Number of analytic functions such that $(f(\frac{1}{n}))^3=\frac{n}{n+1}.$

Question

I want to find the number of analytic functions on the unit disk $\mathbb{E} = \{z:|z|<1\}$ such that $f:\mathbb{E}\ \rightarrow \mathbb{C}$ is analytic and $(f(\frac{1}{n}))^3=\frac{n}{n+1}$ for every $n \in \mathbb{N}$.

I thought of doing the following: We know that $\frac{1}{n}\rightarrow 0$ as $n \rightarrow \infty$, so $\mathbb{E}$ contains an accumulation point of any function $f$. The identity theorem then tells us that there is at most one analytic function $f$ such that the criterium holds.

The problem now, is that we don't have an explicit statement for $f(\frac{1}{n})$, rather we have a statement for $(f(\frac{1}{n}))^3$. I thought that we could just take the cube root to get an explicit statement for $f(\frac{1}{n})= \sqrt[3]{\frac{n}{n+1}}$. The function $g(z) $ that we can make by extending $f(\frac{1}{n})$, would be $g(z)=\sqrt[3]{\frac{1}{z+1}}$ (where we take $z \mapsto \sqrt[3]{z} = \exp\{\frac{1}{3}(\ln(|z|)+i\cdot \text{Arg}(z))\}$. Since this is analytic on our disk, we know that there is just one analytic function such that $(f(\frac{1}{n}))^3=\frac{n}{n+1}$ for every $n \in \mathbb{N}$.

My problem with this method however, is that I don't know whether $f(\frac{1}{n})= \sqrt[3]{\frac{n}{n+1}}$ is the only function for $f$ that satisfies $(f(\frac{1}{n}))^3=\frac{n}{n+1}$. It feels weird to assume that there is just one answer to this function. My question is now, is my suspicion grounded?

What would we do if we had $(f(\frac{1}{n}))^k=\frac{n}{n+1}$ for $k \in \mathbb{N} \backslash\{1\}$? I know that $k=2$ gives two analytic functions, Does this extend to the following: For every even $k$ we have 2 answers, for every odd $k$ we have 1?

Thanks for your time,

K. Kamal

Answer 1

My question is now, is my suspicion grounded?

YES it is. In fact, there are, for any nonzero complex number $w$, exactly $k$ solutions to the equation $z^k=w$. This is because you can multiply any solution $z_1$ by $e^{\frac{2\pi}{k}i}$ (that is, rotate the solution around $0$ by an angle of $\frac{2\pi}{k}$) and still get a solution.

In your case, you could also have $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{1\cdot 2\pi}{3})}$$ or $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{2\cdot 2\pi}{3})}$$

and still get a solution.

Answer 2

Note that$$\frac n{n+1}=\frac1{1+\frac1n}$$and that therefore, by the identity theorem, if such a function exists, then we have $(\forall z\in\mathbb{E}):f^3(z)=\frac1{1+z}$. Besides, in $\mathbb E$ the map $z\mapsto\frac1{1+z}$ has no zeros and $\mathbb E$ is simply connected. Therefore, yes, there is such a function. In fact, there are $3$ such functions, because if $\omega=e^{\frac{2\pi i}3}$ and $f$ is one such function, then the other $2$ will be $\omega f$ and $\omega^2f$.

Answer 3

Since $f(1/n)^3 = n/(n+1) = 1/(1+1/n)$ for all $n,$ we have $f(z)^3 = 1/(1+z)$ for all $z\in \mathbb E$ by the identity principle. So how many analytic functions on $\mathbb E$ satisfy this?

Note $1/(1+z)$ maps $\mathbb E$ into the open right half plane, so one solution is

$$f_0(z)= \exp[\frac13 \log 1/(1+z)],$$

where $\log$ denotes the principal branch. If we let $a=e^{2\pi i/3},$ then $f_1=af_0, f_2=a^2f_0$ are also solutions. Are these all the solutions? Yes, but this needs proof.

Define the sets $A_0 = \{(1/(1+1/n))^{1/3}:n\in \mathbb N\}, A_1 = aA_0, A_2 = a^2A_0.$ Here $(1/(1+1/n))^{1/3}$ denotes the ordinary real cube root, the one that goes along with $f_0.$ Suppose $g$ is a solution. Then for all $n,$

$$g(1/n) \in A_0\cup A_1\cup A_2.$$

It follows that $g(1/n)$ lies in one of these sets for infinitely many $n.$ Suppose WLOG it's $A_1.$ Note that this infinitely many $1/n$ constitute a sequence $\to 0.$ By the identity principle, $g=f_1$ in $\mathbb E.$

Thus each $f_j$ is a solution, and these are the only solutions.