- UnkleRhaukus

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

- jamiebookeater

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- UnkleRhaukus

\[S\left(\tfrac\pi2\right)
=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\
=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(-1)^{r-1}}{2r-1}\]

- anonymous

looks to me a series of log(1+x) and log(1-x)

- UnkleRhaukus

how

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## More answers

- anonymous

also r should start in 1 and not 0

- UnkleRhaukus

hmm

- anonymous

see this
http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29

- UnkleRhaukus

but that has even terms in it

- anonymous

eliminate using series for log (1-x)
try it
if it not works use ix instead of x

- UnkleRhaukus

im not sure how to do that

- anonymous

see that
\[\frac{1}{2}\frac{\log(1+x)}{\log(1-x)}=x+\frac{x^3}{3}+\frac{x^5}{5}...\]
Replace x by ix

- anonymous

now do you see it

- anonymous

Signs will now alternate

- UnkleRhaukus

is there another way to do this?

- UnkleRhaukus

- experimentX

|dw:1355650436133:dw|

- UnkleRhaukus

ln 2 ?

- experimentX

let me check for generating function

- experimentX

probably the one by @NotSoBright works just put x=i and get the sum of series ... it should work out.

- experimentX

\[ {1 \over 1 - x } = 1 + x + x^2 + ....\\
{1 \over 1 +x} = 1 - x + x^2 - ... \\
------\text{ Adding }------\\
{1 \over 1 - x} + {1 \over 1 +x} = {2 \over 1 -x^2}= 2( 1 +x^2 + x^4 + ...) \\
\text{Integrating both sides} \\
{1 \over 2} \log \left( 1 + x \over 1- x\right) = x+\frac{x^3}{3}+\frac{x^5}{5}... \\
\text{Replace x by i you get} \\
\log 2 - \frac{\log \sqrt 2} 2 + i\frac \pi 2 = i \left( 1 - {1 \over 3} + \frac 15 - ... \right)
\]
comparing imaginary terms, you get that sum is pi/2

- experimentX

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

- experimentX

|dw:1355652797759:dw|
putting that value you would get
1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1

- experimentX

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}]

- UnkleRhaukus

do you think that is the answer they want for question 3.?

##### 1 Attachment

- experimentX

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.

- experimentX

lol .. it seems there are quite a lot of questions!! I would put the values in the function ..

- UnkleRhaukus

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

- experimentX

i guess not, it's quite long

- UnkleRhaukus

can i just assume that we know the sum of the alternating harmonic series is ln2,
(in the context of Fourier series)

- experimentX

yes you can!! it's pretty well known

- UnkleRhaukus

how do i get the \(\pi/4\) bit?

- experimentX

Do you know complex logarithm?
\[ \log (1+i) = \log \sqrt 2 + i{\pi \over 4}\]

- UnkleRhaukus

i have not studied complex logarithms , and they are not in the book the question has come from

- experimentX

this shouldn't be difficult
\[ \large z = |z|e^{i \arg(z)} \\
\log(z) = \log |z| + i\arg(z)\]
as you see z = 1+i, |z| = \sqrt(2) and arg(z) = pi/4

- UnkleRhaukus

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..

- UnkleRhaukus

\[\begin{align*} % S(x)
S(x)
&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1}\\
\\
\\% S(π/2)
S\left(\tfrac\pi2\right)
&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\
1&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\
\frac\pi4&=\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\
\\
\end{align*}\]

- UnkleRhaukus

##### 1 Attachment

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