Discover the Arc Tan Sum Formula for Math Help in Just a Few Steps

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Homework Help Overview

The discussion revolves around finding the sum of a series involving the arctangent function, specifically the expression Arc(tan(1/2)) + Arc(tan(1/8)) + ... + Arc(tan(1/2*n^2). The subject area pertains to mathematical series and trigonometric identities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the notation used in the problem, questioning the definition of "Arc" versus "Arctan." There is an inquiry about whether the sum is infinite and what it means to "find" the sum. Suggestions are made regarding potential simplifications and connections to known identities.

Discussion Status

The discussion includes various interpretations of the problem and attempts to clarify the notation. Some participants have provided insights into similar problems and identities that may aid in approaching the original question. However, there is no explicit consensus on a method or solution at this stage.

Contextual Notes

There is a mention of homework rules that discourage providing complete solutions without the original poster showing effort. The discussion also highlights the potential for confusion regarding the notation used in the problem statement.

hadi amiri 4
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Homework Statement


Find the sum
Arc(tan1/2)+Arc(Tan1/8)+...+Arc(Tan1/2*n^2)

Homework Equations



nothing

The Attempt at a Solution

 
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People still haven't gotten this part yet? :smile:

Hadi, what have you attempted so far? As always, we don't do homework for you; you must show some effort before we do :wink:

Is this an infinite sum?
 


And don't you mean Arctan(1/2), etc. rather than Arc(tan(1/2))- else you will need to define "Arc" for me!
 


How can I edit this one
 


hadi amiri 4 said:
How can I edit this one

You can't. Just post again this time with the correct sum.
 


Presumably the sum is
[tex]\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{1}{8}\right) + \cdots + \arctan\left(\frac{1}{2n^2}\right).[/tex]

If so, then what exactly does it mean to "find" this sum? If the goal is to simplify it, then this problem is similar to an old, well-known one that asks for a simplification of the sum
[tex]\sum_{k=1}^{n} \arctan\left(\frac{1}{1+k+k^2}\right).[/tex]

One of the ways of doing this is to first note that [itex]\arctan(k+1) - \arctan(k) = \arctan(1/(1+k+k^2))[/itex]*, and then telescope.

If you can figure out how to get this identity, then you can play around to come up with a similar one that will solve your problem.

It's also interesting to try to evaluate
[tex]\sum_{k=1}^{\infty} \arctan\left(\frac{1}{2k^2}\right).[/tex]

(* What's up with [itex]?)[/itex]
 


thanks a lot my question is solved
 

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