what i do to do this?
sage: var('t w')
sage: f(t) = sin(w*t)
t |--> sin(t*w)
sage: w = 2
sage: f
t |--> sin(2*w)
Without doing f(w=2)!!! imagine that the function is f->f(a,b,c,d,f....,t) like doing this:
...
...
sage: aw1=aw1(m1=1,m2=0.5,l1=1,l2=0.5,g=9.8)
sage: aw2=aw2(m1=1,m2=0.5,l1=1,l2=0.5,g=9.8)
sage: aw4=aw4(m1=1,m2=0.5,l1=1,l2=0.5,g=9.8)
...
...
etc
The w in sin(t*w) does not refer to whatever the object with identifier w in the current scope. There isn't a way to get Python to work like this. You need to do something like:
sage: var('t, w')
(t, w)
sage: params = {w: 2}
sage: f(a,b,c,d,e,f,g,t) = sin(t*w)
sage: f.subs(params)
(a, b, c, d, e, f, g, t) |--> sin(2*t)
You just have to define the replacements you want once in a dictionary. You still have to call subs once for every function you want to change.
I've updated my answer. Could you explain the annoying part? Do you mean having to do the ".function(*f.arguments())" part?
I think I'm just explaining Mike's answer a little, but I think that you should be able to define a dictionary
sage: params = {m1:1,m2:0.5,l1:1,l2:0.5,g:9.8}
and then do f.subs(params) as above. In his example,
var('t, w')
params = {w: 2,d:1,c:1}
f(a,b,c,d,e,f,g,t) = sin(t*w*d-c)
g=f.subs(params)
g
should do it. Mike, is that extra thing necessary, given that the arguments are the same in both cases?
Actually, you don't need. However, if you did `f(w=2) then that doesn't return a CallableSymbolicExpression, but `f.subs(params)` does return one. I'll update my answer.
Asked: 2010-08-25 19:04:25 +0100
Seen: 970 times
Last updated: Oct 23 '12
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