Unit - 1

Roots of Equation and Simultaneous Equations

Q1) Explain bisection method.

A1)

Bisection method consists of finding the root of the equation    which lies between a and b (say).

The function is continuous function between a and b and f (a) and f (b) are of opposite signs then there is a least one root between a and b.

Suppose f (a) is negative and f (b) is positive, then the first approximate value of the root is

If, then the correct root is .But if, then the root either lies between a and or and b according as is positive or negative, we again bisect the interval as above and the process is repeated the root is found to desired accuracy.

Q2) Find a real root of using bisection method correct to five decimal places.

A2)

Let then by hit and trial we have

Thus .So the root of the given equation should lies between 1 and 2.

Now,

I.e. positive so the root of the given equation must lies between

Now,

i.e., negative so the root of the given equation lies between

Now,

i.e., positive so the root of the given equation lies between

Now,

i.e., negative so that the root of the given equation lies between

Now,

i.e., positive so that the root of the given equation lies between

Now,

i.e., positive so that the root of the given equation lies between

Now,

i.e., negative so that the root of the given equation lies between

Now,

i.e., negative so that the root of the given equation lies between

Hence the approximate root of the given equation is 1.32421

Q3) What do you understand by Regula-falsi method?

A3)

This is the oldest method of finding the approximate numerical value of a real root of an equation.

In this method we suppose that and are two points where and are of opposite sign. Let

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

The Regula Falsi formula 

Find is positive or negative. If then root lies between and or if then root lies between and  similarly we calculate

Q4) Find the real root of the equation

By the method of false position correct to four decimal places

A4)

Let   

By hit and trail method

0.23136 > 0

So, the root of the equation lies between 2 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.72101 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.74020 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.74063 and 3 and also

By Regula   Falsi Method

Hence the root of the given equation correct to four decimal places is 2.7406

Q5) Apply Regula Falsi Method to solve the equation

A5)

Let 

By hit and trail

And

So, the root of the equation lies between and also

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

So, 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 place is 0.60710.

Q6) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.

A6)

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 corrects to five decimal place 2.79838.

Q7) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A7)

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

Q8) Solve the equation by Gauss Elimination Method:

A8)

Given

Rewrite the given equation as

                 … (i)

                             ….(ii)

                             ….(iii)

                                 …(iv)

(I)                We eliminate x from (ii), (iii) and (iv) we get

Apply eq(ii) + 6eq(i), eq(iii) -3eq(i), eq(iv)-5eq(i) we get

                 …(i)

                             ….(v)

                       ….(vi)

                             …(vii)

(II)             We eliminate y   from (vi) and (vii) we get

Apply 3.8 eq(vi)-3.1eq(v) and 3.8eq(vii)+5.5eq(v) we get

                 …(i)

                             ….(v)

                         …(viii)

                             …(ix)

(III)           We eliminate z from eq (ix) we get

Apply 9.3eq (ix) + 8.3eq (viii), we get

                 … (i)

                             ….(v)

                         …(viii)

350.74u=350.74

Or u = 1

(IV)          Back Substitution

From eq(viii)

Form eq(v), we get

From eq(i),

Hence the solution of the given equation is x=5, y=4, z=-7 and u=1.

Q9) Apply Gauss Elimination Method to solve the following system of equation:

A9)

Given                    … (i)

              … (ii)

              … (iii)

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

Apply we get

                … (i)

              … (iv)

             … (v)

(II)             We eliminate y from (v)

Apply we get

                … (i)

              … (vi)

                      … (vii)

(III)           Back substitution 

From (vii)  

From (vi)

From (i)

Hence the solution of the equation is

Q10) Use Jacobi’s method to solve the system of equations:

A10)
Since

So, we express the unknown with large coefficient in terms of other coefficients.

Let the initial approximation be

2.35606

0.91666

1.932936

0.831912

3.016873

1.969654

3.010217

1.986010

1.988631

0.915055

1.986532

0.911609

1.985792

0.911547

1.98576

0.911698

Since the approximation in ninth and tenth iteration is same up to three decimal places, hence the solution of the given equations is

Q11) Use Gauss –Seidel Iteration method to solve the system of equations

A11)

Since

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

Rewrite the above system of equations

Let the initial approximation be

3.14814

2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Q12) Solve the following equations by Gauss-Seidel Method

A12)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Q13) Solve the system of equations

A13)

The given system of equation is a tri-diagonal system

    ….(1)

  ….(2)

   ….(3)

        …(4)

Here the main diagonal elements are , super diagonal elements and the sub diagonal elements are. Also.

We know that

Again, we have   

For

For

For

Hence the system will be

On putting values, we get

Hence

Putting value of in equation (3’) we get

Putting values of in equation (2’) we get

Putting the value of in equation (1’) we get

Hence the solution is.

Q14) Solve the system

A14)

The given system is a tri-diagonal system

    …..(2)

    ….(3)

          …..(4)

Here  the main diagonal elements are , the super diagonal elements are , the sub diagonal elements are and the right side coefficient are .

We know that

Again, we have   

For

For

For

Hence the system will be

On putting values, we get

Hence

Putting value of in equation (3’) we get

Putting values of in equation (2’) we get

Putting the value of in equation (1’) we get

Hence the solution is.

Q15) Solve the system

A15)

The given system is a tri-diagonal system

     ….(3)

Here  the main diagonal elements are , the super diagonal elements are , the sub diagonal elements are and the right side coefficient are .

We know that

Again, we have   

For

For

For

Hence the system will be

On putting values, we get

Hence

Putting value of in equation (3’) we get

Putting values of in equation (2’) we get

Putting the value of in equation (1’) we get

Hence the solution is

.