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root_finder.py
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"""
Simple zero/root finder for functions with one variable.
Finds all the zeroes within a range of values, using user-defined searchstep and tolerance.
Utilizes the scipy.optimize.fsolve: https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html
LICENCE: MIT
"""
import numpy as np
import scipy
from scipy.optimize import fsolve
from matplotlib import pyplot as plt
class RootFinder:
def __init__(self, start, stop, step=0.01, root_dtype="float64", xtol=1e-9):
self.start = start
self.stop = stop
self.step = step
self.xtol = xtol
self.roots = np.array([], dtype=root_dtype)
def add_to_roots(self, x):
if (x < self.start) or (x > self.stop):
return # outside range
if any(abs(self.roots - x) < self.xtol):
return # root already found.
self.roots = np.append(self.roots, x)
def find(self, f, *args):
current = self.start
for x0 in np.arange(self.start, self.stop + self.step, self.step):
if x0 < current:
continue
x = self.find_root(f, x0, *args)
if x is None: # no root found.
continue
current = x
self.add_to_roots(x)
return self.roots
def find_root(self, f, x0, *args):
x, _, ier, _ = fsolve(f, x0=x0, args=args, full_output=True, xtol=self.xtol)
if ier == 1:
return x[0]
return None
def f(u, V=90, ell=5):
# Example from https://stackoverflow.com/q/14878110/3015186
w = np.sqrt(V ** 2 - u ** 2)
jl = scipy.special.jn(ell, u)
jl1 = scipy.special.yn(ell - 1, u)
kl = scipy.special.kn(ell, w)
kl1 = scipy.special.kn(ell - 1, w)
return jl / (u * jl1) + kl / (w * kl1)
if __name__ == "__main__":
r = RootFinder(1, 20, 0.01)
args = (90, 5)
roots = r.find(f, *args)
print("Roots: ", roots)
# plot results
u = np.linspace(1, 20, num=600)
fig, ax = plt.subplots()
ax.plot(u, f(u, *args))
ax.scatter(roots, f(np.array(roots), *args), color="r", s=10)
ax.grid(color="grey", ls="--", lw=0.5)
plt.show()