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Homework Help: Finding a fourier series for a cosine function

  1. Sep 6, 2009 #1
    Hi I have been working on a example and have worked it out as this

    1. The problem statement, all variables and given/known data

    [tex]f(x) = \( \left( \left \begin{array}{ccc}0 & \mathrm{for} & -\pi < x \leq 0 \\ cos(x) & \mathrm{for} & 0 < x < \pi\end{array}[/tex]

    where f is defined on the interval [tex]]-\pi,\pi[[/tex].

    Find the corresponding fourier series for f.

    and writ the sum for [tex]x = p \cdot \pi[/tex] where [tex]p \in \mathbb{Z}[/tex]

    2. Relevant equations

    Complex Fourier series is defined as

    [tex]\sum_{n=-\infty}^{\infty} c_{n} \cdot e^{i\cdot n \cdot t} [/tex]

    where [tex]c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \cdot e^{-i \cdot n \cdot t} dt[/tex]
    3. The attempt at a solution

    First I find [tex]c_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} cos(t) dt = [\mathrm{sin(t)}]_{\mathrm{t} = -\pi}^{\pi} = 0 [/tex]

    Next I find

    [tex]c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} (cos(t) \cdot cos(n \cdot t) - cos(t) \cdot sin(n \cdot t) \cdot i) dt[/tex]

    [tex]c_{n} = [\frac{{sin(n-1)\cdot t}}{2 \cdot (n-1)} + \frac{{sin(n11)\cdot t}}{2 \cdot (n+1)}-(\frac{{cos(n-1)\cdot t}}{2 \cdot (n-1)} - \frac{{cos(n+1)\cdot t}}{2 \cdot (n+1)}) \cdot i]_{t=-\pi}^{\pi} = \frac{sin(n+1)\cdot \pi}{2\cdot(n+1)}[/tex]

    which I then multiply by [tex]\frac{1}{2\pi}[/tex] from which I get

    [tex]c_{n} = \frac{sin(n+1)}{2(n+1)}[/tex]

    Where I end up with the fourier series

    [tex]\sum_{n=-\infty}^{\infty} \frac{sin(n+1)}{2(n+1)} \cdot e^{i \cdot n \cdot x}[/tex]

    Have I got it right? I fairly new to fourier series so I would be very happy if someone would reflect on my work and comment it please.

    p.s. Not sure about question two. Do I insert the value of into the sum?

    Thank you

    Best Regards

  2. jcsd
  3. Sep 6, 2009 #2
    remember that [itex]c_0=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) dt[/itex]

    but [itex]f(t)[/itex] is not constnat on the interval [itex] [-\pi,\pi] [/itex]....

    and so [itex]c_0=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) dt=\frac{1}{2 \pi} \int_{-\pi}^0 0 dt + \frac{1}{2 \pi} \int_0^{\pi} \cos{t} dt=\frac{1}{2 \pi} \int_0^{\pi} \cos{t} dt=\frac{1}{2 \pi} \sin{t} |_{0}^{\pi}=0[/itex] which is what you got.

    but you'll have to do the same when calculating [itex]c_n[/itex] which will probably change things.
  4. Sep 6, 2009 #3

    following your line of thought I then get

    [tex]c_{n} = \frac{1}{2\pi} \int_{\pi}^{pi} cos(t) \cdot cos(n \cdot t) - (cos(t) \cdot sin(n \cdot t) \cdot i) dt = \frac{1}{2\pi} \int_{\pi}^{0} cos(t) \cdot cos(n \cdot t) - (cos(t) \cdot sin(n \cdot t) \cdot i) dt + \frac{1}{2\pi} \int_{0}^{pi} cos(t) \cdot cos(n \cdot t) - (cos(t) \cdot sin(n \cdot t) \cdot i) dt [/tex]


    [tex]c_{n} = (\frac{-1}{2(n+1)\cdot \pi}-\frac{1}{2(n-1)\cdot \pi}) \cdot sin(n \cdot \pi) [/tex]

    does that look better?

    Best Regards
  5. Sep 6, 2009 #4
    [itex]c_n=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) e^{-int} dt=\frac{1}{2 \pi} \int_{-\pi}^{0} 0 dt + \frac{1}{2 \pi} \int_{0}^\pi} \cos{t} e^{-int} dt[/itex]

    do you understand why that is?

    so [itex]c_n=\frac{1}{2 \pi} \int_{0}^\pi} \cos{t} e^{-int} dt=\frac{1}{2 \pi} \int_{0}^\pi} \cos{t} \( \cos{nt}-i \sin{nt} \)[/itex]
    as the first term disappears entirely
  6. Sep 6, 2009 #5
    Hi again yes the reason why that term disappears it do to basic calculus.

    I have tried the new version of [tex]c_{n}[/tex]

    and I get

    [tex]c_n = (\frac{-1}{4(n+1)\cdot \pi} - \frac{1}{4(n-1)\cdot \pi}) \cdot sin(n\pi) + \frac{cos(n\cdot \pi) -1}{2 \cdot n \cdot \pi} \cdot i[/tex]

    Hopefully this looks better?

    Best Regards
    Last edited: Sep 6, 2009
  7. Sep 6, 2009 #6
    first of all i meant to put a bracket in my last line:

    [itex]c_n=\frac{1}{2 \pi} \int_0^{\pi} \cos{t} (\cos{nt}-i \sin{nt}) dt[/itex]

    then it's just integrating by parts.

    i'll assume you managed that ok.

    you have [itex]c_n=(\frac{-1}{4(n+1) \cdot \pi}-\frac{1}{4(n-1) \cdot \pi}) \cdot \sin{(n \pi)} + \frac{\cos{(n \cdot \pi)}-1}{2 \cdot n \cdot \pi} \cdot i[/itex]

    but [itex]\sin{n \pi}=0[/itex] and [itex]\cos{n \pi}=(-1)^n[/itex] which will simplify things i hope...
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