#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Jan 30 07:38:00 2019

@author: nemenman
"""

# -*- coding: utf-8 -*-
"""
Created on Wed Feb  3 08:00:09 2016

@author: nemenman
"""
import numpy as np

#first let's check if vector arithmetic is faster than non-vector one
#and also check on how arrays work
L = 10000000
a = np.arange(L)
aa = a
a[1] = 2
 
c = np.arange(L)
for i in np.arange(L):
#    b[i] = a[i]
    c[i] = a[i]

d = np.copy(a)

dx = 1.0
x0 = 7
Tolerance = 1e-5
x = x0
while(np.abs(dx) > Tolerance):
    F = x**3 + 5    
    G = 3*x**2
    dx = -F/G
    x = x+dx
    print(x)
    

p1 = 0 # position of three fixed charges
p2 = 10 
p3 = 12
q1 = 1.0 # values of three charges
q2 = 2.0
q3 = 0.5
x0 = 7 # initial condition
Tolerance = 1e-5 # tolerance for change before terminationx = x0 #initial value for starting the loop
dx = 1.0 # fake "last change in x"
while(np.abs(dx) > Tolerance):
    F = q1*(x-p2)**2*(x-p3)**2 - q2*(x-p1)**2*(x-p3)**2 - q2*(x-p2)**2*(x-p1)**2
    G = 2*q1*(x-p2)*(x-p3)**2 + 2*q1*(x-p2)**2*(x-p3) - 2*q2*(x-p1)*(x-p3)**2  - 2*q2*(x-p1)**2*(x-p3) - 2*q3*(x-p2)*(x-p1)**2 - 2*q3*(x-p2)**2*(x-p1)
    dx = -F/G
    x = x+dx
    print(x)
    
print("The zero of the function is ", x, ".", sep="")