Re: Re: Replace in operators
- To: mathgroup at smc.vnet.net
- Subject: [mg102987] Re: [mg102922] Re: [mg102869] Replace in operators
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 3 Sep 2009 07:10:38 -0400 (EDT)
- References: <200909010750.DAA18607@smc.vnet.net> <200909020803.EAA03297@smc.vnet.net> <1251889441.21165.1332823679@webmail.messagingengine.com>
In this particular case it makes no significant difference. I don't
even remember why I used :>, probably it was just to make the answers
a little more varied ;-)
Andrzej Kozlowski
On 2 Sep 2009, at 13:04, Elton Kurt TeKolste wrote:
> Andrzej
>
> You and I have similar solutions for this problem in your answer #2
>
> f[x] + D[f[x], x] /. {f[x] :> g[x], Derivative[1][f][x] :>
> Derivative[1][g][x]}
>
> except that I used the static replace (->) rather than the dynamic
> replace (:>).
>
> I suspect that I have not fully understood the distinction between ->
> and :> and, in particular, what advantage or difference :> provides in
> this case over ->.
>
> Why did you select :> over -> in this example?
>
> Kurt
>
> On Wed, 02 Sep 2009 04:03 -0400, "Andrzej Kozlowski" <akoz at mimuw.edu.pl
> >
> wrote:
>> I think, you can do on of three things:
>>
>> 1.
>> Unevaluated[f[x] + D[f[x], x]] /. f[x] -> g[x]
>>
>> Derivative[1][g][x] + g[x]
>>
>> or
>>
>> 2.
>>
>> f[x] + D[f[x], x] /. {f[x] :> g[x], Derivative[1][f][x] :>
>> Derivative[1][g][x]}
>>
>> Derivative[1][g][x] + g[x]
>>
>> 3.
>>
>> f[x] + D[f[x], x] /. f -> g
>>
>> Derivative[1][g][x] + g[x]
>>
>> (There are also some variants of 1 and 2 ).
>>
>> From what you say, it seems that the first choice (wrapping
>> Unevaluated around the input) is the most suitable.
>>
>> Andrzej Kozlowski
>>
>>
>>
>> On 1 Sep 2009, at 09:50, did wrote:
>>
>>> I can't figure out how to force Mathematica to replace
>>> f[x] by g[x] in expressions involving operators.
>>> For example:
>>>
>>> f[x] + D[f[x],x] /. f[x] -> g[x]
>>>
>>> f is not replaced in the derivative. I found
>>> somewhere in the manual that the replacement
>>> does not work on operators, but then it does
>>> not indicate how to do it.
>>>
>>> In my real problem, f[x]->g[x] is a (long)
>>> list of complicated transformations resulting
>>> from the resolution of many equations.
>>> Thus, changing f[x]->g[x] by f->g is not
>>> an option.
>>>
>>> f[x] + D[f[x],x] is actually a complicated
>>> expression involving many operators.
>>>
>>> There are no ways I can do the
>>> substitution by hand, bit by bit.
>>>
>>> Any suggestions ?
>>> Thanks
>>>
>>> PS: it may help to precise the context.
>>> I'm trying to solve a system of PDEs by
>>> a small parameter expansion. Solving the
>>> system at order N, I want to re-inject the
>>> results into the equations at order N+1.
>>>
>>
>>
> Regards,
> Kurt Tekolste
>
- References:
- Replace in operators
- From: did <didier.oslo@hotmail.com>
- Re: Replace in operators
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Replace in operators