Which function (s) is(are) the k-th derivative of itself?

As I attempted to solve for a function that is the $$k$$-th derivative of itself where $$k$$ is some natural number, and generalise the result, I realised it gets very hard to prove anything. For $$k= 1$$, apart from trivial corner cases, it's the exponential function with any constant multiplied.

For $$k=2$$, along with the exponential function and it's reciprocal, it works for their sum along with any mixture by constants multiplied...

For any $$k$$, I guessed it should be of the form $$e^{x w}$$ where w is the $$k$$-th root of unity. And we can add all these $$k$$ terms for each root with constants multiplied.

While it is easy to verify it's satisfying the condition, it is beyond my scope to prove this is the only family of functions that satisfy it.

I am hoping for some interesting math at work here, and want to know if this is an already established fact. Also, how do I prove anything here?

• This might be easier to do if you can define the function as a power series. – robjohn Nov 6 '18 at 11:26
• Could you please elaborate, seems like a good idea... – Dhvanit Nov 6 '18 at 11:28
• – lulu Nov 6 '18 at 11:29
• It's quite easy to see that the question is only relevant when $k$ is prime. Looking at the power series, it's given that the $x^n$-term becomes the $x^{n-1}$-term when derived (and there are no other terms that can contribute to that), but as I can see the question @lulu points to contains the details. – Henrik Nov 6 '18 at 11:34
• Thanks a lot for the help – Dhvanit Nov 6 '18 at 11:40

So, your unknown function satisfies $$y^{(k)}(t) \equiv y(t)$$. This is called a linear differential equation of $$k$$-th order. There is an established method for solving such equations. It is based on observation that you can pick any function $$y_j (t) = \exp{\omega_j t}$$, where $$\omega_j = \exp(\frac{2\pi i}{k}j)$$ and it would be a solution. Moreover, any solution of this equation can be expressed as some linear combination $$\sum\limits_{p=1}^{k} \alpha_p y_p(t)$$, $$\alpha_p \in \mathbb{C}$$. The resulting solution is a complex valued function (i.e., a mapping $$\mathbb{R} \mapsto \mathbb{C}$$), and if you want to get only real-valued functions, you can do this by appropriate choice of $$\alpha_p$$. Namely, if $$\omega_j \not \in \mathbb{R}$$ you put $$\alpha_{k-j} = \overline{\alpha_j}$$, then a pair of terms $$\alpha_j y_j(t) + \alpha_{k-j} y_{k-j}(t)$$ transforms to $$\alpha_j y_j(t) + \overline{(\alpha_j y_j(t))}$$ and basically to $$2 \mathrm{Re}\,(\alpha_j y_j(t)) = 2(\mathrm{Re}\,\alpha_j \cos{\omega_j t} + \mathrm{Im}\,\alpha_j \sin{\omega_j t}).$$ Thus, you can also get any linear combination of $$\cos \omega_j t$$ and $$\sin \omega_j t$$ as a solution of such equation.