Reputation: 161
I did this little test program in python to see how solve
and nsolve
work.
from sympy import *
theta = Symbol('theta')
phi = Symbol('phi')
def F(theta,phi):
return sin(theta)*cos(phi)+cos(phi)**2
def G(phi):
return ((1 + sqrt(3))*sin(phi) - 4*pi*sin(2*phi)*cos(2*phi))
solution1 = solve(F(pi/2,phi),phi)
solution2 = solve(G(phi),phi)
solution3 = nsolve(G(phi),0)
solution4 = nsolve(G(phi),1)
solution5 = nsolve(G(phi),2)
solution6 = nsolve(G(phi),3)
print solution1, solution2, solution3, solution4, solution5, solution6
And I get this output:
[pi/2, pi] [] 0.0 -0.713274788952698 2.27148961717279 3.14159265358979
The first call of solve gave me two solutions of the corresponding function. But not the second one. I wonder why? nsolve
seems to work with an initial test value, but depending on that value, it gives different numerical solutions. Is there a way to get the list all numerical solutions with nsolve
or with another function, in just one line?
Upvotes: 5
Views: 9513
Reputation: 58521
The first call of solve gave me two solutions of the corresponding function. But not the second one. I wonder why?
In general, you cannot solve an equation symbolically and apparently solve
does exactly that. In other words: Consider yourself lucky if solve
can solve your equation, the typical technical applications don't have analytic solutions, that is, cannot be solved symbolically.
So the fall-back option is to solve the equation numerically, which starts from an initial point. In the general case, there is no guarantee that nsolve
will find a solution even if exists one.
Is there a way to get the list all numerical solutions with nsolve or with another function, in just one line?
In general, no. Nevertheless, you can start nsolve
from a number of initial guesses and keep track of the solutions found. You might want to distribute your initial guesses uniformly in the interval of interest. This is called multi-start method.
Upvotes: 6