We are probably all familiar with the following identities from basic calculus:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}e^x &= e^x, & \frac{\mathrm{d}}{\mathrm{d}x}\sin(x) &= \cos(x), & \frac{\mathrm{d}}{\mathrm{d}x}\cos(x) &= -\sin(x). \end{aligned}

A consequence of these facts is that, at least for the exponential, sine and cosine functions, if you differentiate them often enough you end up with the original function again:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}e^x &= e^x, & \frac{\mathrm{d}^2}{\mathrm{d}x^2}e^{-x} &= e^{-x}, & \frac{\mathrm{d}^4}{\mathrm{d}x^4}\sin(x) &= \sin(x), & \frac{\mathrm{d}^4}{\mathrm{d}x^4}\cos(x) &= \cos(x). \end{aligned}

We can represent these identities as a chain, where each link indicates differentiation and we stop once we reach the original function again:

\begin{aligned} e^x &\to e^x, & \sin(x) &\to \cos(x) \to -\sin(x) \to -\cos(x) \to \sin(x),\\ e^{-x} &\to -e^{-x} \to e^x, & \cos(x) &\to -\sin(x) \to -\cos(x) \to \sin(x) \to \cos(x). \end{aligned}

So the first-order derivative of $$e^x$$ is $$e^x$$ again; the chain has length one. The chain for $$e^{-x}$$ has length two, because you have to differentiate $$e^{-x}$$ twice before you get $$e^{-x}$$ again. For the sine function, the chain has length four because the fourth order derivative of $$\sin(x)$$ is $$\sin(x)$$ again. Similarly, for the cosine function, we also have a chain of length four.

This suggests the following little recreational math problem:

For any $$n$$, which functions $$f$$ satisfy $$f^{(n)} = f$$?

Let’s introduce the following set:

$\mathcal{F}_n = \{ f: \mathbb{C} \to \mathbb{C} \mid f^{(n)} = f \}.$

Then obviously $$e^x \in \mathcal{F}_n$$ for any $$n \geq 1$$, $$e^{-x} \in \mathcal{F}_{2n}$$ and $$\sin(x), \cos(x) \in \mathcal{F}_{4n}$$. But can we be more general than this?

It is actually very easy to find at least one element of $$\mathcal{F}_n$$ for every $$n$$. Consider the function

$f(x) = \alpha\exp(\beta x).$

where $$\alpha$$ and $$\beta$$ are non-zero complex numbers. Note that

$f^{(k)}(x) = \alpha\beta^k\exp(\beta x).$

Therefore, if we let $$\beta$$ be an $$n$$th root of unity, we have $$\beta^n = 1$$ and we obtain the desired result:

Define the following function for any $$n > 0$$:

$F_n(x) = \exp\left(\exp\left(\frac{2\pi i}{n}\right)x\right).$

This function satisfies

$\frac{\mathrm{d}^k}{\mathrm{d}x^k}F_n(x) = \exp\left(\frac{2k\pi i}{n}\right)\exp\left(\exp\left(\frac{2\pi i}{n}\right)x\right).$

In particular, $$F_n^{(n)} = F_n$$ and so $$F_n \in \mathcal{F}_n$$.

We recognize a few particular cases. For $$n = 1$$, we find the exponential function, as expected:

$F_1(x) = \exp\left(\exp\left(2\pi i\right)x\right) = e^x.$

For $$n = 2$$, we find the negative exponential:

$F_2(x) = \exp\left(\exp\left(\pi i\right)x\right) = e^{-x}.$

For $$n = 4$$, we have

$\exp\left(\exp\left(\frac{\pi i}{2}\right)x\right) = e^{ix} = \cos(x) + i\sin(x).$

Hence the real part of $$F_4(x)$$ equals the cosine function and the imaginary part is the sine. However, we also already know that the sine and cosine are individual members of $$\mathcal{F}_4$$, so this suggests a broader solution than that given by our formula for $$F_n$$.

For other values of $$n$$ I don’t really recognize familiar functions, but there is the following nice result:

For every $$n$$, the function $$F_n$$ can be written as

$F_n(x) = e^{x\alpha}\left(\cos(x\beta) + i\sin(x\beta)\right)$

where

\begin{aligned} \alpha &= \cos\frac{2\pi}{n}, & \beta &= \sin\frac{2\pi}{n}. \end{aligned}

This characterizes a class of functions $$F_n$$ where the $$n$$th derivative of $$F_n$$ is the derivative of smallest order so that $$F_n^{(n)} = F_n$$. Furthermore, because of the properties of the derivative, we also have the following observations:

1. If $$f \in \mathcal{F}_n$$ then $$af \in \mathcal{F}_n$$ for every $$a \in \mathbb{C}$$.

2. If $$g \in \mathcal{F}_n$$ and $$h \in \mathcal{F}_n$$, then $$g + h \in \mathcal{F}_n$$ as well.

From the above observations it follows that $$\mathcal{F}_n$$ actually forms a vector space, which means we can completely characterize it by finding a basis. For $$n = 1$$, the only functions in $$\mathcal{F}_1$$ are known to be multiples of $$e^x$$. This is because, if $$f$$ is any function with $$f' = f$$, then we can define $$g(x) = f(x)e^{-x}$$ and observe that

$g'(x) = f'(x)e^{-x} - f(x)e^{-x} = 0.$

Therefore, by simple integration, $$g$$ must be constant, $$g(x) = c$$, which implies $$f(x) = ce^x$$. Hence

$\mathcal{F}_1 = \mathrm{span}\{ e^x \},$

which also shows that $$\mathrm{dim}(\mathcal{F}_1) = 1$$. It is instructive to investigate this property from the point of view of linear independence. Specifically, if we claim that $$\mathcal{F}_1$$ is spanned by two linearly independent functions,

$\mathcal{F}_1 = \mathrm{span}\{ f_1, f_2 \},$

then this is tantamount to the claim that all solutions of the following differential equation,

$\label{eq:f1}\tag{1} F' - F = 0$

are linear combinations of $$f_1$$ and $$f_2$$:

$F = af_1 + bf_2.$

However, we already know that all solutions of $$\eqref{eq:f1}$$ are given by

$F = Ce^x.$

And so it follows that $$\mathcal{F}_1$$ has only one basis function.

This reasoning generalizes to $$\mathcal{F}_n$$. For if we claim that $$\mathcal{F}_n$$ is spanned by $$m$$ basis functions, then this is equivalent to saying that all solutions of the differential equation

$\label{eq:fn}\tag{2} F^{(n)} - F = 0$

are linear combinations of the $$m$$ basis functions:

$F = a_1f_1 + \dots + a_mf_m.$

It is a basic result that the solutions of a differential equation of the form $$\eqref{eq:fn}$$ can be found via the characteristic polynomial, which in this case takes the special form

$\chi_n(\lambda) = \lambda^n - 1.$

The roots of $$\chi_n$$ are given by the roots of unity,

$\lambda_n = \exp\left(\frac{2\pi i}{n}\right) = \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right).$

The solutions of $$\eqref{eq:fn}$$ are therefore all linear combinations of the functions

\begin{aligned} &x^ke^{\alpha x}\cos(\beta x), & &x^ke^{\alpha x}\sin(\beta x) \end{aligned}

where

\begin{aligned} \alpha &= \cos\frac{2\pi}{n}, & \beta &= \sin\frac{2\pi}{n} \end{aligned}

and $$k = 0, \dots, n-1$$. This finally completes our picture:

Every function $$f \in \mathcal{F}_n$$ takes the form

$f(x) = \sum_{k=0}^{n-1}x^ke^{x\alpha}\left(a_k\cos(x\beta) + ib_k\sin(x\beta)\right)$

for constants $$a_0, \dots, a_{n-1} \in \mathbb{C}$$ and $$b_0, \dots, b_{n-1} \in \mathbb{C}$$ where

\begin{aligned} \alpha &= \cos\frac{2\pi}{n}, & \beta &= \sin\frac{2\pi}{n}. \end{aligned}

In particular, this implies

$\mathrm{dim}(\mathcal{F}_n) \leq 2n.$

The functions $$F_n$$ which formed our initial guess for $$\mathcal{F}_n$$ are clearly special cases, with $$a_0 = b_0 = 1$$ and $$a_k = b_k = 0$$ for $$k > 0$$.