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Fourier Series

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the fourier series f(x) = [tex]\sum[/tex]ansin(nx) + bncos(nx) can be written as [tex]\sum[/tex]kn(cos(nx+[tex]\vartheta[/tex])) and define kn and [tex]\vartheta[/tex]

    where the summation is from 0 to [tex]\infty[/tex]

    2. Relevant equations
    sin [tex]\vartheta[/tex] = cos (90 - [tex]\vartheta[/tex]) ??

    3. The attempt at a solution
    Well what I originally did was replace the sin term by cos (90 - nx), put cosine in terms of complex exponentials, and then try to solve the equation, but I only got what I was given in the first place and not the solution (i.e. I went in a circle).

    Is there some kind of property of sin or cos I could use?
     
  2. jcsd
  3. Apr 11, 2010 #2

    LCKurtz

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    Think about something like this:

    [tex] a \cos x + b \sin x = \sqrt{a^2+b^2}\ \left(\frac a {\sqrt{a^2+b^2}}\cos x +\frac b {\sqrt{a^2+b^2}}\sin x\right )[/tex]

    and then think about what the expansion of

    [tex]\cos{(x -\phi)}[/tex]

    looks like.
     
  4. Apr 11, 2010 #3

    vela

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    Try expanding [itex]k_n\cos(nx+\theta_n)[/itex] using the angle addition trig identity.
     
  5. Apr 19, 2010 #4
    Thank you for the responses - I was able to derive the proof exactly.
     
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