math

posted by chuck

I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it? integral (sin(u)sin(t-u)) du from 0 to t.

  1. MathMate

    If f(t)=g(t)=sin(t)
    The convolution would be
    ∫ sin(u)sin(t-u) du from 0 to t

    Use the identity:
    sin(x)sin(y)=(1/2)(cos(x-y)-cos(x+y))
    where x=u, y=t-u

    ∫ sin(u)sin(t-u) du from 0 to t
    =∫ (1/2)(cos(u-t+u)-cos(u+t-u))du
    =∫ (1/2)(cos(2u-t)-cos(t))du
    =(1/2)[(1/2)sin(2u-t)-ucos(t)] (from 0 to t)
    =(1/2)[(1/2)sin(t-(1/2)sin(-t)-tcos(t)]
    =(1/2)(sin(t)- t*cos(t))

  2. chuck

    I figured it out before I got your answer, and I used the trig identity sin(t-u)=sin(t)cos(u)-cos(t)sin(u). I ended up having to use four or five more trig substitutions before finally getting to the answer you have there. Your substitution is much easier to compute. Too bad I didn't check back here before going through all of that! Thank you!

  3. MathMate

    You're most welcome!
    Glad that it helped, and thank you for your feedback.

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