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I want to express $\delta^n(\mathbf{f}(x))$ as a sum of delta functions shifted to roots of $f$ (similar to this) where function $f:\mathbb{R}\rightarrow\mathbb{R}^n$ and $f(x^\ast_j) = [0, \dots, 0]^\top, \ \forall j=1,\dots,n_\text{roots}$. I tried deriving something by writing down the integral as

$$ \int_{\mathbb{R}^n}\delta^n(\mathbf{f}(x))g(x)(dx)^n = \sum_{j = 1}^{n_\text{roots}}\int_{x^\ast_j - \epsilon}^{x^\ast_j + \epsilon}\delta^n(\mathbf{f}(x))g(x)(dx)^n $$ for some $\epsilon$ around $x^\ast_j$ and then substituting $\mathbf{u} = \mathbf{f}(x)$ to get $$ \sum_{j = 1}^{n_\text{roots}}\int_{x^\ast_j - \epsilon}^{x^\ast_j + \epsilon}\delta^n(\mathbf{f}(x))g(x)(dx)^n = \sum_{j = 1}^{n_\text{roots}}\int_{\mathbf{f}(x^\ast_j - \epsilon)}^{\mathbf{f}(x^\ast_j + \epsilon)}\delta^n(\mathbf{u})g(\mathbf{f}^{-1}(u))dV_{u} $$ where $dV_{u}$ is the new volume element. I am not able to express this volume element in terms of $x$ and $\nabla f$. Any help will be greatly appreciated.

I feel that the solution will probably be of the form $$ \delta^n(\mathbf{f}(x)) = \sum_{j}\frac{\delta(x - x^\ast_j)}{\lvert\prod_{k} f^{'}_k(x^\ast_j)\rvert} $$ but I am not sure.

apsri
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  • You have written your integral with $f(x)$ where the integral is over $x\in \mathbb{R}^n$, but also said that $f:\mathbb{R}\to \mathbb{R}^n$. Which is it? Is the argument of $f$ a single variable, or $n$ variables? – Joshua Tilley Oct 05 '23 at 23:41
  • Hi @JoshuaTilley, $x\in\mathbb{R}$. I have written $(dx)^n$ to replicate the volume element we will have in an $n$-dimensional space (Since we have an $n$-dimensional $\delta$-functions). – apsri Oct 08 '23 at 01:20

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