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I Dirac Delta δ(sin(ωx))?

  1. Oct 16, 2017 #1
    If we were to replace δ(x), the orginal Dirac Delta, with δ(sin(ωx)), what would be the result?

    Would we have an infinite spike everywhere on the graph of sinx where x is a multiple integer of π/ω? and 0 everywhere else?

    I apologize in advance if I had posted in the wrong category.
     
  2. jcsd
  3. Oct 16, 2017 #2

    Mark44

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    Seems reasonable that ##\delta(\sin(\omega x))## would be infinite wherever ##\omega x## is an odd multiple of ##\pi/2##, and zero everywhere else. In other words, where ##x = \frac{(2k + 1)\pi}{2\omega}##, with k in the integers.
     
  4. Oct 16, 2017 #3

    Orodruin

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    Yes. For a fuction ##f(x)## with a countable set of zeroes ##x_i##, it holds that
    $$
    \delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{|f’(x_i)|}
    $$
     
  5. Oct 16, 2017 #4

    Orodruin

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    The zeroes of the sine function are integer multiples of pi.
     
  6. Oct 16, 2017 #5
    Thank you guys for answering my question. Very much appreciated.
     
  7. Oct 16, 2017 #6

    andrewkirk

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    Speaking very loosely, the answer is yes.

    Speaking more carefully, the dirac delta is a distribution, not a function, and only gives a value when included in an integral. That is
    ##\int_a^b\delta(x)f(x)dx## is equal to ##f(0)## if ##a\leq 0 \leq b## and to zero otherwise. So your question could be expressed more precisely as:
    what is the value of ##\int_a^b \delta(\sin\omega x)f(x)dx##?

    This can be answered by substitution. Set ##u=\sin\omega x## so that ##du=\omega\cos x\,dx=\omega\sqrt{1-u^2}\,dx##. Then the Dirac delta part of the integrand becomes just ##\delta(u)##.

    We'll end up with an answer that is equal to the sum of the values of ##f(x)## for all values of ##x## in ##[a,b]## that are multiples of ##\pi/\omega##.
     
  8. Oct 17, 2017 #7

    Mark44

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    Doh!
     
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