Unit-5

Numerical Methods

Question-1: Find the root of the equation, using the bisection method.

Solution: Let then by hit and trial we have

Thus .So the root of the given equation should lie between 2 and 3.

Now,

i.e. negative so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. negative so the root of the given equation must lie between

Now,

i.e. negative so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. negative so the root of the given equation must lie between

Hence the root of the given equation correct to two decimal places is 2.67965.

Question-2: Using the Secant Method find the root of the equation correct to three decimal places

Solution:

Let

By Secant Method

Let the initial approximation be

For n=1, the first approximation

 Now,

For n=2, the second approximation

563839

 Now,

For n=3, the third approximation

56717

 Now,

For n=4, the fourth approximation

567143

Hence the root of the given equation correct to four decimal places is 0.5671.

Question-3: Using the Secant Method find the root of the equation correct to four decimal place

Solution: Let

By Secant Method

Let the initial approximation be

For n=1, the first approximation

Now,

So the root of the equation lies between 2 and 1.92857

For n=2, the second approximation,

Now,

So the root of the equation lie between 2 and 1.96590

For n=3, the third approximation

Now,

So the root of the equation lie between 2 and 1.96600

For n=4, the fourth approximation

Now,

So the root of the equation lie between 2 and 1.96687

For n=5, the fifth approximation

Now,

So the root of the equation lie between 2 and 1.96690

For n=6, the sixth approximation

Now,

Hence the root of the given equation correct to four decimal places is 1.9669.

Question-4: Find a real root of the equation near, correct to three decimal places by the Regula Falsi method.

Solution: Let

 Now,

And also

Hence the root of the equation lie between and and so,

By Regula   Falsi Method

     Now,

So the root of the equation lie between 1 and 0.5 and so

By Regula Falsi  Method

Now,

So the root of the equation lie between 1 and 0.63637 and so

By Regula Falsi  Method

Now,

So the root of the equation lie between 1 and 0.67112 and so

By Regula Falsi  Method

Now,

So the root of the equation lie between 1 and 0.63636 and so

  By Regula Falsi  Method

Now,

So the root of the equation lie between 1 and 0.68168 and so

 By Regula Falsi  Method

Now,

Hence the approximate root of the given equation near 1 is 0.68217

Question-5: Apply Regula Falsi Method to solve the equation

Solution: Let 

By hit and trail

  And

So the root of the equation lie between and also

By Regula  Falsi Method

     Now,

So, the root of the equation lies between 0.60709 and 0.61 and also

By Regula  Falsi Method

     Now,

So, the root of the equation lies between 0.60710 and 0.61 and also

By Regula  Falsi Method

Hence the root of the given equation correct to five decimal places is 0.60710.

Question-6: Using the Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.

Solution: Given

By Newton- Raphson Method

                   = 

                  =

   The initial approximation is in radian.

      For n =0, the first approximation

For n =1, the second approximation

For n =2, the third approximation

For n =3, the fourth approximation

Hence the root of the given equation correct to five decimal places 2.79838.

Question-7: Using the Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

  Solution:  Let

 By Newton Raphson Method

Let the initial approximation be

 For n=0, the first approximation

   For n=1, the second approximation

For n=2, the third approximation

Since therefore the root of the given equation correct to four decimal places is -2.9537

For n=2, the third approximation

Since therefore the root of the given equation correct to four decimal places is -2.9537

Question-8: Find the real  root  of the polynomial correct to three decimal places?

Solution: Given equation    ….(1)

      Here

Also

Therefore the root of the equation lies between .

Again

    ….(2)

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 0.755.

Question-9: Apply Gauss Elimination method to solve the equations:

Solution: Given

Check Sum (sum of coefficient and constant)

                        -1                            …. (I)

                      -16                          …. (ii)

                          5 …. (iii)

(I)We eliminate x from (ii) and (iii)

Apply eq(ii)-eq(i) and eq(iii)-3eq(i) we get

                        -1                            ….(i)

                      -15                          ….(iv)

                          8            ….(v)

(II) We eliminate  y from eq(v)

Apply

                        -1                            ….(i)

                      -15                          ….(iv)

                      73            ….(vi)

(III) Back Substitution we get

                 From (vi) we get

                  From (iv) we get

                   From (i) we get 

Hence the solution of the given equation is

Question-10: Solve the equations-

Solution: Let

So that-

3.    

4.    

5.    

So

Thus-

Writing UX = V,

The system of given equations become-

By solving this-

We get-

Therefore the given system becomes-

Which means-

By back substitution, we have-

Question-11: Solve by Jacobi’s Method, the equations

Solution: Given equation can be rewritten in the form

     … (i)

    ..(ii)

    ..(iii)

Let the initial approximation be

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

0.90025

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Hence   solution approximately is

Question-12: Use the Gauss-Seidel Iteration method to solve the system of equations

Solution: Since

So, we express the unknown of larger coefficients in terms of the unknowns with smaller coefficients.

Rewrite the above system of equations

      (1)

Let the initial approximation be

3.14814

2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Question-13: Solve the following equations by Gauss-Seidel Method

Solution: Rewrite the above system of equations

          (1)

Let the initial approximation be

Hence the required solution is