Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2012

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: function manipulation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg127831] Re: function manipulation
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Sat, 25 Aug 2012 04:26:01 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: l-mathgroup@wolfram.com
  • Delivered-to: mathgroup-newout@smc.vnet.net
  • Delivered-to: mathgroup-newsend@smc.vnet.net
  • References: <20120824005115.06CE5687C@smc.vnet.net>

gl2 = Module[
   {eqn, z = Subscript[\[Tau], xy]''[x], coef},
   eqn = Subscript[g,
      K] (Subscript[e, 1] Subscript[t, 1] +
        Subscript[e, 2] Subscript[t, 2]) Subscript[\[Tau], xy][x] ==
     Subscript[e, 1] Subscript[e, 2] Subscript[t, 1] Subscript[t, 2]
       Subscript[t, K] (Subscript[\[Tau], xy]^\[Prime]\[Prime])[x];
   (z /. Solve[eqn, z][[1]]) - z == 0];

{gl3, repl} = Module[
   {coef = Coefficient[gl2[[1]], Subscript[\[Tau], xy][x]]},
   {gl2 /. Numerator[coef] ->
      \[Omega]^2*Denominator[coef],
    \[Omega] -> Sqrt[coef]}];

gl3

\[Omega]^2*Subscript[\[Tau], xy][x] -
     Derivative[2][Subscript[
           \[Tau], xy]][x] == 0

(gl3 /. repl) === gl2

True


Bob Hanlon


On Fri, Aug 24, 2012 at 6:39 AM, Andreas Talmon l'Arm=E9e
<talmon at fsm.tu-darmstadt.de> wrote:
> Hi
>
> Thanks, this did exactly the job. But unfortunately I got another issue. I
> wanted to replace the fraction by the parameter omega^2 in gl3. So far it
> worked well. But then I tried to define omega for further calculations. At
> the end it did not matter if I solved gl2 or gl3 with DSolve. omega^2 was
> always replaced by its definition. Can I force mathematica to use the much
> shorter parameter omega^2 for a clear arrangement. I attached the file where
> I tried to do the calculation :
>
> Best regards Andreas Talmon
>
>
>



  • Prev by Date: Re: Ambiguity of "Plot"
  • Next by Date: Re: Find the value of x & y
  • Previous by thread: Re: function manipulation
  • Next by thread: Getting the Derivative of an HoldForm Expression