In an answer to this question a user writes, $\frac{|B(x,2\delta)|}{|B(y,\delta)|} = 2^n$. I am not sure how the user arrived at this. Can someone kindly help me ? Thanks
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1$B(y,\delta)$ is the ball in $\mathbb R^n$ of radius $\delta$, centered at $y$, right? – kimchi lover Jun 10 '21 at 15:54
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1In that answer is link to https://en.wikipedia.org/wiki/Volume_of_an_n-ball#The_volume from where is $2^n$ clear. – zkutch Jun 10 '21 at 16:19
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@zkutch That was added later. Thanks though. – gaufler Jun 10 '21 at 16:51
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$$ |B(x,2\delta)|=\int_{x_1^2+\ldots+x_n^2\leqslant (2\delta)^{2}}dx=\int_{\left(\frac{x_1}{2}\right)^2+\ldots\left(\frac{x_n}{2}\right)^2\leqslant\delta}dx $$ Now substitute $u_i=\frac{x_i}{2}$, then $\det{\rm Jac}((u_i)_{1\leqslant i\leqslant n})=\det{\rm diag}\left(\frac{1}{2},\ldots,\frac{1}{2}\right)=\frac{1}{2^n}$, therefore $$ |B(x,2\delta)|=2^n\int_{u_1^2+\ldots+u_n^2\leqslant\delta^2}du=2^n|B(x,\delta)| $$ More generally, for any linear map $T$ and measurable set $A\subset\mathbb{R}^n$, $|T(A)|=|\det(T)||A|$, this is the particular case where $T(x)=2x$.
Tuvasbien
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