secant method
def secant_method(f, x0, x1, iterations):
"""Return the root calculated using the secant method."""
for i in range(iterations):
x2 = x1 - f(x1) * (x1 - x0) / float(f(x1) - f(x0))
x0, x1 = x1, x2
return x2
def f_example(x):
return x ** 2 - 612
root = secant_method(f_example, 10, 30, 5)
print("Root: {}".format(root)) # Root: 24.738633748750722secant method
def secant(f,a,b,N):
'''Approximate solution of f(x)=0 on interval [a,b] by the secant method.
Parameters
----------
f : function
The function for which we are trying to approximate a solution f(x)=0.
a,b : numbers
The interval in which to search for a solution. The function returns
None if f(a)*f(b) >= 0 since a solution is not guaranteed.
N : (positive) integer
The number of iterations to implement.
Returns
-------
m_N : number
The x intercept of the secant line on the the Nth interval
m_n = a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))
The initial interval [a_0,b_0] is given by [a,b]. If f(m_n) == 0
for some intercept m_n then the function returns this solution.
If all signs of values f(a_n), f(b_n) and f(m_n) are the same at any
iterations, the secant method fails and return None.
Examples
--------
>>> f = lambda x: x**2 - x - 1
>>> secant(f,1,2,5)
1.6180257510729614
'''
if f(a)*f(b) >= 0:
print("Secant method fails.")
return None
a_n = a
b_n = b
for n in range(1,N+1):
m_n = a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))
f_m_n = f(m_n)
if f(a_n)*f_m_n < 0:
a_n = a_n
b_n = m_n
elif f(b_n)*f_m_n < 0:
a_n = m_n
b_n = b_n
elif f_m_n == 0:
print("Found exact solution.")
return m_n
else:
print("Secant method fails.")
return None
return a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))Copyright © 2021 Codeinu
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