为什么 SymPy 没有为我的方程找到任何解？

Question

I want to calculate roots of my graphs I plotted (through matplotlib).

I tried with SciPy but it needs a rough estimate of the coordinates of the roots which is too impractical. Now I'm trying it with SymPy but it can't find a solution and prints empty brackets. Am I doing something wrong?

Here's my code:

import matplotlib as plt
import numpy as np
from sympy import *

h = 1.8
alpha = 0
g = 9.81
K = 0.0004
v_0 = 360


x = np.linspace(0, 10000, 500)
f_x = h + x * np.tan(alpha) - (g / (2 * (K ** 2) * (v_0 ** 2))) * ((np.e ** (K * x) - 1) ** 2)

*bunch of labels and titles*

x = Symbols("x")
roots = solve(f_x, x)
print(roots)

The variables of the formula are all defined except x of course.

Is there any way to make it work like Geogebra, where you can calculate roots, maxima, etc. of a graph or is Python not fitted for that kind of stuff?

Answer 1

If you investigate your equation symbolically you will find that you can easily solve for the two roots of this equation for the values you have chosen. The value of alpha being 0 makes the equation look like this: 1.8 - 236.55*(exp(0.0004*x) - 1)**2 .

>>> from sympy import exp, tan
>>> from sympy.abc import x, a
>>> h = 1.8
>>> alpha = a
>>> g = 9.81
>>> K = 0.0004
>>> v_0 = 360
>>> f_x = h+x*tan(alpha)-(g/(2*(K**2)*(v_0**2)))*((exp(K*x)-1)**2)
>>> solve(f_x.subs(a, alpha), x)

This is a very important starting point when working with nonlinear equations, because if you can find the solutions at some point you can often find the solution at a nearby point. In your case, a nearby point would be alpha slightly positive or negative. And such an equation for arbitrary alpha can only be solved as an implicit function (ie numerically). But since you have a previously known value you can use that as your initial guess for the solver.

But how easy it is for the solver to use the guess to find the solution depends on the shape of the function near the solution. Your equation has an undesirable maximum near the solution. It looks like your equation was academically constructed to have this property. If you rewrite your equation as two equations (one for each branch)

>>> branches = [exp(K*x) - s for s in solve(f_x, exp(K*x))]

And then solve each of those branches at a non-zero value of alpha it will be much better behaved: eg nsolve(branch[0].subs(a, new alpha value), value from when alpha was 0) .

That's the main idea. Recap:

1. don't use numpy when looking for symbolic answers
2. when working with nonlinear equations see if you can make some x-related term be zero (alpha=0 in your case) so you can solve the rest of the equation for x; use that value when allowing the zero term to have a nonzero value (a non-zero alpha)
3. given f(x) + g(x) = 0 that is ill-behaved near root, try work with x = f^-1(g(x))