## 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 Group Title

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

2. NotSObright Group Title

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

3. UnkleRhaukus Group Title

how

4. NotSObright Group Title

also r should start in 1 and not 0

5. UnkleRhaukus Group Title

hmm

6. NotSObright Group Title
7. UnkleRhaukus Group Title

but that has even terms in it

8. NotSObright Group Title

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

9. UnkleRhaukus Group Title

im not sure how to do that

10. NotSObright Group Title

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 Group Title

now do you see it

12. NotSObright Group Title

Signs will now alternate

13. UnkleRhaukus Group Title

is there another way to do this?

14. UnkleRhaukus Group Title

@mukushla

15. experimentX Group Title

|dw:1355650436133:dw|

16. UnkleRhaukus Group Title

ln 2 ?

17. experimentX Group Title

let me check for generating function

18. experimentX Group Title

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

19. experimentX Group Title

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

20. experimentX Group Title

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 Group Title

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

22. experimentX Group Title

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 Group Title

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

24. experimentX Group Title

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 Group Title

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

26. UnkleRhaukus Group Title

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 Group Title

i guess not, it's quite long

28. UnkleRhaukus Group Title

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

29. experimentX Group Title

yes you can!! it's pretty well known

30. UnkleRhaukus Group Title

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

31. experimentX Group Title

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

32. UnkleRhaukus Group Title

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

33. experimentX Group Title

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 Group Title

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 Group Title

\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 Group Title