Take the 2-minute tour ×

The simplest example of this is $e^x$ which we could say has period 1 (it is its own derivative). $e^{-x}$ would have period 2.

Using similar constructions, I can get a function that has a derivative of period $n$ by doing $e^{x\cdot (1)^{1/n}}$

Is this the only way to get periodic derivatives?

Note: I am treating sin and cos as special cases of this when $n=4$

Is there a proof to this effect?

share|improve this question
The space of functions satisfying $f^{(n)}=f$ has dimension $n$ (dimensions of solution spaces is an important consideration in differential equations), and there are $n$ distinct $n$th roots of unity. –  anon Sep 9 '12 at 7:05
There aren't any. –  André Nicolas Sep 9 '12 at 7:06
@AndréNicolas why not though? –  soandos Sep 9 '12 at 7:07
The answer is presumably being typed by anon. If not, recall how to solve linear homogeneous DE with constant coefficients. –  André Nicolas Sep 9 '12 at 7:10
The other thing to note is that a solution to a differential equation of order $n$ is determined by initial conditions on $y, y', \ldots, y^{(n-1)}$, and therefore the space of solutions has dimension $n$. Thus once you have solutions of the form $e^{\lambda x}$ for $n$ different $\lambda$ (and these are easily seen to be linearly independent), all solutions are linear combinations of these. –  Robert Israel Sep 9 '12 at 7:24

1 Answer 1

up vote 2 down vote accepted

Assume $f^{(n)}(x)=f(x)$ for all $x$. Let $g(x) = \sum_{k=0}^{n-1}\xi^{-k} f^{(k)}(x)$, where $\xi^n=1$. Then $g'(x)=\sum_{k=0}^{n-1}\xi^{-k} f^{(k+1)}(x) = \xi\sum_{k=1}^{n}\xi^{-k} f^{(k)}(x)=\xi g(x)$. We already know all solutions of $y'=c y$ and conclude that $g(x)=a e^{\xi x}$.

If we additionally assume that $\xi$ is primitive, we find for $0\le m<n$ that $$ \sum_{k=0}^{n-1}\xi^{-mk} f^{(k)}(x)= a_m e^{\xi^m x}$$ Adding all equations leads to $$n f(x) = \sum_{m=0}^{n-1} a_m e^{\xi^m x}.$$

share|improve this answer

Your Answer


