Unit - 1

Roots of Equation and Simultaneous Equations

1.1 Roots of Equation: Bracketing method

Bisection method

This 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.

Note: Remember that root of an equation lies between its positive and negative values and we take average of them to come closer to its accurate root.

Example1 Find a real root of using bisection method correct to five decimal places.

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

Example2 Find the root of the equation, using the bisection method.

Let then by hit and trial we have

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

Now,

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

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 must lies between

Now,

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

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 must lies between

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

Example3 Find the root of the equation   between 2 and 3, using bisection method correct to two decimal places.

Let

Where

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

Now,

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

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 must lies between

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Hence the root of the given equation correct to two decimal place is 2.1269

Regula-Falsi method

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

Proceed in this manner until the desired accurate root is found.

Example1 Find a real root of the equation near, correct to three decimal places by the Regula Falsi method.

Let

Now,

And also

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

By Regula   Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

Example 2 Find the real root of the equation

By the method of false position correct to four decimal places

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

Example3 Apply Regula Falsi Method to solve the equation

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.

Key takeaways

1. 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.

2. Root of an equation lies between its positive and negative values and we take average of them to come closer to its accurate root.

3.

1.2 Newton-Raphson method

Let be the approximate root of the equation.

By Newton Raphson formula

In general,

Where n=1, 2, 3…… we keep on calculating until we get desired root to the correct decimal places.         

Example1 Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.

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 place 2.79838.

Example2   Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

Let

The initial approximation

By Newton Raphson Method

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 equation correct to three decimal places is 4.5579

Example3 Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

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

1.3 Solution of simultaneous equations: Gauss Elimination Method with Partial pivoting

In this method we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to upper triangular system. At last with help of back substitution we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

To convert the above system into upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

…          ……                …..         … 

…          ……                 ….         …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

Thus .

Then we calculate the values of .

Note: In (i) the coefficient is the pivot element and the equation is called the pivot equation. If then the above method fails and if it is close to zero the round off error may occur.

If or very small compared   to other coefficient of the equation, then we find the largest available coefficient in the column given below the pivot equation and then interchange the two rows to obtain new pivot variable this is known as partial pivoting.

Example1 Apply Gauss Elimination method to solve the equations:

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

Example2 Solve the equation by Gauss Elimination Method:

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.

Example3 Apply Gauss Elimination Method to solve the following system of equation:

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

Example4: Solve the system by Gauss Elimination method using partial pivoting

Given exact solution is

Given equations are

Using partial pivoting we rewrite the given equations as

       (1)

   (2)

Using Gauss elimination method

Multiplying (1) by (-0.0003120/0.5000) + (2) we get

Or 

Substituting value of y in equation (1) we get

Hence

Example5: Solve the system of linear equations

Using partial pivoting by Gauss elimination method we rewrite the given equations as

     (1)

        (2)

     (3)

Apply    and

     (1)

   (4)                       

              (5)

Apply )

     (1)

   (4)

Or     .

Putting value of z in we get .

Putting values of y and z in we get .

Hence the solution of the equation is .

1.4 Gauss-Seidel method

Jacobi’s iteration Method:

Let us consider the system of simultaneous linear equation

The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get

Take the initial approximation we get the values of the first approximation of .

By the successive iteration we will get the desired the result.

Example1 Use Jacobi’s method to solve the system of equations:

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

Example2 Solve by Jacobi’s Method, the equations

Given equation can be rewrite 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

Example3Use Jacobi’s method to solve the system of the equations

Rewrite the given equations

Let the initial approximation be

1.2

1.3

0.9

1.03

0.9946

0.9934

1.0015

Hence the solution of the above equation correct to two decimal places is

Gauss Seidel method:   

This is the modification of the Jacobi’s Iteration.  As above in Jacobi’s Iteration, we take first approximation as and put in the right hand side of the first equation of (2) and let the result be  . Now we put right hand side of second equation of (2) and suppose the result is now put in the RHS of third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to desired accuracy.

Example1 Use Gauss –Seidel Iteration method to solve the system of equations

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

Example2 Solve the following system of equations

  By Gauss-Seidel method.

Rewrite the given system of equations as

Le t the initial approximation be

Thus the required solution is

Example3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

1.5 Thomas algorithm for Tri-diagonal Matrix

Tri-diagonal Matrix

Consider a matrix A =

Which is a matrix and main diagonal elements are ,.The super diagonal elements are , and the sub diagonal elements are .

A tri-diagonal matrix is a matrix in which all the entries are zero expect the main diagonal, super diagonal and sub diagonal. 

Example of tri-diagonal matrix A =

The nth order tri-diagonal matrix is   

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

Also note that  and .

A system of linear equations that can be represented in the form of tri-diagonal matrix is called tri-diagonal system.

The general form of tri-diagonal system with n equation is

Here the main diagonal elements are denoted by , the super diagonal elements by and the sub diagonal elements by for .

The tri-diagonal matrix is applicable to those system only which are diagonally dominate i.e..

…………………………….

The above system can be solved by tri-diagonal matrix algorithm   also known as Thomas Algorithm.

Working Rule:

1. Dividing equation (1)  by   we get

…………………………….

II.                 Now equation (2)   equation  (1) we  get

…………………………….

Dividing equation (2) by we   get

…………………………….

Hence in general we can have

III.               Using   the above formula finally we get

…………………………….

IV.              From the last equation we  get

Substituting the value of in (n-1)’ equation we get the   value of .

Similarly, by back substitution we get the values of .

Formula used:

1. 
2. 
3. Back substitution.

Example1: Solve the system of equations

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.

Example2: Solve the system

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.

Example3: Solve the system

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.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
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