Question

You have been provided with starter code that contains the place holders and some implementations for Bisection and Secant methods. Complete the code where it says #provide this line so that it could be run for the following case:

1) ?(?) = ? 3 + 4, ?? = −3, ?? = 0, ??????? = 100, ?? = 0.2

Starter Code:

def eq(x):

          return x**3+4

def err(xr,xold):

          return abs((xr-xold)/xr)*100

def bisection(xl,xu,maxIter,sc):

    # xl is the lower bound of the initial bracket

    # xu is the upper bound of the initial bracket

    # maxIter is the maximum number of iteration for finding the root

    # sc is the stopping criteria

    

          xm=(xl+xu)/2

          for i in range(maxIter):

                    if eq(xl)*eq(xm)>0:

                        # provide this line

                        # provide this line

                    else:

                         # provide this line

                         # provide this line

                    xn=(xl+xu)/2

                    if  # provide this line

                              return xn

                    xm=xn

def secant(xi,xip1,maxIter,sc):

    # xip1 is xi plus one

    # maxIter is the maximum number of iteration for finding the root

    # sc is the stopping criteria

          for i in range(maxIter):

                    eqDer= # provide this line

                    xi=xip1

                    # provide this line

                    if eq(xip1)=1: # provide this line

                              return xip1

Answer 1

Bisection Method

Step1: First let us understand about bisection method.

> Given a function f(x) on number x.

> Also given two numbers ‘p’ and ‘q’ such that f(p)*f(q) < 0 and f(x) is continuous in [p, q].

Here f(x) represents algebraic or transcendental equation.

> Now we need to find the root of function in interval [p, q] such that f(x) is 0.e

Step 2: Let us have the algorithm for the bisection method

1. Let r=(p+q)/2

2. If f(r)=0, stop and return cc.

3. If sign(f(p))≠sign(f(r)), then set q=r. Else if sign(f(q))≠sign(f(r)), then set p=r.

4. Go to the beginning and repeat until convergence.

Secant Method

Step 1: It is the procedure to find the roots of an equation for two distinct values using an interpolation method.

Step 2: The algorithm for the second method is given as follows:

1. Take the initial values of p, q and r. Here r is the stopping criteria, absolute error or the desired degree of accuracy
2. Compute f(p) and f(q)
3. Compute s = (p*f(q) – q*f(p)) / (f(q) – f(p))
4. Test for accuracy of s
   If | (s – q)/s | > r, then assign p = q and q = s
   goto step 3 Else, go to step 4
5. Display the required root as s.
6. Stop

def eq(x):

          return x**3+4

def err(xr,xold):

          return abs((xr-xold)/xr)*100

def bisection(xl,xu,maxIter,sc):

    # xl is the lower bound of the initial bracket

    # xu is the upper bound of the initial bracket

    # maxIter is the maximum number of iteration for finding the root

    # sc is the stopping criteria

   xm=(xl+xu)/2

for i in range(maxIter):

                    if eq(xl)*eq(xm)>0:

xl=xm# provide this line

xu=xu# provide this line

                    else:

xu=xm# provide this line

xl=xl# provide this line

                    xn=(xl+xu)/2

                    if xn==0:# provide this line

                              return xn

                    xm=xn

def secant(xi,xip1,maxIter,sc):

    # xip1 is xi plus one

    # maxIter is the maximum number of iteration for finding the root

    # sc is the stopping criteria

          for i in range(maxIter):

                    eqDer=xi - (xip1-xi)*f(xi)/( f(xip1) - f(xi) ) # provide this line

                    xi=xip1

xip1=eqDer# provide this line

                    if eq(xip1)==1: # provide this line

                              return xip1