Relates issues:

In my old notes about Lie groups and/or operator calculus, I've encountered the following formulas: $$ e^{\lambda\,x^2\,\frac{d}{dx}}\,f(x) = f\left(\frac{x}{1-\lambda\,x}\right) \\ e^{\lambda\,\frac{1}{x}\,\frac{d}{dx}}\,f(x) = f\left(\sqrt{x^2+2\lambda}\right) \\ e^{\lambda\,x^3\,\frac{d}{dx}}\,f(x) = f\left(\frac{x}{\sqrt{1-2\lambda\,x^2}}\right) $$ I know how to derive the first one, but have no idea how I did the two others at that time.
Please help to refresh my memory.


Let $y = \frac{1}{2}x^2$. Then $\frac{1}{x}\frac{d}{dx} = \frac{d}{dy}$. Also notice that $f(x) = f(\sqrt{2y})$. Thus

$$ e^{\frac{\lambda}{x}\frac{d}{dx}} f(x) = e^{\lambda \frac{d}{dy}}f(\sqrt{2y}) = f(\sqrt{2y+2\lambda}) = f(\sqrt{x^2+2\lambda}).$$

Second one can be solved in a similar way.

  • $\begingroup$ Please notice my generalization of your answer. Hope it is correct. Thanks. $\endgroup$ – Han de Bruijn Oct 5 '15 at 16:07
  • $\begingroup$ Maybe I'm picky, but if $y = \frac12 x^2$, then you have the two solutions $x = \pm \sqrt{2y}$. $\endgroup$ – Wauzl Aug 17 '17 at 13:47

The aim of this answer is to make the question self-contained, and propose a generalization of the answer by Sangchul Lee. The basic formula to be employed is (anticipating with $y$ instead of $x$ and $g$ instead of $f$): $$ e^{\lambda\frac{d}{dy}}\,g(y) = g(y+\lambda) $$ Which is easily proved with Taylor series expansions for differential operators and functions: $$ e^{\lambda\frac{d}{dy}}\,g(y) = \sum_{k=0}^\infty \frac{1}{k!}\left(\lambda\,\frac{d}{dy}\right)^k g(y) = \sum_{k=0}^\infty \frac{g^{(k)}(y)}{k!}\lambda^k = g(y+\lambda) $$ Generalization. Substitute $y = \phi(x)$ , then: $$ \frac{d}{dy} = \frac{d}{d\phi(x)} = \frac{dx}{d\phi(x)}\frac{d}{dx} = \frac{1}{\phi'(x)}\frac{d}{dx} $$ Also notice that $g(y) = g(\phi(x)) = f(x) = f(\phi^{-1}(y))$. Thus: $$ e^{\lambda\frac{d}{dy}}\,g(y) = e^{\frac{\lambda}{\phi'(x)}\frac{d}{dx}} f(x) = g(y+\lambda) = f(\phi^{-1}(\phi(x)+\lambda)) $$ Third formula in the question as a specialization of the above generalization: $$ 1/\phi'(x) = x^3 \quad \Longrightarrow \quad \phi(x) = \int \frac{dx}{x^3} = \frac{-1}{2x^2} = y \quad \Longrightarrow \quad x = \sqrt{\frac{-1}{2y}} \quad \Longrightarrow \\ \phi^{-1}(x) = \sqrt{\frac{-1}{2x}} \quad \Longrightarrow \quad \phi^{-1}(\phi(x)+\lambda) = \sqrt{\frac{-1}{2(-1/(2x^2)+\lambda)}} = \frac{x}{\sqrt{1-2\lambda x^2}} $$ Hence it follows that: $$ e^{\lambda\,x^3\,\frac{d}{dx}}\,f(x) = f\left(\frac{x}{\sqrt{1-2\lambda\,x^2}}\right) $$ But, with the general formula, another old puzzle is now solved completely ! Let $g$ be an arbitrary (real-valued, neat) function. Consider a Lie series as in Exponential of a function times derivative : $$ e^{g(x)\partial}f(x) \qquad \mbox{with} \quad \partial = \frac{d}{dx} $$ First solve the differential equation: $$ g(x) = \frac{1}{\phi'(x)} \quad \Longrightarrow \quad \phi(x) = \int \frac{dx}{g(x)} $$ Then we have: $$ \boxed{ e^{g(x)\partial} f(x) = f(\phi^{-1}(\phi(x)+1)) } $$

  • $\begingroup$ This sounds okay for me. :) $\endgroup$ – Sangchul Lee Oct 6 '15 at 16:17

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