Analyzing Analytic Functions: Solving a Complex Analysis Conundrum

[Jorriss]

I came across an interesting problem that I have made no progress on.

Let f be an analytic function on the disc $D = \{z \in C ~|~ |z| < 1\}$ satisfying $f(0) = 1$. Is the following
statement true or false? If $f(a) = f^\prime(a)$ whenever $\frac{1+a}{a}$ and $\frac{1-a}{a}$ are prime numbers then $f(z) = e^z$ for all $z \in D$.

Obviously I know that $f(z) = e^z$ solves $f^\prime = f$, but I don't see how to use that here.

 

[mathwonk]

i guess you want to show your function also satisfies f-f' = 0. since f is analytic, it suffices to show it satisfies this equation on an infinite set with a limit point in the open disc. do you know the principle of analytic continuation?

 

[WWGD]

i guess you want to show your function also satisfies f-f' = 0. since f is analytic, it suffices to show it satisfies this equation on an infinite set with a limit point in the open disc. do you know the principle of analytic continuation?

And notice the a here are all on the Real line.

 

[Infrared]

Are you sure this is a serious problem? It looks to be equivalent to the twin prime conjecture.

Note that if \frac{1+a}{a}=p, then a=\frac{1}{p-1}, which is inside your domain for all odd primes p. f will necessarily be the exponential iff there are infinitely many such values of a (so that they will have a limit point in the domain. We don't need to worry about the limit point being on the boundary |z|=1 since a decreases as p increases), but \frac{1+a}{a} and \frac{1-a}{a} are twin primes so your statement is equivalent to the twin prime conjecture.

 

[Jorriss]

Are you sure this is a serious problem? It looks to be equivalent to the twin prime conjecture.

As it would turn out, no it wasn't. It was the first problem in a joke qualifying exam. The other problems made it far more obvious that it was not serious. Anyhow, thanks for your insight into the joke I suppose!

 

[WWGD]

Kind of embarrassed at how I missed that. I guess I should read problems more carefully before trying to solve them.