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

Question

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?

Answer 1

Questions and Answers in MSE and two other references:

• Ad 1) : How to derive these Lie Series formulas & chain references
• Ad 3) : Ambiguous matrix representation of imaginary unit? idem
• Extra : Where the exponent in the Laplace Transform comes from
• Ad 4) : Gaussian Blur as a sample application of the Extra item

Updates.
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.

Ad 5). In this reference

• Derivation of the Boltzmann Distribution

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.

Answer 2

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).