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UnkleRhaukus

  • 3 years ago

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

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

  2. NotSObright
    • 3 years ago
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    looks to me a series of log(1+x) and log(1-x)

  3. UnkleRhaukus
    • 3 years ago
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    how

  4. NotSObright
    • 3 years ago
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    also r should start in 1 and not 0

  5. UnkleRhaukus
    • 3 years ago
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    hmm

  6. NotSObright
    • 3 years ago
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    see this http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29

  7. UnkleRhaukus
    • 3 years ago
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    but that has even terms in it

  8. NotSObright
    • 3 years ago
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    eliminate using series for log (1-x) try it if it not works use ix instead of x

  9. UnkleRhaukus
    • 3 years ago
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    im not sure how to do that

  10. NotSObright
    • 3 years ago
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    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
    • 3 years ago
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    now do you see it

  12. NotSObright
    • 3 years ago
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    Signs will now alternate

  13. UnkleRhaukus
    • 3 years ago
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    is there another way to do this?

  14. UnkleRhaukus
    • 3 years ago
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    @mukushla

  15. experimentX
    • 3 years ago
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    |dw:1355650436133:dw|

  16. UnkleRhaukus
    • 3 years ago
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    ln 2 ?

  17. experimentX
    • 3 years ago
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    let me check for generating function

  18. experimentX
    • 3 years ago
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    probably the one by @NotSoBright works just put x=i and get the sum of series ... it should work out.

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

  20. experimentX
    • 3 years ago
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    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
    • 3 years ago
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    |dw:1355652797759:dw| putting that value you would get 1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1

  22. experimentX
    • 3 years ago
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    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
    • 3 years ago
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    do you think that is the answer they want for question 3.?

  24. experimentX
    • 3 years ago
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    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
    • 3 years ago
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    lol .. it seems there are quite a lot of questions!! I would put the values in the function ..

  26. UnkleRhaukus
    • 3 years ago
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    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
    • 3 years ago
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    i guess not, it's quite long

  28. UnkleRhaukus
    • 3 years ago
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    can i just assume that we know the sum of the alternating harmonic series is ln2, (in the context of Fourier series)

  29. experimentX
    • 3 years ago
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    yes you can!! it's pretty well known

  30. UnkleRhaukus
    • 3 years ago
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    how do i get the \(\pi/4\) bit?

  31. experimentX
    • 3 years ago
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    Do you know complex logarithm? \[ \log (1+i) = \log \sqrt 2 + i{\pi \over 4}\]

  32. UnkleRhaukus
    • 3 years ago
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    i have not studied complex logarithms , and they are not in the book the question has come from

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

  36. UnkleRhaukus
    • 3 years ago
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