Simplest way to solve mathematical equations in Python

Question

I want to solve a set of equations, linear, or sometimes quadratic. I don\'t have a specific problem, but often, I have been in this situation often.

It is simple to

Answer 1

Try applying Bisection method in py to find the root given an interval:

def f(x, rhs): # f(x) = e^x
    return math.e ** x - rhs # e^x = rhs -> e^x - rhs = 0

def solve(rhs, a = 0, b = 100, tol = 1e-3):
    while True:
        c  = (a + b) / 2.0
        if(f(a, rhs) * f(c, rhs) > 0):
            a = c
        else:
            b = c
        if(abs(f(c, rhs)) < tol):
            break
    return c

y = math.e ** 3.75 # x = 3.75
print(solve(y)) # 3.7499..

Answer 2

You discount the best answer as unacceptable.

Your question is "I want a free Computer Algebra System that I can use in Python."

The answer is "SAGE does that."

Have you looked at maxima/macsyma? SAGE provides bindings for it, and that's one of the more powerful free ones.

http://maxima.sourceforge.net/

Answer 3

I'd use Octave for this but I agree, the syntax of Octave isn't what I'd call thrilling (and the docs always confuse me more than they help, too).

Answer 4

I have just started using GNU Scientific Library, which however is C library. Looks like there are Python bindings too. So, it might be worth looking at.

Answer 5

Here is how to solve your original question using Python (via Sage). This basically clarifies the remark Paul McMillan makes above.

sage: a,b,c = var('a,b,c')
sage: solve([a+b+c==1000, a^2+b^2==c^2], a,b,c)
[[a == 1000*(r1 + sqrt(r1^2 + 2000*r1 - 1000000))/(r1 + sqrt(r1^2 + 2000*r1 - 1000000) + 1000), b == -1/2*r1 - 1/2*sqrt(r1^2 + 2000*r1 - 1000000) + 500, c == r1], [a == 1000*(r2 - sqrt(r2^2 + 2000*r2 - 1000000))/(r2 - sqrt(r2^2 + 2000*r2 - 1000000) + 1000), b == -1/2*r2 + 1/2*sqrt(r2^2 + 2000*r2 - 1000000) + 500, c == r2]]

Answer 6

Have you looked at SciPy?

It has an example in the tutorials on solving linear algebra:

http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html#solving-linear-system