Each Answer to this Q is separated by one/two green lines.

What’s the (best) way to solve a **pair of non linear equations using Python.** (Numpy, Scipy or Sympy)

eg:

- x+y^2 = 4
- e^x+ xy = 3

A code snippet which solves the above pair will be great

for numerical solution, you can use fsolve:

http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html#scipy.optimize.fsolve

```
from scipy.optimize import fsolve
import math
def equations(p):
x, y = p
return (x+y**2-4, math.exp(x) + x*y - 3)
x, y = fsolve(equations, (1, 1))
print equations((x, y))
```

If you prefer sympy you can use nsolve.

```
>>> nsolve([x+y**2-4, exp(x)+x*y-3], [x, y], [1, 1])
[0.620344523485226]
[1.83838393066159]
```

The first argument is a list of equations, the second is list of variables and the third is an initial guess.

# Short answer: use fsolve

As mentioned in other answers the simplest solution to the particular problem you have posed is to use something like `fsolve`

:

```
from scipy.optimize import fsolve
from math import exp
def equations(vars):
x, y = vars
eq1 = x+y**2-4
eq2 = exp(x) + x*y - 3
return [eq1, eq2]
x, y = fsolve(equations, (1, 1))
print(x, y)
```

Output:

```
0.6203445234801195 1.8383839306750887
```

# Analytic solutions?

You say how to “solve” but there are different kinds of solution. Since you mention SymPy I should point out the biggest difference between what this *could* mean which is between *analytic* and *numeric* solutions. The particular example you have given is one that does not have an (easy) analytic solution but other systems of nonlinear equations do. When there are readily available analytic solutions SymPY can often find them for you:

```
from sympy import *
x, y = symbols('x, y')
eq1 = Eq(x+y**2, 4)
eq2 = Eq(x**2 + y, 4)
sol = solve([eq1, eq2], [x, y])
```

Output:

```
?? ? 5 ?17? ?3 ?17? ?17 1? ? ? 5 ?17? ?3 ?17? 1 ?17? ? ? 3 ?13? ??13 5? 1 ?13? ? ?5 ?13? ? ?13 3? 1 ?13??
??-?- ? - ??????? - ????, - ??? - ??, ?-?- ? + ??????? + ????, - ? + ????, ?-?- ? + ????????? + ??, ? + ????, ?-?? - ??????- ??? - ??, ? - ?????
?? ? 2 2 ? ?2 2 ? 2 2? ? ? 2 2 ? ?2 2 ? 2 2 ? ? ? 2 2 ? ? 2 2? 2 2 ? ? ?2 2 ? ? 2 2? 2 2 ??
```

Note that in this example SymPy finds all solutions and does not need to be given an initial estimate.

You can evaluate these solutions numerically with `evalf`

:

```
soln = [tuple(v.evalf() for v in s) for s in sol]
```

```
[(-2.56155281280883, -2.56155281280883), (1.56155281280883, 1.56155281280883), (-1.30277563773199, 2.30277563773199), (2.30277563773199, -1.30277563773199)]
```

# Precision of numeric solutions

However most systems of nonlinear equations will not have a suitable analytic solution so using SymPy as above is great when it works but not generally applicable. That is why we end up looking for numeric solutions even though with numeric solutions:

1) We have no guarantee that we have found all solutions or the “right” solution when there are many.

2) We have to provide an initial guess which isn’t always easy.

Having accepted that we want numeric solutions something like `fsolve`

will normally do all you need. For this kind of problem SymPy will probably be much slower but it can offer something else which is finding the (numeric) solutions more precisely:

```
from sympy import *
x, y = symbols('x, y')
nsolve([Eq(x+y**2, 4), Eq(exp(x)+x*y, 3)], [x, y], [1, 1])
```

```
?0.620344523485226?
? ?
?1.83838393066159 ?
```

With greater precision:

```
nsolve([Eq(x+y**2, 4), Eq(exp(x)+x*y, 3)], [x, y], [1, 1], prec=50)
```

```
?0.62034452348522585617392716579154399314071550594401?
? ?
? 1.838383930661594459049793153371142549403114879699 ?
```

Try this one, I assure you that it will work perfectly.

```
import scipy.optimize as opt
from numpy import exp
import timeit
st1 = timeit.default_timer()
def f(variables) :
(x,y) = variables
first_eq = x + y**2 -4
second_eq = exp(x) + x*y - 3
return [first_eq, second_eq]
solution = opt.fsolve(f, (0.1,1) )
print(solution)
st2 = timeit.default_timer()
print("RUN TIME : {0}".format(st2-st1))
->
[ 0.62034452 1.83838393]
RUN TIME : 0.0009331008900937708
```

FYI. as mentioned above, you can also use ‘Broyden’s approximation’ by replacing ‘fsolve’ with ‘broyden1’. It works. I did it.

I don’t know exactly how Broyden’s approximation works, but it took 0.02 s.

And I recommend you do not use Sympy’s functions <- convenient indeed, but in terms of speed, it’s quite slow. You will see.

An alternative to `fsolve`

is `root`

:

```
import numpy as np
from scipy.optimize import root
def your_funcs(X):
x, y = X
# all RHS have to be 0
f = [x + y**2 - 4,
np.exp(x) + x * y - 3]
return f
sol = root(your_funcs, [1.0, 1.0])
print(sol.x)
```

This will print

```
[0.62034452 1.83838393]
```

If you then check

```
print(your_funcs(sol.x))
```

you obtain

```
[4.4508396968012676e-11, -1.0512035686360832e-11]
```

confirming that the solution is correct.

I got Broyden’s method to work for coupled non-linear equations (generally involving polynomials and exponentials) in IDL, but I haven’t tried it in Python:

scipy.optimize.broyden1`scipy.optimize.broyden1(F, xin, iter=None, alpha=None, reduction_method='restart', max_rank=None, verbose=False, maxiter=None, f_tol=None, f_rtol=None, x_tol=None, x_rtol=None, tol_norm=None, line_search="armijo", callback=None, **kw)[source]`

Find a root of a function, using Broyden’s first Jacobian approximation.

This method is also known as “Broyden’s good method”.

You can use openopt package and its NLP method. It has many dynamic programming algorithms to solve nonlinear algebraic equations consisting:

**goldenSection, scipy_fminbound, scipy_bfgs, scipy_cg, scipy_ncg, amsg2p, scipy_lbfgsb, scipy_tnc, bobyqa, ralg, ipopt, scipy_slsqp, scipy_cobyla, lincher, algencan,** which you can choose from.

Some of the latter algorithms can solve constrained nonlinear programming problem.

So, you can introduce your system of equations to *openopt.NLP()* with a function like this:

`lambda x: x[0] + x[1]**2 - 4, np.exp(x[0]) + x[0]*x[1]`

```
from scipy.optimize import fsolve
def double_solve(f1,f2,x0,y0):
func = lambda x: [f1(x[0], x[1]), f2(x[0], x[1])]
return fsolve(func,[x0,y0])
def n_solve(functions,variables):
func = lambda x: [ f(*x) for f in functions]
return fsolve(func, variables)
f1 = lambda x,y : x**2+y**2-1
f2 = lambda x,y : x-y
res = double_solve(f1,f2,1,0)
res = n_solve([f1,f2],[1.0,0.0])
```

You can use `nsolve`

of `sympy`

, meaning `numerical solver`

.

Example snippet:

```
from sympy import *
L = 4.11 * 10 ** 5
nu = 1
rho = 0.8175
mu = 2.88 * 10 ** -6
dP = 20000
eps = 4.6 * 10 ** -5
Re, D, f = symbols('Re, D, f')
nsolve((Eq(Re, rho * nu * D / mu),
Eq(dP, f * L / D * rho * nu ** 2 / 2),
Eq(1 / sqrt(f), -1.8 * log ( (eps / D / 3.) ** 1.11 + 6.9 / Re))),
(Re, D, f), (1123, -1231, -1000))
```

where `(1123, -1231, -1000)`

is the initial vector to find the root. And it gives out:

The imaginary part are very small, both at 10^(-20), so we can consider them zero, which means the roots are all real. Re ~ 13602.938, D ~ 0.047922 and f~0.0057.