A Python code example to find an approximate value for x in f(x) = 0 using Newton's method.

```
''' Newton_method3.py
find value for x in f(x) = 0 using Newton's method
see:
http://en.wikipedia.org/wiki/Newton%27s_method
tested with Python27, IronPython273 and Python33 by vegaseat 08jan2013
'''
def derivative(f):
def compute(x, dx):
return (f(x+dx) - f(x))/dx
return compute
def newtons_method(f, x, dx=0.000001, tolerance=0.000001):
'''f is the function f(x)'''
df = derivative(f)
while True:
x1 = x - f(x)/df(x, dx)
t = abs(x1 - x)
if t < tolerance:
break
x = x1
return x
def f(x):
'''
here solve x for ...
3*x**5 - 2*x**3 + 1*x - 37 = 0
'''
return 3*x**5 - 2*x**3 + 1*x - 37
x_approx = 1 # rough guess
# f refers to the function f(x)
x = newtons_method(f, x_approx)
print("Solve for x in 3*x**5 - 2*x**3 + 1*x - 37 = 0")
print("x = %0.12f" % x)
''' result ...
Solve for x in 3*x**5 - 2*x**3 + 1*x - 37 = 0
x = 1.722575335786
'''
# optional test (result should be close to zero)
# change dx and tolerance level to make it a little closer
print("Testing with the above x value ...")
print("%0.12f" % (3*x**5 - 2*x**3 + 1*x - 37))
''' result ...
0.000000399251
'''
```

About the Author

Scientist