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UnkleRhaukus

  • 3 years ago

\[\begin{align*} % S(x) S(x) &=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r-1)x)}{1-(2r-1)^2}\right)\\ \\ S(\tfrac\pi2)&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((\pi r-\tfrac\pi2)}{1-(2r-1)^2}\right)\\ 1&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\right)\\ \frac\pi2-1&=2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\\ \frac\pi4-\frac12&=\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\\ \end{align*}\]

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  1. UnkleRhaukus
    • 3 years ago
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    um , i think i made a mistake somewhere

  2. experimentX
    • 3 years ago
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    sin r pi = 0

  3. UnkleRhaukus
    • 3 years ago
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    well sine of πr is going to be zero, so

  4. UnkleRhaukus
    • 3 years ago
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    so the last two lines are not true

  5. UnkleRhaukus
    • 3 years ago
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    and the third line isnt right either

  6. experimentX
    • 3 years ago
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    the second line is not true .. check again.

  7. UnkleRhaukus
    • 3 years ago
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    is the first line right?

  8. experimentX
    • 3 years ago
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    sorry ... it seems i made mistake I didn't see 2r there. I wouldn't now about the first line since it is your question. What is your function??

  9. experimentX
    • 3 years ago
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    why did you put \( S(\pi/2) = 1 \)

  10. UnkleRhaukus
    • 3 years ago
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    the function is f(x)=|sin(x)|

  11. phi
    • 3 years ago
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    I think the summation is mostly zeros, except for r=1 where it is 0/0 it's pi at r=1 ?

  12. UnkleRhaukus
    • 3 years ago
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    hmm i forgot it could do that

  13. UnkleRhaukus
    • 3 years ago
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    so have i made any mistake then?

  14. phi
    • 3 years ago
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    as far as I can see, your very first line, evaluated at x=pi/2 gives 2/pi *(1 + 2*pi/4) = 2/pi +1

  15. phi
    • 3 years ago
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    maybe you don't want that leading 1?

  16. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*} % a_0 a_0&=\frac1\pi\int\limits_{-\pi}^\pi |\sin(x)| \,\text dx\\ &=\frac2\pi\int\limits_{0}^\pi \sin(x) \,\text dx\qquad\text{(even integrand)}\\ &=\frac{2}\pi\left(-\cos(x)\Big|_0^\pi\right)\\ &=\frac{2(-(-1-1))}\pi \\ &=\frac{4}\pi \end{align*}\] \begin{align*} % a_n a_n&=\frac1\pi\int\limits_{-\pi}^\pi|\sin(x)|\cos(nx)\,\text dx\\ &=\frac2\pi\int\limits_{0}^\pi\sin(x)\cos(nx)\,\text dx\qquad\text{(even integrand)}\\ &=\frac2\pi\int\limits_{0}^\pi\frac{\sin(x+nx)-\sin(x-nx)}2\,\text dx\\ &=\frac1\pi\int\limits_{0}^\pi{\sin\left((1+n)x\right)-\sin\left((1-n)x\right)}\,\text dx\\ &=\frac1\pi\left(\frac{-\cos((1+n)x)}{1+n}-\frac{\cos((1-n)x))}{1-n}\Big|_0^\pi\right)\\ &=\frac1\pi\left(\frac{1-\cos((1+n)\pi)}{1+n}+\frac{1-\cos((1-n)\pi))}{1-n}\right)\\ &=\frac1\pi\left(\frac{1-(-1)^{n+1}}{1+n}+\frac{1-(-1)^{n-1})}{1-n}\right)\\ \\&=\frac1\pi\left(\frac{1+(-1)^{n}}{1+n}+\frac{1+(-1)^{n})}{1-n}\right)\\ \end{align*} \begin{align*} % b_n b_n&=\frac1\pi\int_{-\pi}^\pi|\sin(x)|\sin(nx)\,\text dx\\ &=0\qquad\text{(odd integrand)} \end{align*} \begin{align*} % S(x) S(x)&=\frac{2}\pi+\frac1\pi\sum\limits_{n=1}^\infty\left(\frac{1+(-1)^{n}}{1+n}+\frac{1+(-1)^{n})}{1-n}\right)\cos(nx)\\ &=\frac{2}\pi+\frac2\pi\sum\limits_{n=2,4,6\dots}^\infty\left(\frac{1}{1+n}+\frac{1}{1-n}\right)\cos(nx)\\ &=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1-n^2}\right)\\ &=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r-1)x)}{1-(2r-1)^2}\right)\\ \\&=\frac4\pi\left(\frac12-\frac{\cos(2x)}{3}+\frac{\cos(4x)}{15}+\frac{\cos(6x)}{35}+\dots\right) \end{align*} \]

  17. experimentX
    • 3 years ago
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    a_0 = 2/pi http://www.wolframalpha.com/input/?i=integrate+ |sin%28x%29|%2Fpi+from+0+to+pi

  18. experimentX
    • 3 years ago
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    it seems that you are taking 2pi as period, although it works, pi should also work fine.

  19. experimentX
    • 3 years ago
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    Plot[2/Pi - Sum[2/Pi * 1/(4 n^2 - 1) Cos[n x], {n, 1, 20}], {x, 0, 4 Pi}] however the period seems to be 2pi

  20. experimentX
    • 3 years ago
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    sorry the result was correct .. i forgot it was for even case. On Mathematica Plot[2/Pi - Sum[2/Pi * 1/(4 n^2 - 1) Cos[2 n x], {n, 1, 20}], {x, 0, 2 Pi}]

  21. experimentX
    • 3 years ago
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    so the final series is \[ |\sin x| = {2 \over \pi} +{2 \over \pi} \sum_{n=1}^\infty \frac 1{1 - 4 n^2} \cos (2nx) \]

  22. phi
    • 3 years ago
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    In case you didn't find your mistake, it happens when you change the index from n=2,4,6... to r=1,2,3... you should substitute n=2r (so that r= n/2 giving 1,2,3...) you then get exper's result

  23. UnkleRhaukus
    • 3 years ago
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    ah yes \[\begin{align*} % S(x) S(x)&=\frac{2}\pi+\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1+(-1)^{n}}{1-n^2}\right)\cos(nx)\\ &=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1-n^2}\right)\\ &=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1-(2r)^2}\right)\\ \\&=\frac4\pi\left(\frac12-\frac{\cos(2x)}{3}-\frac{\cos(4x)}{15}-\frac{\cos(6x)}{35}-\dots\right) \end{align*}\]

  24. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*} % S(x) S(x) &=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1-(2r)^2}\right)\\ \\ S(\tfrac\pi2)&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(\pi r)}{1-4r^2}\right)\\ 1&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{(-1)^{r}}{1-4r^2}\right)\\ \frac\pi4&=\frac12+\sum\limits_{r=1}^\infty\frac{(-1)^{r}}{1-4r^2}\\ \end{align*}\] thats better thanks to both of you

  25. UnkleRhaukus
    • 3 years ago
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