# I Generalizing the translation operator

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1. Jan 25, 2017

### BlackHole213

If I have the operator, $e^{a\partial_p}$ acting on $f(p)$, I know that $$e^{a\partial_p}f(p)=f(p+a)\,.$$
If I have $e^{a\partial_p^2}f(p)$, this is just the Weierstrass transform of $f(p)$. However, what happens if I have a general operator, $e^{g(p)\partial_p}$ or $e^{g(p)\partial_p^2}$. How would I know what $e^{g(p)\partial_p}$ or $e^{g(p)\partial_p^2}$ does to $f(p)$?

2. Jan 29, 2017

### DCN

Write it out in a Taylor series.