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UnkleRhaukus
 3 years ago
Evaluate the sum at \(x=\tfrac\pi2\)
\[S(x)
=\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r1) x})}{2r1}\\
\]
UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[S\left(\tfrac\pi2\right) =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r\tfrac\pi2)}{2r1}\\ =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(1)^{r1}}{2r1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0looks to me a series of log(1+x) and log(1x)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see this http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0eliminate using series for log (1x) try it if it not works use ix instead of x

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see that \[\frac{1}{2}\frac{\log(1+x)}{\log(1x)}=x+\frac{x^3}{3}+\frac{x^5}{5}...\] Replace x by ix

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Signs will now alternate

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

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

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

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

experimentX
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.2i guess not, it's quite long

UnkleRhaukus
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2yes you can!! it's pretty well known

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

experimentX
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years 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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