I'm trying to understand how the algebraic properties of the Dirac delta function might be passed onto the argument of the delta function.(adsbygoogle = window.adsbygoogle || []).push({});

One way to go from a function to its argument is to derive a Taylor series expansion of the function in terms of its argument. Then you are dealing with the argument and powers of it.

Normally, the Taylor series expansion forf(x)is:

[tex]f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}({x_0})}}{{n!}} \cdot {{(x - {x_0})}^n}} [/tex]

But how does this change for a composite function? How do you expand [itex]f(g(x))[/itex]? Do you just replacexin the sum withg(x)andx_{0}withg(x? Or do you differentiat_{0})f(g(x))with respect toxin each term before settingxtox_{0}?

But I suspect the even harder question is how do you take the derivatives of the Dirac delta function for each term in the sum. And if you have to evaluate it atx_{0}, won't that make each term infinity? Is there an easy expression for [itex]{\delta ^{(n)}}({x_0})[/itex]? Or better yet, how do we evaluate [itex]{\delta ^{(n)}}(g(x))[/itex]? Thank you.

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# Taylor series expansion of Dirac delta

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