Re: Re: replacing expressions
- To: mathgroup at smc.vnet.net
- Subject: [mg78590] Re: [mg78564] Re: replacing expressions
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Thu, 5 Jul 2007 03:56:44 -0400 (EDT)
- Reply-to: hanlonr at cox.net
f[t_] = a^2 (1 + Csc[(1 + 1/c^2) t]
Sin[t] (-2 Cos[t/c^2] +
Csc[(1 + 1/c^2) t] Sin[t]));
g[t_] = -Sin[(1 + 1/c^2) t]/
(a Sin[t/c^2]);
f[t] /. Solve[x == g[t], a][[1]] //
FullSimplify
1/x^2
f[t] == 1/g[t]^2 // FullSimplify
True
Bob Hanlon
---- Constantinos <physicist_gr at yahoo.gr> wrote:
>
> "dimitris" <dimmechan at yahoo.com> wrote in message
> news:f6da0c$2mk$1 at smc.vnet.net...
> >
> > Constantinos :
> >> Hi,
> >>
> >> I got a 'complicated' expression after FullSimplify which is a function
> >> of t
> >> variable. Let's say f[t].
> >> We know that x=g[t]. After some algebra (by hand) you find that
> >> f[t]=1/xo^2.
> >> Is there any way to do such simplifications by Mathematica?
> >>
> >> Thanks
> >>
> >> Constantinos
> >
> > Geia hear Kosta.
> >
> > Can you show us exactly your expression?
> > It will be very helpful.
> >
> > Cheers
> > Dimitris
>
>
> The expression f[t] is
>
> f[t] = a^2 (1 + Csc[(1 + 1/c^2) t] Sin[t] (-2 Cos[t/c^2] + Csc[(1 + 1/c^2)
> t] Sin[t]))
>
> and the expression g[t]=x is
>
> x=g[t] = -Sin[(1 + 1/c^2) t]/(a Sin[t/c^2]).
>
> When you do FullSimplify[f[t]*x^2] you get 1 which means that f[t]=1/x^2.
>
> Thanks
> Constantinos
>