# Which derivatives are eventually periodic?

Which derivatives are eventually periodic?

I have noticed that is $a_{n}=f^{(n)}(x)$, the sequence $a_{n}$ becomes eventually periodic for a multitude of $f(x)$.

If $f(x)$ was a polynomial, and $\operatorname{deg}(f(x))=n$, note that $f^{(n)}(x)=C$ if $C$ is a constant. This implies that $f^{(n+i)}(x)=0$ for every $i$ which is a natural number.

If $f(x)=e^x$, note that $f(x)=f'(x)$. This implies that $f^{(n)}(x)=e^x$ for every natural number $n$.

If $f(x)=\sin(x)$, note that $f'(x)=\cos(x), f''(x)=-\sin(x), f'''(x)=-\cos(x), f''''(x)=\sin(x)$.

This implies that $f^{(4n)}(x)=f(x)$ for every natural number $n$.

In a similar way, if $f(x)=\cos(x)$, $f^{(4n)}(x)=f(x)$ for every natural number $n$.

These appear to be the only functions whose derivatives become eventually periodic.

What are other functions whose derivatives become eventually periodic? What is known about them? Any help would be appreciated.

• Well, polynomials are periodic. Eventually you have $f^{(n)}(x) = 0$ for sufficiently large $n$.
– D_S
Feb 9, 2016 at 1:25
• @D_S I did point that out in my question. Feb 9, 2016 at 1:26
• A rational function with neg degrees is not. Nor is a composition of rational with other function. cos(1/x) is does not have perioditic derivative. Feb 9, 2016 at 1:27
• @Improve I am asking about properties of these kinds of functions, and other functions whose derivatives become eventually periodic which I did not point out in my question.(polynomials, $e^x$, $\sin(x)$,$\cos(x)$. Normally the two would be seperate questions, but since they seemed very similar, I asked them in the same question. Feb 9, 2016 at 1:29
• Technically, for any periodic sequence $a_n$, $\sum_{n=0}^\infty \frac{a_n}{n!} x^n$ is a well defined real analytic function, whose derivatives at zero are periodic. I don't know any other examples where this is an elementary function, though. I'm also not sure if you mean that $f^{(n)}(c)$ is periodic for some $c$, or whether $f^{(n)}(x)$ is a periodic sequence of functions. The latter is a much stronger requirement: it amounts to requiring $f=f^{(m)}$ for an integer $m$. All these differential equations can be solved explicitly.
– Ian
Feb 9, 2016 at 1:32

The sequence of derivatives being globally periodic (not eventually periodic) with period $m$ is equivalent to the differential equation

$$f(x)=f^{(m)}(x).$$

All solutions to this equation are of the form $\sum_{k=1}^m c_k e^{\lambda_k x}$ where $\lambda_k$ are solutions to the equation $\lambda^m-1=0$. Thus $\lambda_k=e^{2 k \pi i/m}$. Details can be found in any elementary differential equations textbook.

If you merely want eventually periodic, then you can choose an index $n \geq 1$ at which the sequence starts to be periodic and solve

$$f^{(n)}(x)=f^{(m+n)}(x).$$

The characteristic polynomial in this case is $\lambda^{m+n}-\lambda^n$. This has a $n$-fold root of $0$, and otherwise has the same roots as before. This winds up implying that it is a sum of a polynomial of degree at most $n-1$, plus a solution to the previous equation. Again, the details can be found in any elementary differential equations textbook.

• So, it appears that the question I have asked if well known. Unfortunately, I must not have access to these elementary differential equations textbook, because the book that I am currently using contains nothing that you have pointed out in your answer. Feb 9, 2016 at 1:45
• @Δαμον Do they talk about at least second order linear differential equations with constant coefficients? The methods in the second order case basically apply to any order (provided you can find the roots of the relevant polynomials).
– Ian
Feb 9, 2016 at 1:46
• No, I don't think so. I would like to point out I am not a university student. Can these details be found in high school books? Also, I am not a native English speaker, so it might be in a different language. I am going to instead, google what you have said. Feb 9, 2016 at 1:48
• @Δαμον In the US, elementary differential equations is a second year undergraduate course. So I don't think this would be covered explicitly in any "high school" calculus reference.
– Ian
Feb 9, 2016 at 1:51
• @Δαμον However, the basic idea is very simple: given an equation of the form $\sum_{k=0}^n a_k f^{(k)}(x) = 0$, assume that $f(x)=e^{\lambda x}$ (where $\lambda$ is an unknown constant). Then substitute and divide out the common factor of $e^{\lambda x}$ to find the equation $\sum_{k=0}^n a_k \lambda^k = 0$. Then this polynomial has some roots which give you some solutions. The more difficult matters are 1. showing that if the roots are distinct, then all solutions are combinations of these and 2. if the roots are not distinct, how do you find the remaining solutions?
– Ian
Feb 9, 2016 at 1:51

Let's also look at it upside down. You can define analytical (infinitely differentiable) functions with their Taylor series $\sum \frac{a_n}{n!}x^n$. Taylor series are simply all finite and infinite polynomials with coefficient sequences $(a_n)$ that satisfy the series convergence criteria ($a_n$ are the derivatives in the chosen origin point, in my example, $x=0$). Compare this to real numbers (an infinite sequence of non-repeating "digits" - cardinality of continuum). On the other hand, a set of repeating sequences is comparable to rational numbers (which have eventually repeating digits). So... the "fraction" of all functions which have repeating derivatives, is immeasureably small - it's only a very special class of functions that satisfy this criterion (see other answers for appropriate expressions).

EDIT: I mentioned this to illustrate the comparison of how special and rare functions with periodic derivatives are. The actual cardinality of the sets depends on the field over which you define the coefficients. If $a_n\in \mathbb{R}$, then recall that set of continuous function has $2^{\aleph_0}$, the cardinality of a continuum, so cardinalities are the same in this case. If coefficients are rational, then we have $\aleph_0^{\aleph_0}=2^{\aleph_0}$ for infinite sequences and $\aleph_0\times\aleph_0^n=\aleph_0$ for periodic ones.

Not only that, but you can generate all the functions with this property. Just plug any periodic sequence $(a_n)$ into the expression. It's guaranteed to converge for $x\in \mathbb{R}$, because a periodic sequence is bounded, and $n!$ dominates all powers.

A simple substitution can demonstrate that if the coefficients are periodic for a series around one origin point, they are periodic for all of them.

• And not only do most analytic functions lack periodic derivatives, the analytic functions already form but a subset of the infinitely differentiable functions. For e.g. the important compactly-supported functions, periodicity is right out. Feb 9, 2016 at 14:53

if $f(x)$ is analytic at $0$ and for every $n$ : $f^{(n+m)}(0) = f^{(n)}(0)$ then we get by discrete Fourier transform :

$$f^{(n)}(0) = \sum_{k=0}^{m-1} e^{2 i \pi n k /m} \left(\sum_{l=0}^{m-1} e^{-2 i \pi k l /m} \frac{f^{(l)}(0)}{m}\right)$$

hence :

$$f(x) = \sum_{n=0}^\infty x^n \frac{f^{(n)}(0)}{n!} = \sum_{k=0}^{m-1} e^{xe^{2 i \pi k / m}} \left(\sum_{l=0}^{m-1} e^{-2 i \pi k l /m} \frac{f^{(l)}(0)}{m}\right)$$

which confirms that the only functions whose $n$th derivatives at $0$ are $m$ periodic are the sum of complex exponentials $e^{xe^{2 i \pi k / m}}$ where $k \in \mathbb{Z}$.

finally, add any polynomial to those and you'll get that the $n$th derivative at $0$ is periodic for $n \ge N$ (and shift $x \to x-a$ to get it periodic elsewhere than at $0$) .

• If you're just requiring periodicity at zero, then $\sum_{n=1}^\infty \frac{a_n}{n!} x^n$ is an analytic function with periodic derivatives at zero for any periodic sequence $a_n$.
– Ian
Feb 9, 2016 at 1:42
• @Ian : yes so it is a sum of complex exponentials as I showed Feb 9, 2016 at 1:43
• Oh, I see, you're just doing a change of basis essentially. Never mind, carry on.
– Ian
Feb 9, 2016 at 1:44
• @Ian : nobody said it was complicated, that's why many people solved it in $10$ min, and your differential equation implies the function is analytic, so our methods are equivalent Feb 9, 2016 at 1:45

The solutions of the differential equation $f^{(n)} = f$ are the linear combinations of $\exp(\omega t)$ where $\omega$ is an $n$'th root of unity, i.e. $\omega = \exp(2 i \pi k/n)$ for some integer $k$.
The real solutions are linear combinations of $\exp(\cos(2\pi k/n) t) \cos(\sin(2\pi k/n) t)$ and $\exp(\cos(2\pi k/n) t) \sin(\sin(2\pi k/n) t)$. The "eventually periodic" functions are linear combinations of these and polynomials.

If the derivatives are eventually periodic (as OP defines), this means that the function satisfies a linear differential equation with constant coefficients. This restricts the form of the function.

Use the following: $$t_m(x)=\frac{1}{m}\sum_{k=0}^{m-1} \exp( e^{i\frac{2k+1}{m}\pi}x )=\sum_{k=0}^\infty \frac{(-1)^k x^{km}}{(km)!}.$$ It is obvious that for $m=2$ you'll get $\displaystyle\frac{e^{i\pi x}+e^{-i\pi x}}{2}=\cos(x)$. It is also easy to see that $\displaystyle\frac{d(t_m)^m}{dx^m}=-t_m$. So the functions are periodic for $2m$. And they are ...

...known as Mittag-Leffer function $t_m(x) = E_{m}(-x^m)$.

There is a very nice article on ArXiv, "Mittag-Leffler functions and their applications", arXiv:0909.0230

(Last part cite from here...)