Using MATLAB, I am trying to solve the equation using Newtons Method but keep printing the result "None". I am not sure where the mistake is as I do not want to tamper with the formula.
def newt(p):
maxIt = 1000
tol = 10^(-5)
for i in range(maxIt):
fp = 2*p**3 + p**2 - p + 1 #double * for exponent
fprimep = 6*p**2 + 2*p - 1 #f'p
p_new = p - fp/fprimep
if abs (p - p_new) < tol:
return p
p = p_new
#initial values
p0 = -1.2
p = newt(p0)
print(p)
The error in your code is due to a partial conversion from Matlab. You define tol = 10^(-5), but this is not exponentiation in Python, it is bitwise xor. Correcting that, you get the proper result:
def newt(p):
maxIt = 1000
tol = 1e-5 # Error was here
for i in range(maxIt):
fp = 2*p**3 + p**2 - p + 1 #double * for exponent
fprimep = 6*p**2 + 2*p - 1 #f'p
p_new = p - fp/fprimep
if abs (p - p_new) < tol:
return p
p = p_new
#initial values
p0 = -1.2
p = newt(p0)
print(p)
# -1.23375
As for the return value, your function returns None when the method does not converge. I'm not sure this was a intentional decision, but it is a good convention anyway, since it allows the user to know the method did not converge. A more robust method would be to throw an exception, maybe adding the guess at the time the iteration stopped and the convergence ratio.