A question about Dirac Delta Function

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The discussion centers on the application of the Dirac Delta function in the context of the equation involving g(x). It explains the use of Taylor expansion, where only the first two terms are retained because higher-order terms become negligible as x approaches a. The integral is decomposed into sums over small intervals around the zeros of g(x), allowing for the approximation of g(x) near these points. The proof utilizes the property of the Dirac Delta function, specifically that δ(αx) = (1/α)δ(x), to validate the transformation of the integral. This approach clarifies why higher-order terms in the Taylor expansion can be ignored in this context.
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For proving this equation:

<br /> \delta (g(x)) = \sum _{ a,\\ g(a)=0,\\ { g }^{ &#039; }(a)\neq 0 }^{ }{ \frac { \delta (x-a) }{ \left| { g }^{ &#039; }(a) \right| } } <br />

We suppose that
g(x)\approx g(a) + (x-a)g^{&#039;}(a)

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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