3906.sagews

"""
Ubiquitous Newton-Raphson algorithm for solving
    f(x) = 0
where a root is repeatedly estimated by
    x = x - f(x)/f'(x)
until |dx|/(1+|x|) < TOL is achieved.  This termination condition is a
compromise between 
    |dx| < TOL,  if x is small
    |dx|/|x| < TOL,  if x is large
"""
from scipy import exp

def f(x):
	return (3.0-x)*exp(x) - 3.0
	
def fprime(x):
	return (2.0-x)*exp(x)
	
def newton(func, funcd, x, TOL=1e-6):	# f(x)=func(x), f'(x)=funcd(x)
    f, fd = func(x), funcd(x)
    count = 0
    while 1:
	dx = f / float(fd)
	if abs(dx) < TOL * (1 + abs(x)): return x - dx
	x = x - dx
	f, fd = func(x), funcd(x)
	count = count + 1
	print "newton(%d): x=%s, f(x)=%s" % (count, x, f)
	
	
print newton(f,fprime,5,TOL=1e-3)

newton(1): x=4.32659538633, f(x)=-103.404917649 newton(2): x=3.73937134331, f(x)=-34.1064860902 newton(3): x=3.27329578708, f(x)=-10.2145165658 newton(4): x=2.969407321, f(x)=-2.40404327646 newton(5): x=2.84210428177, f(x)=-0.291801085581 newton(6): x=2.82190151635, f(x)=-0.00638132870494 2.82143960878