Maxima evaluation is one-ply (as I would call it--I don't know the correct
term).
I think you need to use ev:
...
omega:1/(sqrt(tau[1]*tau[2]));
abs(h(s));
ev(%);
Simple example of the one-ply evaluation rule:
(%i1) (a:b, b : c, c : d)$
(%i2) a;
(%o2) b
(%i3) b;
(%o3) c
(%i4) c;
(%o4) d
Notice that a does not evaluate to d.
--Barton
maxima-bounces at math.utexas.edu wrote on 08/13/2010 09:51:53 AM:
> [image removed]
>
> [Maxima] variable substitution unsuccessfull [newbie]
>
> BVBA NuKey Music
>
> to:
>
> maxima
>
> 08/13/2010 09:51 AM
>
> Sent by:
>
> maxima-bounces at math.utexas.edu
>
> When using the following sequence of commands, the sixth line
> doesn't have the value of omega substituted as can be seen in the
> output of line 7 which still displays omega and not the value I want
> it to have in line 6. Can anyone tell me what I'm doing wrong and
> how to fix it?
>
> h(s):=tau[1]*s/(1+(tau[1]+tau[2]+r[1]*c[2])*s+tau[1]*tau[2]*s^2);
> s:%i*omega;
> abs(h(s))^2;
> diff(%,omega);
> solve(%,omega);
> omega:1/(sqrt(tau[1]*tau[2]));
> abs(h(s));
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