## UnkleRhaukus Group Title Evaluate the sum at $$x=\tfrac\pi2$$ $S(x) =\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r-1) x})}{2r-1}\\$ one year ago one year ago

1. UnkleRhaukus

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

2. NotSObright

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

3. UnkleRhaukus

how

4. NotSObright

also r should start in 1 and not 0

5. UnkleRhaukus

hmm

6. NotSObright
7. UnkleRhaukus

but that has even terms in it

8. NotSObright

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

9. UnkleRhaukus

im not sure how to do that

10. NotSObright

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

11. NotSObright

now do you see it

12. NotSObright

Signs will now alternate

13. UnkleRhaukus

is there another way to do this?

14. UnkleRhaukus

@mukushla

15. experimentX

|dw:1355650436133:dw|

16. UnkleRhaukus

ln 2 ?

17. experimentX

let me check for generating function

18. experimentX

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

19. 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

20. 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

21. experimentX

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

22. 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}]

23. UnkleRhaukus

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

24. 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.

25. experimentX

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

26. 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

27. experimentX

i guess not, it's quite long

28. UnkleRhaukus

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

29. experimentX

yes you can!! it's pretty well known

30. UnkleRhaukus

how do i get the $$\pi/4$$ bit?

31. experimentX

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

32. UnkleRhaukus

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

33. 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

34. 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..

35. 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*}

36. UnkleRhaukus