Analyzing Analytic Functions: Solving a Complex Analysis Conundrum

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Discussion Overview

The discussion revolves around a problem in complex analysis concerning analytic functions defined on the unit disc. Participants explore whether a specific condition involving prime numbers leads to the conclusion that the function must be the exponential function.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving an analytic function on the unit disc and a condition related to prime numbers, questioning if this implies the function is the exponential function.
  • Another participant suggests that to show the function satisfies the condition, it is sufficient to demonstrate that it holds on an infinite set with a limit point in the open disc, referencing the principle of analytic continuation.
  • A different participant raises a concern that the problem may be equivalent to the twin prime conjecture, noting that the conditions on the variable a relate to prime numbers.
  • One participant later clarifies that the problem was part of a joke qualifying exam, indicating that it was not intended to be taken seriously.
  • A participant expresses embarrassment for initially taking the problem seriously and acknowledges the need for careful reading.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the seriousness of the problem, with some treating it as a legitimate mathematical inquiry while others recognize it as a joke. The equivalence to the twin prime conjecture remains a point of contention.

Contextual Notes

The discussion highlights the complexity of the problem, including the implications of the conditions on prime numbers and the analytic properties of the function, but does not resolve the underlying mathematical question.

Jorriss
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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.
 
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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?
 
mathwonk said:
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.
 
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.
 
HS-Scientist said:
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!
 
Kind of embarrassed at how I missed that. I guess I should read problems more carefully before trying to solve them.
 

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