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Community Discussion Question: Summation

Forum Index -> Trignometry

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8 May 2008 14:45:18 IST

Subject: Summation

nasa_hs (0)

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The sum of th infinte series cot^-12+cot^-18+cot^-118+cot^-132....is equal to

pie/4

pie/2

3pie/4

none

ans is pie/4

[Post New]

8 May 2008 15:01:11 IST

Subject: Re:Summation

sandeepramesh (1090)

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this is nothing but $\sum {cot}^{-1} (2{r}^2) = tan^{-1} (\frac {1}{1+(2r^{2}-1)})$

i think you can handle it from here now! :D

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8 May 2008 15:29:47 IST

Subject: Re:Summation

sboosy (2860)

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$\mbox{Continuing from where sandeep has left} \\ \\ \tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right) \\ \\ = \tan^{-1}(2r+1) - \tan^{-1} (2r-1) \\ \\ \mbox{Summing from 1 to} \ \infty \ \mbox{we get} \\ \\ \tan^{-1}(3) - \tan^{-1}(1) +\tan^{-1}(5) - \tan^{-1}(3)+......\tan^{1}(\infty) \\ \\ = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

this reply: 15 points (with 3 Olaaa!! Perrrfect answer.

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