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MathGroup Archive 2006

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Re: Patterns_ to define linear operators?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71407] Re: Patterns_ to define linear operators?
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Fri, 17 Nov 2006 04:30:30 -0500 (EST)
  • References: <ejgvpf$8tm$1@smc.vnet.net>

Something like?

myruleoflin = FourierTransform[(a_.)*(f_), x_, s_] :>
a*FourierTransform[f, x, s] /; FreeQ[a, x]
FourierTransform[(a_.)*(f_), x_, s_] :> a*FourierTransform[f, x, s] /;
FreeQ[a, x]

FourierTransform[a*f[x], x, s] /. myruleoflin
a*FourierTransform[f[x], x, s]

FourierTransform[f[x], x, s] /. myruleoflin
FourierTransform[f[x],x,s]

FourierTransform[a x f[x], x, s] /. myruleoflin
a FourierTransform[x f[x],x,s]

Keep in mind that the rule is implemented in Mathematica. E.g.

FourierTransform[a*(1/x)*Sin[x], x, s]
(1/2)*a*Sqrt[Pi/2]*(Sign[1 - s] + Sign[1 + s])

Regards
Dimitris

W. Craig Carter wrote:
> 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


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