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

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

NotSObright
 2 years ago
Best ResponseYou've already chosen the best response.1looks to me a series of log(1+x) and log(1x)

NotSObright
 2 years ago
Best ResponseYou've already chosen the best response.1also r should start in 1 and not 0

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0but that has even terms in it

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0im not sure how to do that

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

NotSObright
 2 years ago
Best ResponseYou've already chosen the best response.1Signs will now alternate

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0is there another way to do this?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1355650436133:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2let me check for generating function

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

experimentX
 2 years ago
Best 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2woops 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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2this 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}]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0do you think that is the answer they want for question 3.?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2well ... 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.

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2i guess not, it's quite long

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

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

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

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

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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.2this 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

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

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