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    Hi all

    I've got a question on Fourier series, I know the answer i should be getting, however i'm getting the opposite sign. I've checked my work and just cant see where i have gone wrong, please help. Ill put my working below.
    If someone could inform me why im getting 12/(pi)(n) instead of -12/(pi)(n), i would be very grateful

    Thanks in advance
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    (Original post by Ramjam)
    Hi all

    I've got a question on Fourier series, I know the answer i should be getting, however i'm getting the opposite sign. I've checked my work and just cant see where i have gone wrong, please help. Ill put my working below.
    If someone could inform me why im getting 12/(pi)(n) instead of -12/(pi)(n), i would be very grateful

    Thanks in advance
    I'm a little puzzled as to where you got your expression for b_n. . .
    Should it not be b_n=\dfrac{1}{\pi} \displaystyle\int_{-\pi}^{\pi} f(t) \sin(nt) \ dt?
    With this expression the result you want follows.

    Edit: Also what is \omega here? You've written it in some places, and not others. . .
    An image of the question itself might be useful.
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    (Original post by joostan)
    I'm a little puzzled as to where you got your expression for b_n. . .
    Should it not be b_n=\dfrac{1}{\pi} \displaystyle\int_{-\pi}^{\pi} f(t) \sin(nt) \ dt?
    With this expression the result you want follows.

    Edit: Also what is \omega here? You've written it in some places, and not others. . .
    An image of the question itself might be useful.
    It came from 'Modern Engineering Mathematics by Glyn James', it states that i can use that formula specifically for odd function. However I will work through using the original formula for bn and see what answer I achieve.

    The question is on the working, in black pen, it is to find the fourier series of f (t) = 3 (-pi < t < 0) -3 (0 < t < pi
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    (Original post by Ramjam)
    It came from 'Modern Engineering Mathematics by Glyn James', it states that i can use that formula specifically for odd function. However I will work through using the original formula for bn and see what answer I achieve.

    The question is on the working, in black pen, it is to find the fourier series of f (t) = 3 (-pi < t < 0) -3 (0 < t < pi
    I can't say I've seen that formula before, so I can't comment as to what's gone on there.
    All I can see is that for an odd function f on [-\pi,\pi] one can rewrite: b_n=\dfrac{2}{\pi} \displaystyle\int_0^{\pi} f(t) \sin(nt) \ dt.

    Yeah, I figured that was the question, but then I don't see what \omega has to do with anything.
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    (Original post by joostan)
    I can't say I've seen that formula before, so I can't comment as to what's gone on there.
    All I can see is that for an odd function f on [-\pi,\pi] one can rewrite: b_n=\dfrac{2}{\pi} \displaystyle\int_0^{\pi} f(t) \sin(nt) \ dt.

    Yeah, I figured that was the question, but then I don't see what \omega has to do with anything.
    The \omega issue is just me making mistakes again (sorry)

    I've reworked the question and reached a point in which i'm unsure on where to go from here. Whats do i have to do next? The point at which you assign odd and even values of n, although sounds simple always confuses me.
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    (Original post by Ramjam)
    The \omega issue is just me making mistakes again (sorry)

    I've reworked the question and reached a point in which i'm unsure on where to go from here. Whats do i have to do next? The point at which you assign odd and even values of n, although sounds simple always confuses me.
    There seems to be a mistake prior to that: f(t) changes sign in the interval, so you'll want to write:
    \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt .
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    (Original post by joostan)
    There seems to be a mistake prior to that: f(t) changes sign in the interval, so you'll want to write:
    \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt .
    Whats happens after that? do you integrate between limits separately for both of those?
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    (Original post by Ramjam)
    Whats happens after that? do you integrate between limits separately for both of those?
    You could do that, or you could send t \mapsto -t in either integral to obtain a single integral, it doesn't really matter.
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    (Original post by joostan)
    You could do that, or you could send t \mapsto -t in either integral to obtain a single integral, it doesn't really matter.
    Well i'm getting no where quick, i just cant seem to get my head around this question, i must be making mistakes somewhere as it keeps going completly wrong
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    (Original post by Ramjam)
    Well i'm getting no where quick, i just cant seem to get my head around this question, i must be making mistakes somewhere as it keeps going completly wrong
    Spoiler:
    Show
    \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt

\Rightarrow \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt=\displaystyle\int_{-\pi}^{0} 3\sin(nt) \ dt - \displaystyle\int_{0}^{\pi} 3\sin(nt) \ dt
    These integrals are then simple to evaluate.
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    (Original post by joostan)
    Spoiler:
    Show
    \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt = \displaystyle\int_{-\pi}^{0} f(t)\sin(nt) \ dt + \displaystyle\int_{0}^{\pi} f(t)\sin(nt) \ dt

\Rightarrow \displaystyle\int_{-\pi}^{\pi} f(t)\sin(nt) \ dt=\displaystyle\int_{-\pi}^{0} 3\sin(nt) \ dt - \displaystyle\int_{0}^{\pi} 3\sin(nt) \ dt
    These integrals are then simple to evaluate.
    from here would you integrate them to get it in terms of cos?
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    (Original post by Ramjam)
    from here would you integrate them to get it in terms of cos?
    Well I actually used the fact that \sin(nt) is odd to write\displaystyle\int_{-\pi}^{0} 3\sin(nt) \ dt=-\displaystyle\int_{0}^{\pi} 3\sin(nt) \ dt.
    But you could just as well integrate now.
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    Hint: \displaystyle b_n=\frac{2}{\pi} \int_0^{\pi} f(t)\sin (nt)~dt=-\frac{6}{\pi} \int_0^{\pi} \sin (nt)~dt (Since the integrand is an even function.)

    Then note \displaystyle \cos(n\pi)=(-1)^n to tidy up your answer.
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    (Original post by poorform)
    Hint: \displaystyle b_n=\frac{2}{\pi} \int_0^{\pi} f(t)\sin (nt)~dt=\frac{6}{\pi} \int_0^{\pi} \sin (nt)~dt (Since the integrand is an even function.)

    Then note \displaystyle \cos(n\pi)=(-1)^n to tidy up your answer.
    You dropped a minus in your expression for b_n.
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    (Original post by joostan)
    You dropped a minus in your expression for b_n.
    Will update now thanks!

    I (wrongly) assumed that f(t) was positive for positive t and negative for negative t.

    Since I did a very similar question the other day that was defined that way haaaaa.
 
 
 
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