Dear Scott,
Thanks for posting.
On Oct 15, 6:46 pm, Scott <spectre...@gmail.com> wrote:
> I know this is really basic, but for some reason I cannot figure this
> out and my searching (here, sage website, google, etc) has not yielded
> me any results! I just started using sage, literally, yesterday.
>
> I am using solve() to solve an equation for q1 in terms of q2. The
> solution of this is saved, but when I print it, it prints as an exact
> answer, but I want it in decimal/approximate form. I have tried using
> the n() function, but I get an error when trying it.
>
> Here is an example:
> --------------------------------------------------
> q1, q2 = var('q1, q2')
> r = 0.60
> t1 = 0.005
> t2 = 0.007
> d = 2.0
> A1 = (1/2)*pi*r*r
> A2 = (1/2)*d*(2*r)
> theta1 = 1/(2*A1)*(q1*2*pi*r/t1+(q1-q2)*2*r/t1)
> theta2 = 1/(2*A2)*(q2*d/t2+q2*sqrt(d^2+(2*r)^2)/t2+(q2-q1)*2*r/t1)
> sol_q1 = solve([theta1==theta2], q1)
> print sol_q1
> print sol_q1[0].substitute(q2=1).right().n()
> --------------------------------------------------
>
> When this is run, the result for sol_q1 is: q1 = (223317 pi + 416000)/
> ((223317 pi + 416000) q2)*q2
> The last line does work to give me the desired value (0.58239....),
> but this is definitely not an ideal way of getting it! And of course
> this wouldn't work for a more complicated result (since I use q2=1).
>
Why not? You should be able to put whatever you want in for q2.
Maybe I am misunderstanding the question.
> I've tried many other combinations of sol_q1.n(), sol_q1[0].n(), etc,
All these will do nothing if you haven't specified q2, because you are
trying to approximate something which has no approximation; you have
to substitute all free variables before approximating. I suppose it
would be possible to make Sage do everything on coefficients, but I
don't think that is the usual functionality. Your solution is
actually pretty good; even doing .substitute(q2=1.0) wouldn't help,
because we prefer to leave pi unapproximated until necessary. I do
agree that it's annoying! But I'm not sure how to make it less so,
short of writing another function that evaluated a symbolic equation
at a given point - without checking that you had something like
sage: solve([sin(x-y)==x*y],y)
[y == -sin(-x+y)/x]
where it would cause trouble.
Hope this helps!
- kcrisman
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