 2015 at 10:00
• @uniquesolution: Your suggestion has been used in my (updated) answer. Thanks. Nov 13, 2015 at 10:07

Questions and Answers in MSE and two other references:

Ad 2). The Fourier transform is a special case of the double-sided Laplace transform: $$F(p) = \int_{-\infty}^{+\infty} e^{-pt}\,f(t) dt \quad \Longrightarrow \quad F(i\omega) = \int_{-\infty}^{+\infty} e^{-i\omega t} f(t)\,dt$$ The Fourier transform, in turn, is a generalization of the complex Fourier series: start with equation (20) in the Wolfram reference and read until the end.

it is argued on page 23 (with obviously a typo in it) that the Boltzmann probability distribution $f(E)$ must have the following form: $$f(E_1) \times f(E_2) = h(E_1+E_2)$$ Let's elaborate on this a little bit: $$f(E_1) \times f(E_2) = h(E_1+E_2) \quad \Longrightarrow \quad h(E) = h(E+0) = f(0)f(E)$$ Derivative: $$h'(E) = f(0)f'(E) = \lim_{\delta\to 0} \frac{h(E+\delta)-h(E)}{\delta} = \lim_{\delta\to 0} \frac{f(E)f(\delta)-f(0)f(E)}{\delta} =\\ f(E)\,\lim_{\delta\to 0} \frac{f(\delta)-f(0)}{\delta} = f(E) f'(0) \quad \Longrightarrow \quad f'(E) = \frac{f'(0)}{f(0)} f(E)$$ Continuing in terms of the article: $$\frac{df(E)}{f(E)} = \frac{-dE}{E_c} \quad \Longrightarrow \quad f(E) = A e^{-E/E_c}$$ Which is the Ansatz for the Boltzmann distribution.

Ad 3). According to Wikipedia, De Moivre's formula is: $$\left[cos(x)+i\sin(x)\right]^n = \cos(nx) + i\sin(nx)$$ And this can be proved for any integer $n$ , quite independent of Euler's formula. It's rather the other way around: because of the pattern $\;f(x)^n = f(nx)\;$ , de Moivre's formula can be considered as a heuristics for Euler's formula.
But uniquesolution is quite right: <quote> Probably the most fundamental fact about it is that it is the only measurable function for which $f(x+y)=f(x)f(y)$ for all $x,y$ </quote> Can we mimic this behavior of $\,e^x$ with the function $f(x) = \cos(x)+i\sin(x)$ ? From trigonometry we know that: $$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)\\ \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$ Hence: $$f(x+y) = \cos(x+y) + i \sin(x+y) =\\ \left[\cos(x)\cos(y) - \sin(x)\sin(y)\right] + i \left[\sin(x)\cos(y) + \cos(x)\sin(y)\right] =\\ \left[\cos(x) + i \sin(x)\right]\left[\cos(y) + i \sin(y)\right] = f(x)f(y)$$ So our $f(x)$ behaves like an exponential function.

• Thanks for the awesome answer to this old question! It's going to take me some time to get through all of these links, but in the meantime I wanted to mention that my favorite derivation of the Boltzmann distribution in physics is directly related to symmetry, the basic idea being that entropy must be an additive integral of the motion, and hence must depend only on energy. It can be found in Landau vol 5 Nov 12, 2015 at 16:38
• @MattCalhoun: Alas, as a pensionado I have no easy access to a decent library these days. So I can't lay my hands on Landau & Lifschitz. Would it be possible to send me a copy of the relevant page(s)? An email address can be found in my (updated) profile. Nov 13, 2015 at 10:05
• the words "Landau vol 5" at the end of my previous comment are a link to the relevant pages, in google books. Nov 13, 2015 at 16:39

The most important traits of the exponential function that make $e^x$ occur in that much applications are

a) Euler's identity $e^{i\pi} + 1 = 0$

and its implications for the definition of complex trigonimetric functions, i.e.

$$e^{i \cdot x} = \cos x + i \sin x$$

(like Fourier transformation, unit roots)

b) the fact that $\frac{d}{dx}e^x = e^x$

which lets $e^x$ naturally occur in many solutions of differential equations, which are often the most basic formulation of natural laws (growth/decay, gaussian distribution).

• And using b) $\frac{d}{dx}e^x = e^x$ in taylor series expansion you can get the usual series expression for $e^x$
– 911
Jan 9, 2011 at 13:11