Let $\mu(n)$ the Möbius function, we define $F:[0,1]\to[0,1]$ as $$F(x)=\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n.\tag{1}$$

For a function of this kind (I presume that this continuous function has image $[0,1]$) was defined, for example in last paragraph of page 986, what is a periodic point, and its corresponding order of periodicity.

Question. Can you find examples of periodic points of $(1)$? Is it possible or feasible to prove or refute that our function has periodic points of order $3$? Many thanks.

Since I think that these questions are very difficult feel free to add comments about your thoughts and reasonings, heuristics or conjectures. I believe that this question concerning the function $(1)$ isn't in the literature.


[1] Tien-Yien Li, James A. Yorke, Period Three Implies Chaos, The American Mathematical Monthly, Vol. 82, No. 10 (1975).


I claim $F(x)<x$ for all $0<x<1$. From there it follows that the only periodic point is the fixed point $0$.

Let $m(x)=\sum_{n\leq x}\frac{\mu(n)}{n}$. We have (see e.g. equation (5) here) $m(x)\leq 1$ for all $x$, with strict inequality e.g. for $n=2$ (indeed, for all $n>1$, but it's not needed). Using summation by parts, we get $$\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n=\sum_{n=1}^\infty(m(n)-m(n-1))x^n\\=\sum_{n=1}^\infty m(n)(x^n-x^{n+1})<\sum_{n=1}^\infty(x^n-x^{n+1})=x$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy