Re: Backwards?
- To: mathgroup at smc.vnet.net
- Subject: [mg3496] Re: Backwards?
- From: wagner at bullwinkle.cs.Colorado.EDU (Dave Wagner)
- Date: Mon, 18 Mar 1996 00:57:13 -0500
- Organization: University of Colorado, Boulder
- Sender: owner-wri-mathgroup at wolfram.com
In article <4ibbb7$fep at ralph.vnet.net>,
Artur KUO <artur at galilei.pi.tu-berlin.de> wrote:
>I would like to ask a question please.
>How can I let Mathematica do this:
>
>1) c[x,y]*a[x]*D[b[x],x]+c[x,y]*b[x]*D[a[x],x]+d[x,y]
> --->c[x,y]*D[a[x]*b[x],x]+d[x,y] ?
> It's backward-chainrule.
This exact question came up in a Mathematica workshop that I
taught recently. What you want to do is back-substitution.
Unfortunately, rule substitution using "/." is completely syntactic,
not algebraic. (But see AlgebraicRules in the Mma book). You can
get around this problem by a tricky use of Solve. Here's a simple
example:
(Local2) In[1]:=
g[x_] := 1/(1+Exp[x])
(Local2) In[2]:=
g'[x]
(Local2) Out[2]=
x
E
-(---------)
x 2
(1 + E )
We'd like to rewrite this expression in terms of g[x]. Here's how to do it:
(Local2) In[3]:=
Solve[{dummy==%, gx == g[x]}, dummy, x]
(Local2) Out[3]=
{{dummy -> (-1 + gx) gx}}
The second argument to Solve gives the name of the variable(s) to solve for,
and the little-used third argument gives the names of the variable(s) to
attempt to eliminate.
With a bit of fooling around you can probably apply this technique
to your own problem
>2) If I let u^2+v^2:=1, how cann I do:
>
> a*u^2+a*v^2==a*1==a ?
(Local2) In[4]:=
Solve[{dummy==a*u^2 + a*v^2, u^2+v^2==1}, dummy]
(Local2) Out[4]=
{{dummy -> a}}
Dave Wagner
Principia Consulting
(303) 786-8371
dbwagner at princon.com
http://www.princon.com/princon
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