Write a Python program to re-implement the secant method using recursion. Use yo

Question

Write a Python program to re-implement the secant method using recursion. Use your program to find a root of the function with tolerance 0.0001 Requirement: define a function with name "Secant recursion Q3 ### Solution to Question 3 f(x)2cos(x) 2sin(z). r + def Secant.recursion.Q30 codes calls Secant.recursionQ30 recursively. For the starting points in secant's method, one may use 1 and 1.02. Note for efficiency, it is important that in each recursion call, only 1 function evaluation f(x) is done.

Explanation / Answer

Solution:

code:

from math import cos, sin

tolerance = 0.0001

def non_poly_func(x):

    return x**2 + cos(x)**7 + 2*sin(x)

def non_poly_func_der(x):

    return 2*x - 7 * cos(x)**6 * sin(x) + 2 * cos(x)

def Secant_recurssion_Q3(f, x0, x1, tolerance, num_steps):

    initial_x0 = x0

    initial_x1 = x1

    y0 = f(x0)

    y1 = f(x1)

    if(abs(y1) <= tolerance):

        print (num_steps, "steps and approx. root is ", x1)       

        return x1

    else:

        # y1 / (x1 - x0) / (y1 - y0)

        x2 = x1 - y1 * (x1-x0) / (y1-y0)

        # updates

        x0 = x1

        x1 = x2

        #y0 = f(x0) # <-- important that this NOT needed (to save running-time)

        y0 = y1

        y1 = f(x1)  

        num_steps = num_steps + 1

       

        return Secant_recurssion_Q3(f, x0, x1, tolerance, num_steps)

Root = Secant_recurssion_Q3 (non_poly_func, 1, 1.02, tolerance, 0)

print(Root)

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