A question about Dirac Delta Function

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The discussion focuses on the Dirac Delta Function and its application in proving the equation: δ(g(x)) = ∑_{a, g(a)=0, g'(a)≠0} δ(x-a) / |g'(a)|. The participants explain the use of Taylor Expansion, specifically why only the first two terms are retained while higher-order terms are neglected due to their insignificance in the limit. The integral decomposition ∫_{-∞}^{+∞} f(x) δ(g(x)) dx is discussed, emphasizing the approximation of g(x) near its zeros and the application of the property δ(αx) = (1/α) δ(x).

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For proving this equation:

[tex] \delta (g(x)) = \sum _{ a,\\ g(a)=0,\\ { g }^{ ' }(a)\neq 0 }^{ }{ \frac { \delta (x-a) }{ \left| { g }^{ ' }(a) \right| } } [/tex]

We suppose that
[tex]g(x)\approx g(a) + (x-a)g^{'}(a)[/tex]

Why for Taylor Expansion we just keep two first case and neglect others? Are those expressions so small? if yes how we can explain it?
 
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OK. Let me explain it to you. We start from decomposing the integral

##\int_{+\infty}^{-\infty} f(x)\ \delta(g(x))\,dx = \sum_{a} \int_{a + ε}^{a - ε} f(x)\ \delta((x - a) g^{'}(a)) ##

into a sum of integrals over small intervals containing the zeros of g(x). In these intervals, since x is supposed to be very near to the a we can approximate g(x) as g(a) + (x - a)##g^{'}(a)## (note also that it is just the definition of derivative of g(x) when x goes toward a). Now we have proved it if we employ the equation ##\delta(αx) = 1/α\ \delta(x)## on the right-hand side of the integral.
 
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