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Say I have a Radon transform of $f:\mathbb R^2 \to \mathbb R^+$ and denote it by $\mathcal Rf$. I now read that the Radon-Transform of $f_a(x,y):=a^2f(ax,ay)$ is then given by $$\mathcal Rf_a(t,\theta)=a\mathcal Rf(at,\theta), \quad t\in \mathbb R, \theta \in [0,\pi].$$

How come? If we set $x\cos\theta + y\sin\theta = t $, then from $ax\cos\theta + ay\sin\theta = t \iff x\cos\theta+y\sin\theta = \frac{t}{a}$, wouldn't we get

$$\mathcal Rf_a(t,\theta)=a^2\mathcal Rf(\frac{t}{a},\theta)$$ or what property that I can make us of am I missing?

user26872
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Tesla
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1 Answers1

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$\def\CR{\mathcal{R}} \def\th{\theta} \def\RR{\mathbb{R}}$We have \begin{align*} \CR f_a(t,\th) &= \int_\RR a^2 f(a(t\cos\th-s\sin\th),a(t\sin\th+s\cos\th))ds \\ &= a\int_\RR f(at\cos\th-as\sin\th,at\sin\th+as\cos\th)a ds \\ &= a\int_\RR f(at\cos\th-u\sin\th,at\sin\th+u\cos\th)du & \textrm{let }u=a s \\ &= a\CR f(at,\th). \end{align*} In terms of derived properties of the Radon transform, recall that if $h(x,y) = f(ax,ay)$ that $\CR h(t,\th) = \frac{1}{a}\CR f(at,\th)$. Then \begin{align*} \CR f_a(t,\th) &= a^2 \CR h(t,\th) = a^2\frac{1}{a}\CR f(at,\th) = a \CR f(at,\th). \end{align*} We assume in the work above, as is implicit in the question statement, that $a>0$.

user26872
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