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Homework Help: Proof I don't even know how to start

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]f(x) = a_{1}\sinx + a_2\sin(2x) + ... + a_Nsin(Nx)[/tex] where N[tex]\geq[/tex]1 is an integer and [tex]a_1, ... , a_N \in\Re[/tex]. Prove that for every [tex]n = 1, ... , N[/tex] we have
    [tex]a_n = \frac{1}{\pi}\int{f(x)\sin(nx)dx}[/tex]
    with the integral going from -[tex]\pi[/tex] to [tex]\pi[/tex] (sorry I don't know how to write definite integrals in LaTeX)

    For some reason, it's not showing the integral sign. Before [tex]\sin(nx)dx[/tex] there should be an integral sign followed by [tex]f(x)[/tex], but it's not showing up.

    2. Relevant equations

    3. The attempt at a solution

    I have no idea how to even start it. I'm not looking for the solution, just a push in the right direction. Your answers to all of my other questions of late have been spot on and for that I thank everyone who has responded. Thanks in advance for your help with this problem!
    Last edited: Oct 1, 2009
  2. jcsd
  3. Oct 1, 2009 #2


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    It shows up OK on my screen. Try refreshing your browser.

    Try substituting the expression for [tex]f(x)[/tex] into

    [tex]\frac{1}{\pi} \int f(x) sin(nx) dx[/tex]

    and using trig identities. You will save yourself some work if you keep in mind that

    [tex]\int_{-\pi}^{\pi} sin(ax) dx = 0[/tex]

    for any real number [tex]a[/tex].
  4. Oct 1, 2009 #3


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    First is f(x) really equal to what you have? [itex]f(x)= a_1 sin(x)+ a_2 sin(2x)+ \cdot\cdot\cdot a_N sin(Nx)[/itex] would make more sense!

    Assuming that is what it should be, look at
    [tex]\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(nx)dx= \frac{1}{\pi}\int_{-\pi}^\pi (a_1+a_2sin(2x)+ \cdot\cdot\cdot+ a_Nsin(Nx))sin(nx) dx[/tex]
    [tex]= \frac{1}{\pi}a_1\int_{-\pi}^\pi} sin(x)sin(nx)dx+ \frac{1}{\pi}a_2\int_{-\pi}^\pi}sin(2x)sin(nx)dx+ \cdot\cdot\cdot+ \frac{1}{\pi}a_N \int_{-\pi}^\pi} sin(Nx))sin(nx)dx[/tex]

    And also consider the value of
    [tex]\int_{-\pi}^\pi} sin(mx)sin(nx)dx[/itex] for m= n and for [itex]m\ne n[/tex].
  5. Oct 1, 2009 #4
    Oh, I proved the second part of your post in part a of the problem and did not realize that it was involved in any way; thanks!
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