Have I found the only solution to this curious-looking DE? $f[f(x+1)]=\prod_{i=0}^{\infty} f\left[ \frac{f^{(k)}(x)}{k!} \right]$?

53 Views Asked by At

This was just a curiosity I discovered while fooling around with the product function: One solution to the (admittedly weird looking) DE

\begin{equation*} f[f(x+1)]=\prod_{k=0}^{\infty} f\left[ \frac{f^{(k)}(x)}{k!} \right] \end{equation*}

is the simple (I've hidden it in case you want to try to solve it yourself):

$f(x)=e^{\lambda x}$, since \begin{align} \prod_{i=0}^{\infty}\exp \left(\frac{\lambda}{k!} \frac{d^{(k)}}{dx} e^{\lambda x}\right)&=\exp \left( \lambda \left[ \frac{e^{\lambda x}}{0!}+\lambda \frac{e^{\lambda x}}{1!} +... \right] \right) \\ &=\exp \left( \lambda e^\lambda e^{\lambda x} \right) \\ & =\exp \left( \lambda e^{\lambda (x+1)} \right) \\ &=f[f(x+1)] \end{align}


$Q$: Is this the only (non-trivial) solution? For instance

$f(x)=\mu e^{\lambda x}$

didn't work because the $\mu$ made the product on the RHS blow up.

$Q2$: This is more on the soft side, but do you know of any other "similar" DEs (that have solutions, preferably); this could be either DEs where the solution takes itself as a variable (do such equations have a common name?) or DEs involving the product or summation function?

Thanks!