Secant Method Python With Desired Accuracy

Introduction:

We are given a function and one interval a to b we have to find the solution of function f(x) for which f(x0) = 0.

Why is Secant Method: 

We know some equations like quadratic equations and cubic equations are solvable if a solution exists. but for some complex equations solution exists but it is complex to solve them using normal maths.

Secant Method:

Let, f(x) is given, and interval a to b is given. we want to find value of x = x0 where f(x0) = 0.

1) If f(a) * f(b) >= 0:

        Solution doesn't exists.

2) Find x using this x = a - f(a) * (b-a) / (f(b)-f(a))

        find f(x) for this x.

3) if f(a) * f(x) < 0:

        Update b = x

    Else if f(x) * f(b) < 0:

        Update a = x

    

    Else if f(x) == 0:

        x is solution.

    Else:

        The solution doesn't exist.

4) Repeat steps 2) and 3) until we get desired accuracy.

f(x) = exp(x)+2^(-x) + 2*cos(x)-6 = 0

# f() function which return value of given function.

def f(x):

    # function given in question.

    return exp(x)+2**(-x) + 2*cos(x)-6

secent() function.

# secent() function which takes 4 arguments

# f = function, a and b interval, accuracy for accuracy of result

# return solution to given function.

def secant(f, a, b, accuracy=-1e-5):

    # if f(a)*f(b) >= 0 return None.

    if f(a)*f(b) >= 0:

        return None

    # set an = a.

    an = a

    # set bn = b.

    bn = b

    # index to count iterations.

    i = 0

    # create lest of x variables to store answers for each iterations.

    # which we use to find accuracy of 1e-5 or given.

    x = [-1e10, 1e10]

    # use while loop with this Condition

    # absolute differences of last two x value is greater or equal accuracy.

    # if not then while loop ends.

    while abs(x[-1]-x[-2]) >= accuracy:

        # find xn value.

        xn = an - f(an) * (bn-an) / (f(bn)-f(an))

        # find f value for xn.

        fxn = f(xn)

        # if fan*fxn<0 set bn = xn.

        if f(an)*fxn < 0:

            an = an

            bn = xn

        # elif fxn*fbn<0 set an = xn.

        elif fxn*f(bn) < 0:

            an = xn

            bn = bn

        # if fxn is 0 return xn.

        elif fxn == 0:

            return xn

        # else return None.

        else:

            return None

        # xn append to x.

        x.append(xn)

        # increment i.

        i += 1

    # return xn based on an and bn.

    return an - f(an) * (bn-an) / (f(bn)-f(an))

Printing result:

# print solution.

print("solving : exp(x)+2^(-x) + 2*cos(x)-6 = 0")

for i in range(16):

    # find accuracy

    accuracy = 10**(-i)

    # x value for fx = 0.

    x = secant(f, 1, 2, accuracy)

    # find fx.

    fx = f(x)

    print("x = {:<19}, with {:<6}, givens f(x) = {:<25}".format(x, accuracy, fx))

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Output:

Image:

Comment if you have any questions...