Re: How to substitute a function?
- To: mathgroup at smc.vnet.net
- Subject: [mg94102] Re: How to substitute a function?
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Fri, 5 Dec 2008 05:28:46 -0500 (EST)
- Organization: Uni Leipzig
- References: <gh8hmh$r2t$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
expr=a*y[x]^2 + y'[x];
and
expr /. Thread[{#, D[#, x]} & /@ (y[x] -> Exp[-x]*f[x])]
work fine. Where is the problem ?
Regards
Jens
Alexei Boulbitch 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=EF=BF=BD[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 substitutio=
n
> everywhere?
>
> Thank you in advance, Alexei
>