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Nov 14, 2011 at 9:12 comment added B. Bischof I see now what you meant, I was thinking about this in the wrong way.
Nov 14, 2011 at 7:40 comment added zroslav @Bischof: No, $F(x)$ is assumed to be an arbitrary function. This expression was seen by me in my mind: for example $(F(x)d/dx)^2=F^2d^2/dx^2+FF'd/dx$. The general case can be proved by induction by $n$
Nov 14, 2011 at 7:33 vote accept zroslav
Nov 14, 2011 at 7:08 answer added Gjergji Zaimi timeline score: 9
Nov 14, 2011 at 5:04 comment added B. Bischof Are you assuming F(x) are polynomials? Also, where did you see this expression, it would help direct me to what you are looking for. If you don't assume polys, I'd expect the commutator to suck pretty bad if you take the Taylor polynomial approach then assume all of F's derivs exist(reasonable in this setting) but then idk how to interpret the infinite sums in the commutator.
Nov 13, 2011 at 16:58 history asked zroslav CC BY-SA 3.0