Solve Gamelin's XIII.3 15: Infinite Product Meromorphicity

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Discussion Overview

The discussion revolves around demonstrating the meromorphicity of the function $$\frac{1}{z}\prod_{n=1}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$$ as presented in exercise XIII.3 problem 15 from Gamelin's Complex Analysis. Participants are exploring the use of logarithms, bounding functions, and applying the Weierstrass M-test in the context of complex analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests taking logarithms to transform the product into a sum, which is a common technique in complex analysis.
  • Another participant expresses difficulty in bounding the logarithmic expression and applying the Weierstrass M-test, specifically focusing on the convergence of the series involving logarithms.
  • A participant shares a link to a proof they posted elsewhere and invites critique, indicating a desire for feedback on their approach.
  • There is a discussion about the properties of complex logarithms and the necessity of choosing an analytic branch for the logarithm to ensure holomorphicity.
  • One participant proposes using a specific branch of the logarithm where the argument is between -π and π, asserting that this condition holds for certain terms in the expression.
  • Another participant questions the validity of an equality involving logarithms, pointing out that it is only correct up to a multiple of $$2\pi i$$ and asks for justification on why this multiple can be disregarded in this context.
  • A response indicates that the participant is not concerned with the multiple since the goal is to analyze the expression before exponentiation.

Areas of Agreement / Disagreement

Participants express differing views on the handling of complex logarithms and the implications of choosing an analytic branch. There is no consensus on the correctness of certain logarithmic manipulations, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants highlight limitations related to the properties of complex logarithms and the need for careful handling of branches, which may affect the analysis of meromorphicity.

xiavatar
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How do you show that $$\frac{1}{z}\prod_{n=1}^\infty \frac{n}{z+n}(\frac{n+1}{n})^z$$ is meromorphic? Any hints would be helpful, I'm having trouble bounding the functions and their logarithms. This is exercise XIII.3 problem 15 in Gamelin's Complex Analysis.
 
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Take logarithms - it transforms the product into a sum.
 
I did take logarithms, but the problem is bounding it and applying Weirstrass M-test. So what were reduced to showing is that $$\sum_{n=k}^\infty \text{log}(\frac{n}{z+n}(\frac{n+1}{n})^z)\leq \sum_{n=k}^\infty M_n<\infty$$ on a disk $$D_r$$ centered at zero where $$k>r$$.
 
Last edited:
log(\frac{n}{z+n}(\frac{n+1}{n})^{z})=log(n)-log(z+n)+zlog(n+1)-zlog(n)=(1-z)log(n)-log(z+n)+zlog(n+1).
 
But the problem is that complex logarithms don't preserve the properties were used to with real valued logarithms. In order for them to be holomorphic we have to choose an analytic branch for the logarithm, and once we've done that we can't just rewrite the log as you have written. Or have I misunderstood something?
 
xiavatar said:
we have to choose an analytic branch for the logarithm
Yes, but we have some help. I propose that we go for the branch where the argument is between -π and π. This is trivially fulfilled for log(n) and log(n+1). For sufficiently large n, the argument for log(z+n) is even between -π/2 and π/2.
 
Last edited:
Okay, so we can choose an analytic branch, but your above equality with the logarithm
Svein said:
log(\frac{n}{z+n}(\frac{n+1}{n})^{z})=log(n)-log(z+n)+zlog(n+1)-zlog(n)=(1-z)log(n)-log(z+n)+zlog(n+1).
is only correct up to a multiple of $$2\pi i$$. What allows you to assume that the multiple vanishes in this case?
 
xiavatar said:
What allows you to assume that the multiple vanishes in this case?
Well, since I am only doing logarithms in order to get a grip on the expression, I don't much care. The original expression is not in logarithms, so we need to take the exponent anyway.
 

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