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numerical_approx()

asked 2018-12-04 09:28:23 -0600

thetha gravatar image

updated 2018-12-10 01:55:17 -0600

FrédéricC gravatar image

from sympy.solvers import solve

from sympy import Symbol

x = Symbol('x')

C=solve((x+86)/x-95/58,x)

How its possible to use numerical_approx() on this https://ask.sagemath.org/question/992... this solutions is not working,

C[0].numerical_approx()

numerical_approx(C[0])

also not

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For instance, typing

If we define `f` by

    def f(x, y, z):
        return x * y * z

then `f(2, 3, 5)` returns `30` but `f(2*3*5)` gives:

    TypeError: f() takes exactly 3 arguments (1 given)

produces:

If we define f by

def f(x, y, z):
    return x * y * z

then f(2, 3, 5) returns 30 but f(2*3*5) gives:

TypeError: f() takes exactly 3 arguments (1 given)

Please edit your question to do that.

slelievre gravatar imageslelievre ( 2018-12-04 11:22:12 -0600 )edit

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answered 2018-12-04 10:34:20 -0600

updated 2018-12-04 10:34:55 -0600

Sage's and SymPy's expressions and numbers have different types, but you can convert between them as explained in A Sample Session using SymPy.

In your case:

sage: C[0]._sage_().numerical_approx()
134.810810810811
sage: numerical_approx(C[0]._sage_())
134.810810810811
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Asked: 2018-12-04 09:28:23 -0600

Seen: 57 times

Last updated: Dec 04