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anonymous
 one year ago
Another trig sum/integral:
\[\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}\sin(2n1)x\,dx\]
anonymous
 one year ago
Another trig sum/integral: \[\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}\sin(2n1)x\,dx\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Admittedly, the integral part isn't that hard, but the result is interesting.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Wait when you say it's not that hard, is the x inside the sine function or outside it? Like is it _that_ easy? xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Inside, \(\sin((2k1)x)\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Some plots with \(n=1,5,10,100,1000\).

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[\int_0^{\pi/2}\sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}\sin(2n1)x\,dx\]\[\sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}\int_0^{\pi/2}\sin(2n1)x\,dx\]\[\sum_{n=1}^\infty \frac{(1)^n}{(2n1)^2}\cos((2n1)x)_0^{ \pi/2}\]\[\sum_{n=1}^\infty \frac{(1)^n}{(2n1)^2}[\sin(n \pi)  1]\]\[\sum_{n=1}^\infty \frac{(1)^{n+1}}{(2n1)^2}\] At first I was like, "Ahh I've seen this before, that's just \(\frac{\pi}{4}\)" But then I realized there's a square in the numerator. I'm guessing it's gonna be \(\frac{\pi}{2}\) since you gave the hint but I still haven't officially got it yet!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually the answer isn't in terms of \(\pi\)! But a good start nonetheless.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ooooh right of course that 1.57... was the upper limit, >_> the thing you're looking at is the area under that curve which looks suspiciously like a logarithm. Well I guess it's time to play around and figure this out haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've also been playing around with the idea that \[\sum_{n=1}^\infty \frac{(1)^{n+1}}{2n1}\sin(2n1)x=\sin x\frac{1}{3}\sin3x+\frac{1}{5}\sin5x\frac{1}{7}\sin7x\cdots\] almost resembles a Fourier sine expansion of something, but no progress with that so far...

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Oooh yeah that's a good idea interesting.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately, I don't see any way to extract the closed form without prior knowledge of the closed form: http://mathworld.wolfram.com/CatalansConstant.html
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