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Question: Show there exists a unique bijection $f:\mathbb R^+\to\mathbb R^+$ such that $\frac{d}{dx}f(x)=f^{-1}(x)$, where the right-hand side is the functional inverse.

I figured I would start by finding a trivial example of existence, but 1) I can't think of one and 2) I don't know how I'd prove uniqueness.

user156190
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  • Try differentiating $f\circ f$. – Marra Jun 28 '14 at 01:52
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    This is related: http://math.stackexchange.com/questions/279332/inverse-of-a-bijection-f-is-equal-to-its-derivative Someone suggested $f(x) = \phi^{1-\phi}x^{\phi}$ where $\phi = \tfrac{1+\sqrt{5}}{2}$ is the golden ratio. – JimmyK4542 Jun 28 '14 at 01:53

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For an example, try $f$ of the form $f(x) = c x^p$. See what $c$ and $p$ must be to make this work.

Robert Israel
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