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In prove of the above claim I need prove: $\cot (\pi\cdot z)$ is bounded for $\{y \geq 1 , 0.5 \geq x \geq -0.5\}$ In stein book a function has defined:
$$\Delta ( z ) = \pi \cot \pi z - \sum _{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$$ It has proved that $\Delta ( z ) $ is entire and it's sufficient to $\Delta ( z ) $ be abounded.

Daniel W. Farlow
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Hoseyn Heydari
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    The content and title are asking for separate things, which one is the question? – jimjim Jan 05 '13 at 11:52
  • What has to do what you ask with what you say you have to prove, anyway? Can you give some background, ideas, self effort, etc.? – DonAntonio Jan 05 '13 at 11:52
  • Well, you changed completely the question, yet there still remains one doubt: what has to do the proof of the equality in the title with the boundness of $,\cot \pi z,$ wherever? Couldn't it be that you need to prove the equality and then using it you have to prove the boundness? – DonAntonio Jan 05 '13 at 12:09
  • Can you exhibit a value of $z$ such that $\sum_{n=1}^\infty (1-\frac{z^2}{n^2})$ converges? – Hagen von Eitzen Jan 05 '13 at 12:26
  • @Hagen von Eitzen for all $z \notin \mathbb{Z}$ it converges. – Hoseyn Heydari Jan 05 '13 at 12:36
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    The expression in the - now modified - title may converge as $\sum_{n=-N}^N \frac1{z+n}=\frac 1z+\sum_{n=1}^N\frac{2z}{z^2-n^2}$. But $\sum_{n=1}^\infty (1-\frac{z^2}{n^2})$ (occuring in the definition of $\Delta$ and in the original question title) will never converge as the summands tend to $1$ – Hagen von Eitzen Jan 05 '13 at 13:44
  • You're right. I write it wrong and edit the question. – Hoseyn Heydari Jan 05 '13 at 13:55

1 Answers1

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For the boundedness away from the real axis, $$ \begin{align} |\cot(\pi z)| &=\left|\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi z}}\right|\\ &=\left|\frac{e^{i\pi x-\pi y}+e^{-i\pi x+\pi y}}{e^{i\pi x-\pi y}-e^{-i\pi x+\pi y}}\right|\\ &\le\frac{e^{\pi|y|}+e^{-\pi|y|}}{e^{\pi|y|}-e^{-\pi|y|}}\\[6pt] &=\coth(\pi|y|) \end{align} $$ which decreases monotonically to $1$ as $|y|\to\infty$.

For the question in the title, see this answer.

robjohn
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