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Re: How to substitute a function?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg94131] Re: How to substitute a function?
  • From: Raffy <raffy at mac.com>
  • Date: Fri, 5 Dec 2008 05:34:00 -0500 (EST)
  • References: <gh8hmh$r2t$1@smc.vnet.net>

On Dec 4, 4:16 am, Alexei Boulbitch <Alexei.Boulbi... at iee.lu> wrote:
> Dear crew,
>
> I faced a difficulty when trying to substitute a newly represented
> function into an expression containing a sum of differential and
> algebraic terms. The difficulty is namely, that Mathematica 6
> substitutes the new representation into the algebraic, but not into the
> differential part. For example:
>
> (* This is the definition of a simple example of such an expression *)
>
> expr = \!\(
> \*SubscriptBox[\(\[PartialD]\), \(x\)]\ \(y[x]\)\) + a*y[x]^2;
>
> (* Here I substitute a new representation of the function  into the \
> above expression *)
> (* What I expect to get is the following expression: -Exp[-x]*f[x] + \
> Exp[-x]*f=B4[x] + a \[ExponentialE]^(-2 x) f[x]^2  *)
> (* Instead I get something else. Please have a look and check:  *)
> expr /. y[x] -> Exp[-x]*f[x]
>
> (* This is another approach I could think of with even worse result. *)
> \
>
> (* Note however, that within this approach a miracle happened: once \
> it *)
> (* worked as expected (i.e. the substitution of both terms has been \
> performed as desired, but only once *)
>
> FullSimplify[expr, y[x] == Exp[-x]*f[x]]
>
> Do you have idea of how to instruct Mathematica to make the substitution
> everywhere?
>
> Thank you in advance, Alexei
>
> --
> Alexei Boulbitch, Dr., Habil.
> Senior Scientist
>
> IEE S.A.
> ZAE Weiergewan
> 11, rue Edmond Reuter
> L-5326 Contern
> Luxembourg
>
> Phone: +352 2454 2566
> Fax:   +352 2454 3566
>
> Website:www.iee.lu
>
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Apply FullForm to your initial expression, you notice that: D[y[x],x]
evaluates to Derivative[1][y][x].

Wrap your initial expression in Hold, then do your replacement, and
follow with a ReleaseHold.




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