What happens when we replace the Dirac Delta function with a sine function?

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Discussion Overview

The discussion centers around the implications of replacing the Dirac Delta function with a sine function, specifically examining the behavior of the modified delta function, δ(sin(ωx)). Participants explore the mathematical consequences and properties of this substitution, including its effects in integrals and the locations of spikes in the resulting function.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant questions the result of replacing δ(x) with δ(sin(ωx)), suggesting that it would create infinite spikes at integer multiples of π/ω and be zero elsewhere.
  • Another participant agrees, stating that δ(sin(ωx)) would be infinite at points where ωx is an odd multiple of π/2, specifically at x = (2k + 1)π/(2ω) for integer k.
  • A third participant introduces a general formula for δ(f(x)) in terms of its zeroes, indicating that the behavior of δ(sin(ωx)) can be analyzed using this approach.
  • One participant emphasizes that the Dirac delta is a distribution, not a function, and clarifies that its value is determined within the context of an integral, prompting a more precise formulation of the original question.
  • Another participant reiterates the locations of the zeroes of the sine function as integer multiples of π.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical behavior of δ(sin(ωx)) at specific points, but the discussion remains open regarding the implications and interpretations of this behavior in the context of integrals.

Contextual Notes

The discussion includes assumptions about the properties of the Dirac delta function and its application in integrals, which may not be universally accepted or fully explored.

Kyle Nemeth
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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.
 
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Kyle Nemeth said:
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?
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.
 
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Yes. For a fuction ##f(x)## with a countable set of zeroes ##x_i##, it holds that
$$
\delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{|f’(x_i)|}
$$
 
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Mark44 said:
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.
The zeroes of the sine function are integer multiples of pi.
 
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Thank you guys for answering my question. Very much appreciated.
 
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##.
 
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Orodruin said:
The zeroes of the sine function are integer multiples of pi.
Doh!
 

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