3

If we keep differentiating $\sin x$ we eventually arrive back at $\sin x$:

$$ \begin{align} y &= \sin x \\ \frac{dy}{dx} &= \cos x \\ \frac{d^2y}{dx^2} &= -\sin x \\ \frac{d^3y}{dx^3} &= -\cos x \\ \frac{d^4y}{dx^4} &= \sin x \end{align} $$

It has to be differentiated 4 times before it gets back to itself

I was wondering, what function has the longest chain of derivatives before it gets back to itself?

  • This might be of interest: https://math.stackexchange.com/questions/7511/functions-that-are-their-own-nth-derivatives-for-real-n – J126 Jun 26 '17 at 13:11
  • 1
    Note that this is equivalent to solving the differential equation $y^{(n)} = y$ which is known to have solution of the form $\sum_i e^{\zeta_i x}$ where $\zeta_i$ is a the i-th n-th root of unity. – Zubzub Jun 26 '17 at 13:11
  • @skyking: No, that question explicitly asks for fractional derivatives. – hmakholm left over Monica Jun 26 '17 at 13:57

5 Answers5

8

You can get chains as long as you want: For given $n \in \mathbb{N}$ consider $f(x) = e^{e^{\frac{2\pi i}{n}}x}$. Then $f'(x) = e^{\frac{2\pi i}{n}} f(x)$, $f^{(k)}(x) = e^{\frac{2\pi ik}{n}}f(x) \neq f(x)$ and $f^{(n)}(x) = e^{2\pi i} f(x) = f(x)$.

agb
  • 1,269
  • 6
  • 18
5

I guess that you mean commonly used functions with names. In which case, I do not know any.

It is easy to construct functions with loops as long as you want. To get a loop of length $n$, take the power series for $e^x$ and drop all of the terms whose exponent is not a multiple of $n$. This will converge for all $x$ and have a derivative loop of length $n$.

badjohn
  • 8,854
  • How do I notate such a function? –  Jun 26 '17 at 13:10
  • I asked a similar question myself once but just about the $n = 4$ case.
    https://math.stackexchange.com/questions/2111657/solutions-to-f4x-fx Even this case did not have well known names. So, I guess that you can choose some names.     – badjohn Jun 26 '17 at 13:27
2

Consider the function $f(x)=e^{\zeta_n x}$ where $\zeta_n\in\mathbb C$ is a primitive $n$-th root of unity. This function has period $n$.

Mikhail Katz
  • 47,573
  • I've never heard of an $n\mathrm{th}$ root of unity. What's that? –  Jun 26 '17 at 13:11
  • They are complex numbers whose $n$-th power is $1$. You can write them simply as $\cos(2\pi/ n) + i \sin(2\pi /n)$. – Mikhail Katz Jun 26 '17 at 13:13
  • @theonlygusti: An $n$th root of unity is a solution to $z^n=1$ in the complex numbers. Mikhail is here implicitly talking about a primitive $n$th root of unity, which is one where $z^k\ne 1$ when $0<k<n$. – hmakholm left over Monica Jun 26 '17 at 13:13
0

I don't think that there's an upper bound on this. You can reformulate this as $\frac{d^ny}{dx^n} = y$ which can be rearranged to give $\frac{d^ny}{dx^n} - y = 0$. To solve such a differential equation, solve the characteristic equation $\lambda^n - 1 = 0$, but note that smaller loops also satisfy this equation, but there is a solution that achieves a loop of size $n$. To see this, notice that $\lambda = 1$ always satisfies this and this corresponds to $e^x$. The other solutions are the $n$'th roots of unity. Take the smallest such root i.e. $e^\frac{2\pi i}{n}$ and raise it to the $x$'th power i.e. $f(x) = e^\frac{2\pi ix}{n}$. This solution will satisfy the equation.

Osama Ghani
  • 2,962
0

If you want a realvalued function that can be expressed in terms of elementary functions you can take the real part of Mikhail Katz' example: Let $\alpha:={2\pi\over n}$, and put

$$f(x):={\rm Re}\bigl(e^{e^{i\alpha}\,x}\bigr)=e^{\cos\alpha\>x}\>\cos\bigl(\sin\alpha\>x\bigr)\ .$$ Note that the rule ${\displaystyle{d\over dx}e^{\lambda x}=\lambda e^{\lambda x}}$ is also valid for complex $\lambda$. It follows that $$ {d^n\over dx^n}e^{\lambda x}=\lambda^n e^{\lambda x}\ .$$ When $\lambda=e^{i\alpha}$ then $\lambda^n=e^{i\cdot2\pi}=1$, hence $x\mapsto e^{\lambda x}$ "cycles" under $D$ with period $n$, and so does its real part.

Christian Blatter
  • 232,144