Re: How to substitute a function?
- To: mathgroup at smc.vnet.net
- Subject: [mg94116] Re: [mg94082] How to substitute a function?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 5 Dec 2008 05:31:17 -0500 (EST)
- Reply-to: hanlonr at cox.net
expr = a*y[x]^2 + y'[x]
a*y[x]^2 + Derivative[1][y][x]
expr /. NestList[D[#, x] &,
y[x] -> Exp[-x]*f[x], 1]
(a*f[x]^2)/E^(2*x) +
Derivative[1][f][x]/E^x - f[x]/E^x
Note: For higher order derivatives just increase the last number in the Nes=
tList
% // Simplify
(a*f[x]^2 + E^x*Derivative[1][f][
x] - E^x*f[x])/E^(2*x)
Bob Hanlon
---- Alexei Boulbitch <Alexei.Boulbitch 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=C2=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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Bob Hanlon