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Evaluate the sum at \(x=\tfrac\pi2\)
\[S(x)
=\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r1) x})}{2r1}\\
\]
 one year ago
 one year ago
Evaluate the sum at \(x=\tfrac\pi2\) \[S(x) =\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r1) x})}{2r1}\\ \]
 one year ago
 one year ago

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[S\left(\tfrac\pi2\right) =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r\tfrac\pi2)}{2r1}\\ =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(1)^{r1}}{2r1}\]
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
looks to me a series of log(1+x) and log(1x)
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
also r should start in 1 and not 0
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
see this http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
but that has even terms in it
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
eliminate using series for log (1x) try it if it not works use ix instead of x
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
im not sure how to do that
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
see that \[\frac{1}{2}\frac{\log(1+x)}{\log(1x)}=x+\frac{x^3}{3}+\frac{x^5}{5}...\] Replace x by ix
 one year ago

NotSObrightBest ResponseYou've already chosen the best response.1
Signs will now alternate
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
is there another way to do this?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1355650436133:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
let me check for generating function
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
probably the one by @NotSoBright works just put x=i and get the sum of series ... it should work out.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
\[ {1 \over 1  x } = 1 + x + x^2 + ....\\ {1 \over 1 +x} = 1  x + x^2  ... \\ \text{ Adding }\\ {1 \over 1  x} + {1 \over 1 +x} = {2 \over 1 x^2}= 2( 1 +x^2 + x^4 + ...) \\ \text{Integrating both sides} \\ {1 \over 2} \log \left( 1 + x \over 1 x\right) = x+\frac{x^3}{3}+\frac{x^5}{5}... \\ \text{Replace x by i you get} \\ \log 2  \frac{\log \sqrt 2} 2 + i\frac \pi 2 = i \left( 1  {1 \over 3} + \frac 15  ... \right) \] comparing imaginary terms, you get that sum is pi/2
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
woops sorry ... \[ \log 2  \frac{\log \sqrt 2} 2 + i\frac \pi 4 = i \left( 1  {1 \over 3} + \frac 15  ... \right) \] should have been pi/4
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1355652797759:dw putting that value you would get 1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
this is definitely expansion of what i wrote earlier. If you have Mathematica you can try this code Plot[1/2 + 2/Pi *Sum[ Sin[(2 n  1) x]/(2 n  1), {n, 1, 30}], {x, 0, 2 Pi}]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
do you think that is the answer they want for question 3.?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
well ... i think there are lots of series that uses the convergence of Fourier series to find the value of series. like ... for example: \( \sum {1 \over n^2} , \sum {1 \over n^4} \) .. i think it is other way ... probably to show that the summation of series and the value from Fourier are equal.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
lol .. it seems there are quite a lot of questions!! I would put the values in the function ..
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i think your both getting the right result, i just dont follow the method, and id be surprised if this is the method they are looking for
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
i guess not, it's quite long
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
can i just assume that we know the sum of the alternating harmonic series is ln2, (in the context of Fourier series)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yes you can!! it's pretty well known
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
how do i get the \(\pi/4\) bit?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
Do you know complex logarithm? \[ \log (1+i) = \log \sqrt 2 + i{\pi \over 4}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i have not studied complex logarithms , and they are not in the book the question has come from
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
this shouldn't be difficult \[ \large z = ze^{i \arg(z)} \\ \log(z) = \log z + i\arg(z)\] as you see z = 1+i, z = \sqrt(2) and arg(z) = pi/4
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i think i had the completely idea wrong idea of what this question was asking, because in question 1 i know S(1/2) so..
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\begin{align*} % S(x) S(x) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r1)x\big)}{2r1}\\ \\ \\% S(π/2) S\left(\tfrac\pi2\right) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin(\pi r\tfrac\pi2)}{2r1}\\ 1&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{(1)^{r1}}{2r1}\\ \frac\pi4&=\sum\limits_{r=1}^\infty\frac{(1)^{r1}}{2r1}\\ \\ \end{align*}\]
 one year ago
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