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: 24.738633748750722
secant 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))
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