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nissn

  • 2 years ago

find the fourier series of this

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  1. nissn
    • 2 years ago
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    of this

  2. AccessDenied
    • 2 years ago
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    Is there any general form for a Fourier series of a function f(x)? I'm not particularly well-versed in the subject, although I'll try to help... :)

  3. AccessDenied
    • 2 years ago
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    My first observation of the problem is that f(x) goes through three cycles over the interval (-3pi, 3pi)...

  4. nissn
    • 2 years ago
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    yeah i know

  5. nissn
    • 2 years ago
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    the first thing I have to find is the fourier coefficient. I think it is 1/4?

  6. nissn
    • 2 years ago
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    \[1/(2\pi)(\int\limits_{-\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]

  7. nissn
    • 2 years ago
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    which is 1/4

  8. AccessDenied
    • 2 years ago
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    Yep, that appears correct to me.

  9. nissn
    • 2 years ago
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    \[a _{n}=1/\pi \int\limits_{-\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]

  10. nissn
    • 2 years ago
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    I am not sure if I am doing it right on the last one

  11. AccessDenied
    • 2 years ago
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    The resource I am checking indicate that the formula for a_n would be 1/pi * integral from -pi to pi of f(x) cos(nx) dx

  12. AccessDenied
    • 2 years ago
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    So... \( \displaystyle a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \; \textrm{d}x + \int_{0}^{\pi} \frac{1}{\pi} x \cos nx \; \textrm{d}x \right) \)

  13. nissn
    • 2 years ago
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    yeah so then it is correct

  14. nissn
    • 2 years ago
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    so then it is 1/pi?

  15. nissn
    • 2 years ago
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    sin (1/pi)

  16. AccessDenied
    • 2 years ago
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    So I am finding: \( \displaystyle a_n = \frac{\pi n \sin n \pi + \cos \pi n - 1}{\pi^2 n^2} \)

  17. nissn
    • 2 years ago
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    and then I must do a \[b _{n}\]

  18. nissn
    • 2 years ago
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    \[b _{n}=1/\pi(\int\limits_{-\pi}^{0}0dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]

  19. AccessDenied
    • 2 years ago
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    My resource shows an additional sin(nx) there \( \displaystyle \frac{1}{\pi} \int_{0}^{\pi} \frac{1}{\pi} x \sin nx \; \textrm{d}x \)

  20. nissn
    • 2 years ago
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    oh it's the same I just forgot to write the nx

  21. nissn
    • 2 years ago
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    and that -cos (pi/pi)

  22. AccessDenied
    • 2 years ago
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    Sorry, I gotta go for school. I take the calculation of this integral to wolfram... Wolfram evaluates it out as: \( \displaystyle b_n = \frac{\sin \pi n - \pi n \cos \pi n}{\pi^2 n^2} \) http://www.wolframalpha.com/input/?i=integral+from+0+to+pi+of+1%2Fpi%5E2+x+sin%28n+x%29+dx

  23. nissn
    • 2 years ago
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    okey. thank you

  24. AccessDenied
    • 2 years ago
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    You're welcome! :)

  25. AccessDenied
    • 2 years ago
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    and good luck, I think you're on the right track. :) I was using this as my resource: http://mathworld.wolfram.com/GeneralizedFourierSeries.html Just the end bit with the formula.

  26. nissn
    • 2 years ago
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    thank you :)

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