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UnkleRhaukus

  • 3 years ago

8+8/9+8/25+...=

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  1. hartnn
    • 3 years ago
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    8(1/1^2 + 1/3^2+1/5^2 +....)

  2. hartnn
    • 3 years ago
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    pi^2/8, right ?

  3. UnkleRhaukus
    • 3 years ago
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    the eights cancel

  4. hartnn
    • 3 years ago
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    oh yeah...

  5. experimentX
    • 3 years ago
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    use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html

  6. UnkleRhaukus
    • 3 years ago
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  7. hartnn
    • 3 years ago
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    or f(x) = x for 0<x<= pi/2 = pi-x for pi/2 <= x < pi find its half range cosine series, it comes out to be f(x) = pi/4 - (2/pi)[cos 2x/1^2 + cos6x/3^2 + cos 10x/5^2 +...] thn put x=pi/2

  8. hartnn
    • 3 years ago
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    or simply f(x) = [(pi-x)/2 ]^2 in [0,2pi] find F.S expansion

  9. hartnn
    • 3 years ago
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    2nd one is simpler i guess.

  10. experimentX
    • 3 years ago
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    Define a function T(x) this way T(x) = arctan(2/pi)x for -pi/2<=x<pi/2 T(x) = -arctan(2/pi)x for pi/2<=x<3pi/2 \[ T (x) = {8 \over\pi^2 \ }\sum_{n=1,3,5 ..}{(-1)^{n-1 \over 2} \over n^2} \sin (n x) \\ T(\pi/2) = 1 \\ {8 \over \pi^2 } {(-1)^{n-1} \over (2n-1)^2} \sin((2n-1)\pi/2) = 1 \\ {8 \over \pi^2} ({1^2 + 1/3^2 + 1/5^2 + .. .})=1\]

  11. hartnn
    • 3 years ago
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    was the the Q realting Fourier Transform ?? i thot it was....

  12. hartnn
    • 3 years ago
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    *this

  13. UnkleRhaukus
    • 3 years ago
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    i can do the fourier cosines series method i was wondering if there was another way ,

  14. hartnn
    • 3 years ago
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    yeah, as i said the 2nd way i shown is simpler...u want the formulas ? or know it ?

  15. UnkleRhaukus
    • 3 years ago
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    i have worked it out

  16. hartnn
    • 3 years ago
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    ok...

  17. UnkleRhaukus
    • 3 years ago
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    \[\large{\color{grey}{π^2}}\]

  18. experimentX
    • 3 years ago
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    Plot[8/Pi^2 Sum[(-1)^(n - 1)/(2 n - 1)^2 Sin[(2 n - 1) x], {n, 1, 20}], {x, 0, 2 Pi}]

  19. experimentX
    • 3 years ago
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    \[ T(x) = {2 \pi} \arcsin(\sin(x))\] http://www.wolframalpha.com/input/?i=Plot+2%2Fpi+sin^-1%28sin%28x%29%29

  20. mukushla
    • 3 years ago
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    just for fun\[\zeta(2)=\frac{\pi^2}{6}\]\[\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\frac{1}{4}\zeta(2)+A\]where\[A=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}\]and \[A=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}\]so ur expression becomes\[8A=\pi^2\]

  21. experimentX
    • 3 years ago
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    yep!! i realized that :)

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