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If I have the distribution: $\delta'(x^3 + x)$ and I want to write it using only $x$ as argument.

I have tried using $$\delta(f(x)) = \frac{\sum_{i=1}^{k} \delta(x - x_i)}{|f'(x_i)|}$$ where $f(x) = x^3 + x$ and $x_i$ are the simples roots of $f(x)$, so $f(x_i) = 0$ and $x_1 = 0, x_2 = i\sqrt{3}, x_3 = -i\sqrt{3}$.

I tried to differentiate that equation, but did not get the result I wanted, which is $\frac{\delta'(x)}{9} $.

Can anybody help me? Thanks!

Edit: Given that only $x_i \in \mathbb{R} $ can be considered, we have the only point $x_1 = 0$ and we write:

$\delta(x^3 + 3x) = \frac{\delta(x)}{9}$, then we differentiate: $\delta'(x^3+3x) = \frac{\delta'(x)}{9(3x^2 + 1)} $.

Now using the fact that $g(x)\delta(x-a) = g(a)\delta(x-a) $ where $a = 0$ and $g(x) = \frac{1}{3x^2+1} $, we get $\frac{\delta'(x)}{9} $.

Rael
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    The complex zeros don't count : $\delta'(f(x))$ is a distribution on $\mathbb{R}$. What do you obtain from $(\delta(f(x))' = \delta'(f(x))f'(x)$ ? – reuns Aug 29 '17 at 11:18
  • True! I did not pay attention to that detail... If only $x = 0$ is a root than I get the result I want... thanks – Rael Aug 29 '17 at 11:28
  • You didn't answer to my question. How do you simplify $\delta'(x^3+x)$ ? – reuns Aug 29 '17 at 11:53
  • I edited to how I did it. – Rael Aug 29 '17 at 22:21
  • If you meant $f(x) = x^3+3x$ then $\delta(f(x)) = \frac{\delta(x)}{3}$. Differentiating we obtain $$\delta'(f(x))f'(x) = \frac{\delta'(x)}{3}, \qquad \int_{-\infty}^\infty \delta'(f(x)) \phi(x)dx = \int_{-\infty}^\infty \delta'(x)\frac{\phi(x)}{3 f'(x)}dx= \ldots$$ – reuns Aug 29 '17 at 23:01
  • All this being justified by a rigorous definition of $\delta(f(x))$ such as $$\mathcal{G}(x) = e^{-\pi x^2}, \qquad \int_{-\infty}^\infty\delta(x)\phi(x)dx =\lim_{n \to \infty} \int_{-\infty}^\infty n \mathcal{G}(nx )\phi(x)dx \ \int_{-\infty}^\infty\delta(f(x))\phi(x)dx =\lim_{n \to \infty} \int_{-\infty}^\infty n \mathcal{G}(nf(x))\phi(x)dx, \\int_{-\infty}^\infty\delta'(f(x))\phi(x)dx =\lim_{n \to \infty} \int_{-\infty}^\infty n^2 \mathcal{G}'(nf(x)) \phi(x)dx$$ – reuns Aug 29 '17 at 23:06
  • https://math.stackexchange.com/questions/1073579/composition-of-a-dirac-delta-and-a-function-in-higher-dimensions – md2perpe Sep 03 '17 at 14:19

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