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let be the dilation operator $ x \frac{d}{dx} $ i know this operator is invariant under the change $ y=ax$ for any positive 'a' real number

however let be the change $ y= \frac{-1}{x}$ then the operator $x \frac{d}{dx}= y \frac{d}{dy} $ still remains invariant .. is there a mathematical reason to see why this happens ?? , given the group of dilations in one dimension, is there a bigger group containing it ??

Jose Garcia
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    Why only positive reals? Also, it seems to me that $y\frac{d}{dy}=\color{Red}-x\frac{d}{dx}$. – anon Apr 29 '12 at 09:56
  • BTW setting $x=g(y)$ looks like it leads to a simple DE. – anon Apr 29 '12 at 10:02
  • To sum up, (i) $y=-1/x$ does NOT yield $xd/dx=yd/dy$, (ii) solving $xd/dx=yd/dy$ yields $y=ax$ for some nonzero $a$. – Did May 12 '12 at 07:57

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