RE: Patterns_ to define linear operators?

*To*: mathgroup at smc.vnet.net*Subject*: [mg71410] RE: [mg71385] Patterns_ to define linear operators?*From*: "David Park" <djmp at earthlink.net>*Date*: Fri, 17 Nov 2006 04:30:38 -0500 (EST)

Craig, How about this? breakoutFT[x_][expr_] := expr /. FourierTransform[a__?(FreeQ[#, x] &)b_., x, k_] -> a FourierTransform[b , x, k] FourierTransform[a f[x], x, k] % // breakoutFT[x] FourierTransform[a f[x], x, k] a*Sqrt[2*Pi]*DiracDelta[k]*f[x] FourierTransform[a x f[x], x, k] % // breakoutFT[x] FourierTransform[a x f[x], x, k] a*Sqrt[2*Pi]*x*DiracDelta[k]*f[x] David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: W. Craig Carter [mailto:ccarter at mit.edu] Hello, I'd like to implement a rule that will factor out constants from linear operators such as FourierTransform; i.e., FourierTransform[a*f[x], x, k] to a*FourierTransform[f[x],x,k] Here is a method that seems sensible at first, but is potentially dangerous: MyRule1 = FourierTransform[a_ f_[y_], x_, k_] :-> a FourierTransform[f[y],x,k] (which would fail if a was not free of x) An improvement is: MyRule2 = FourierTransform[a_?(FreeQ[#,x]&), f_[y_], x_, k_] :-> a FourierTransform[f[y],x,k] (which would fail if a were a product, say b x) Does anybody have advice on a robust way to implement this rule? Thanks, Craig Carter