Write a program for the secant method using recurssion (tolernace = 0.0001) .Note: it is important that in each recursion call, only 1 function evaluation f(x) is done.

Code below does not use recurssion. Write or add changes to the code below to implement the secant method using recurssion.

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):

initial_x0 = x0

initial_x1 = x1

y0 = f(x0)

y1 = f(x1)

num_steps = 0

while (abs(y1) > tolerance):

# 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

print ('# [Secant] x0 = ', initial_x0, 'x1 = ', initial_x1, ' after', \

num_steps, "steps, approx. root is ", x1)

return x1

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