问题
I'm writing a program in python and in it I need to find the roots of a function that is:
a*x^n + b*x -c = 0
where a and b are constants that are calculated earlier in the program but there are several thousand of them.
I need to repeat this equation twice for all values of a and b once with n = 77/27 and once with n = 3.
How can i do this in python?
I checked numpy.roots(p) and that would work for when n = 3 I think. But for n = 77/27 how would I be able to do that?
回答1:
I think your beast choice is scipy.optimize.brentq():
def f(x, n, a, b, c):
return a * x**n + b * x - c
print scipy.optimize.brentq(
f, 0.0, 100.0, args=(77.0/27.0, 1.0, 1.0, 10.0))
prints
2.0672035922580592
回答2:
Look here and here.
I'm so proud of myself, I still remember the specifics (without reading the link!) :)
If you don't get that, look here.
回答3:
I would use fsolve from scipy,
from scipy.optimize import fsolve
def func(x,a,b,c,n):
return a*x**n + b*x - c
a,b,c = 11.,23.,31.
n = 77./27.
guess = [4.0,]
print fsolve(func,guess,args=(a,b,c,n)) # 0.94312258329
This of course gives you a root, not necessarily all roots.
Edit: Use brentq, it's much faster
from timeit import timeit
sp = """
from scipy.optimize import fsolve
from scipy.optimize import brentq
from numpy.random import uniform
from numpy import zeros
m = 10**3
z = zeros((m,4))
z[:,:3] = uniform(1,50,size=(m,3))
z[:,3] = uniform(1,10,m)
def func(x,a,b,c,n):
return a*x**n + b*x - c
"""
s = "[fsolve(func,1.0,args=tuple(i)) for i in z]"
t = "[brentq(func,0.,10.,args=tuple(i)) for i in z]"
runs = 10**2
print 'fsolve\t', timeit(s,sp,number=runs)
print 'brentq\t', timeit(t,sp,number=runs)
gives me,
fsolve 15.5562820435
brentq 3.84963393211
回答4:
You need a root finding algorithm like Newton's method. All root finding algorithms will work with non-integer powers. They need not even be rational numbers.
来源：https://stackoverflow.com/questions/6519380/find-roots-of-a-function-a-xn-bx-c-0-where-n-isnt-an-integer-with-numpy