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2 \times 3
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2^{34}
234
a_{i-1}
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Comments
If you ask me how to do this by complex analysis , there is a much simpler answer.
i am using the summation theorem, first take a look here if necessary to see how all this happens.
Let the function be f(z)=z+a1 . It has a simple pole at z=−a and by summation theroem we need to evaluate the residue of πcot(πz)f(z) at z=−a
The residue is therefore R=z→−alim(z+a)z+aπcotπz=−πcotπa
The answer is therefore the negative of the residue ,
−∞∑∞n+a1=πcotπa
The series is actually the representation of ψ(a) where ψ(.) is the Digamma Function.
(Assuming you know complex analysis,) do you see why complex analysis is the natural way to approach this problem?
When asking about "what have you tried", part of it is also to understand how much background you have, in order to provide a suitable reply. Your solution didn't give me great confidence in your understanding of calculus/anlysis, which is why I was unable to suggest taking a "high power" approach instead of a "crude summation like Riemann sum".
In future, by providing this context to others (e.g. I got this problem when working through Serge Lang's complex analysis book on this chapter), that will give you better replies.
Thank you so much for the response. I don't know how to deal with it. I know about General Harmonic series. But i am not getting any idea about how to solve a summation from "negative infinity to positive infinity" which also involves a constant "a" .
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
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or__bold__
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[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
If you ask me how to do this by complex analysis , there is a much simpler answer.
i am using the summation theorem, first take a look here if necessary to see how all this happens. Let the function be f(z)=z+a1 . It has a simple pole at z=−a and by summation theroem we need to evaluate the residue of πcot(πz)f(z) at z=−a
The residue is therefore R=z→−alim(z+a)z+aπcotπz=−πcotπa
The answer is therefore the negative of the residue ,
−∞∑∞n+a1=πcotπa
The series is actually the representation of ψ(a) where ψ(.) is the Digamma Function.
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Wow. This is exactly what I was looking for. Thank you so much for such a wonderful reply.
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(Assuming you know complex analysis,) do you see why complex analysis is the natural way to approach this problem?
When asking about "what have you tried", part of it is also to understand how much background you have, in order to provide a suitable reply. Your solution didn't give me great confidence in your understanding of calculus/anlysis, which is why I was unable to suggest taking a "high power" approach instead of a "crude summation like Riemann sum".
In future, by providing this context to others (e.g. I got this problem when working through Serge Lang's complex analysis book on this chapter), that will give you better replies.
What have you tried, where did you get stuck?
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Thank you so much for the response. I don't know how to deal with it. I know about General Harmonic series. But i am not getting any idea about how to solve a summation from "negative infinity to positive infinity" which also involves a constant "a" .