# -*- coding: utf-8 -*-
"""
Created on Thu Dec 18 10:50:10 2014

@author: grivet

Exercice 5-1e : Résolution d'une équation non-linéaire
par la méthode de la sécante
"""

from pylab import *

TOL = 1E-6
def  fn(x):
    return x - 0.2*sin(x) - 0.5

xavder = float(input("abscisse gauche: "))
xder = float(input("abscisse droite: "))
favder = fn(xavder); fder = fn(xder)
i = 1
print( 'iteration      xavder           x            xder')

while abs(xder - xavder) > TOL:
    x = xder - fder*(xder-xavder)/(fder-favder)
    f = fn(x)
    favder = fder; xavder = xder
    xder = x; fder = f
    print('%d \t %12.4f \t %12.4f \t %12.4f' %(i, xavder,x,xder))
    i += 1
x = xder - fder*(xder-xavder)/(fder-favder)
print ('\t x  \t     fn(x)')
print( '%14.6f\t%14.6f' %(x,fn(x)))