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ghdoru11

  • one year ago

Find the Fourier expansion for: f:R->R periodic T=2pi , f(x)=3x, x is between [-pi,pi]

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  1. beginnersmind
    • one year ago
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    Why was this question closed?

  2. ghdoru11
    • one year ago
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    no one solve this and i have another question

  3. beginnersmind
    • one year ago
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    Anyway you can just go back to the definition of fourier coefficients: \[a_n = \int_{-\pi}^{\pi} 3x*sin(nx)dx\] \[b_n = \int_{-\pi}^{\pi} 3x*cos(nx)dx\] then evaluate the integrals. Use the substitution t = nx and integration by parts with u being the linear term and v' the trigonometeric part.

  4. IrishBoy123
    • one year ago
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    yeah, and it's odd so you ignore the cosine term

  5. beginnersmind
    • one year ago
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    Right, that saves a lot of work :)

  6. ghdoru11
    • one year ago
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    so just the first integral is neded?

  7. beginnersmind
    • one year ago
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    Yes, \(b_n\) = 0 for every n.

  8. ghdoru11
    • one year ago
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    thank you so much guys

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