Given the following:
a = var('a')
b = var('b')
c0 = var('c0')
c1 = var('c1')
eq1 = a==exp(b)
eq2 = b==4*c0+2*c1
solve([eq1, eq2], a, b, c0, c1, solution_dict=True)
[{a: e^r5, b: r5, c0: r6, c1: 1/2*r5 - 2*r6}]
I'd like extract a and have something similar to what I would have if I did: view(a==exp(4*c0+2*c1)).
You want the equations to hold for all values of c0, c1, so you can omit them as variables to solve for:
sage: solve([eq1, eq2], a, b)
[[a == e^(4*c0 + 2*c1), b == 4*c0 + 2*c1]]
To extract the solution for a in particular, you can do e.g.
sage: sols = solve([eq1, eq2], a, b, solution_dict=True); sols
[{a: e^(4*c0 + 2*c1), b: 4*c0 + 2*c1}]
sage: a == sols[0][a]
a == e^(4*c0 + 2*c1)
Asked: 2021-05-09 22:19:51 +0200
Seen: 248 times
Last updated: May 10 '21
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What is this supposed to mean ? Could you give us something similar to an example of what you expect ?
With the accepted answer, I can do
view(a == sols[0][a])which is what I meant.