Unit - 1

Numerical analysis

1.1 Solution of algebraic and transcendental Equations: Newton–Raphson method

Introduction of numerical analysis-

Numerical analysis is a branch of mathematics that deal with solving the difficult mathematical problems by using efficient methods

Sometime the mathematical problems are very hard then an approximation to a difficult Mathematical problem is very important to make it more easy to solve numerical approximation has become more popular and a modern tool there are three parts of numerical analysis. The first part of the subject is about the development of a method to a problem. The second part deals with the analysis of the method, which includes the error analysis and the efficiency analysis. Error analysis gives us the understanding of how accurate the result will be if we use the method and the efficiency analysis tells us how fast we can compute the result

The third part of the subject is the development of an efficient algorithm to implement the method as a computer code.

Concept of roots of an equation-

There are two types of equations Linear and Non linear equations. Linear equations are those in which dependent variable y is directly proportional to independent variable x and is of degree one. On the other hand non linear equation are those in which y does not directly proportional to x and of degree more than one.

Ex: +b, where a and b are constant is a linear equation.

is a non linear equation.

Algebraic Equation: If f(x) is a pure polynomial, then the equation is called an algebraic equation in x.

Ex:

Transcendental Equation: If f(x) is an expression contain function as trigonometric, exponential and logarithmic etc. Then is called transcendental equation.

Ex

Non –linear equation can be solved by using various analytical methods. The transcendental equations and higher order algebraic equations are difficult to solve even sometime are impossible. Finding solution of equation means just to calculate its roots.

Numerical methods are often repetitive in nature. They consist of repetitive calculation of the same process where in each step the result of preceding values are used (substitute).  This is known as iteration process and is repeated till the result is obtained to desired accuracy.

The analytical methods used to solve equation; exact value of the root is obtained whereas in numerical method approximate value is obtained.

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.       

Example1Using 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

Example 3: 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

Key takeaways-

1. Newton-Raphson Method-

1.2 Method of false position

Regula - Falsi Method (Method of false position)

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.

Example 1: Find a real root of the equation near, correct to three decimal place by the Regula Falsi method.

Let

Now,

And also

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

By Regula   Falsi Mehtod

Now,

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

By Regula Fasli  Method

Now,

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

By Regula Fasli  Method

Now,

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

By Regula Fasli  Method

Now,

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

By Regula Fasli  Method

Now,

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

By Regula Fasli  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 Mehtod

Now,

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

By Regula   Falsi Mehtod

Now,

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

By Regula   Falsi Mehtod

Now,

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

By Regula   Falsi Mehtod

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

Example 3: 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 Mehtod

Now,

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

By Regula  Falsi Mehtod

Now,

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

By Regula  Falsi Mehtod

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

Key takeaways-

1. Algebraic Equation: If f(x) is a pure polynomial, then the equation is called an algebraic equation in x.
2. Transcendental Equation: If f(x) is an expression contain function as trigonometric, exponential and logarithmic etc. Then is called transcendental equation.
3. Root of an equation lies between its positive and negative values and we take average of them to come closer to its accurate root.
4. The Regula Falsi formula

1.3 Solution of simultaneous linear equations using Gauss-Seidal method and Crout’s method (LU decomposition)

Jacobi’s Iteration method and Gauss-Seidal 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.

Example 1 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

Example 2 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

Example 3 Use 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.

Example 1 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

Example 2 Solve the following system of equations

  By Gauss-Seidel method.

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Example 3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Crout’s method (LU decomposition)

The method is based on the fact that every matrix A can be expressed as the product of a lower triangular matrix and an upper triangular matrix              , provided all the principal minors of A are non-singular.

Which means-

If , then-

Now consider the equations-

We can write it as-

Where-

Let

Where-

Equation (1) becomes-

Writing-

Equation (3) becomes-

which is equivalent to the equations-

Solving these for we know V. Then equation (4) becomes-

From which can be found by back substitution.

We write (2) as to find the matrix L and U-

Multiplying the matrix on the left and equating corresponding elements from both sides, we get-

3.    

4.    

5.    

We compute the elements of L and U in the following manner-

1. First row of U
2. First column of L
3. Second row of U
4. Second column of L
5. Third row of U

Example: Solve the equations-

Sol.

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-

1.4 Numerical solution of ordinary differential equations: Taylor's series method

Taylor’s Series Method:

The general  first  order differential  equation

….(1)

With the initial condition …(2)

Let be the exact solution of equation (1), then the Taylor’s series for around is given by

   (3)

If the values of  are known, then equation (3) gives apowwer series   for y. By total derivatives   we have

,

And other higher derivatives of y.  The method can easily be extended to   simultaneous   and higher –order  differential  equations. In   general,

Putting   in these above results, we can obtain the values of finally, we substitute these values of  in equation (2) and obtain the approximate value of y; i.e. the solutions of (1).

Example 1: Solve, using Taylor’s series method and compute .

Here This implies that .

Differentiating, we get

.

.

.

The   Taylor’s series at ,

(1)

At in equation (1) we get

At in equation (1) we get

Example 2: Using Taylor’s series method, find the solution of

At ?

Here

At implies that or or

Differentiating, we get

implies that or .

implies that or

implies that or

implies that or

The   Taylor’s series at ,

  (1)

At   in equation (1) we get

At in equation (1) we get

Example 3: Solve numerically, start from and carry to using Taylor’s series method.

Here .

We have

Differentiating, we get

implies that or

implies that or .

implies that

implies that

The   Taylor’s series at ,

Or

Here

The   Taylor’s series

.

1.5 Euler’s modified method

Euler’s  method:

In this method the solution is in the form  of a  tabulated  values

Integrating both side of the equation (i) we get

Assuming that  in  this  gives Euler’s formula

In general  formula 

, n=0,1,2,…..

Error estimate for the Euler’s method

Example 1: Use Euler’s method to  find y(0.4) from the differential equation

    with h=0.1

Given  equation  

Here

We  break  the  interval in four steps.

So that

By Euler’s  formula

, n=0,1,2,3   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

.01

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

Hence y(0.4)  =1.061106.

Example 2: Using  Euler’s method  solve the differential equation  for  y at x=1 in five steps

Given equation  

Here 

No. Of steps  n=5 and  so that

So that

Also

By Euler’s  formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we  get

For n=1  in equation (i) we  get

For n=2  in equation (i) we  get

For n=3  in equation (i) we  get

For n=4  in equation (i) we  get

Hence 

Example 3: Given  with  the initial condition  y=1 at   x=0.Find  y for x=0.1 by  Euler’s   method(five  steps).

Given equation is  

Here   

No. Of steps  n=5 and so that

So that

Also

By Euler’s  formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

For n=4 in equation (i) we  get

Hence  

Modified Euler’s  Method:

Instead of  approximating as in Euler’s method.  In  the  modified Euler’s  method we  have the iteration formula

Where is the nth approximation to .The iteration started with the Euler’s  formula

Example1: Use modified  Euler’s method  to compute y  for  x=0.05.  Given that

Result correct to three decimal places.

Given  equation  

Here

Take  h  =  = 0.05

By modified Euler’s  formula the  initial iteration is

)

The iteration  formula by  modified Euler’s method  is

-----(i)

For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

For n=3  in equation  (i)  we get

Since  third and  fourth approximation are equal .

Hence y=1.0526  at x =  0.05  correct  to  three decimal places.

Example 2: Using modified   Euler’s method,  obtain a solution of  the equation

Given  equation 

Here

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(i)

For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

For n=2 in equation  (i)  we get

For n=3  in equation  (i)  we get

Since third and fourth approximation are equal.

Hence y=0.0952 at  x=0.1

To calculate  the value  of  at x=0.2

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(ii)

For n=0  in equation  (ii)  we get

1814

For n=1  in equation  (ii)  we get

1814

Since first and second approximation   are  equal .

Hence  y  = 0.1814  at  x=0.2

To  calculate the value  of  at x=0.3

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(iii)

For n=0  in equation  (iii)  we get

For n=1  in equation  (iii)  we get

For n=2 in equation  (iii)  we get

For n=3 in equation  (iii)  we get

Since third and fourth approximation are same.

Hence y = 0.25936 at x = 0.3

Key takeaways-

1. Euler’s  method:

, n=0,1,2,…..

2.     Modified Euler’s  Method:

1.6 Runge-Kutta fourth order method

Runge-kutta methods-

This method is more accurate than Euler’s method.

Consider the differential equation of first order

Let  be the first interval.

A second order  Runge Kutta formula  

Where

Rewrite  as 

A fourth  order Runge Kutta formula:

Where 

Example 1: Use Runge Kutta method to find y when x=1.2  in step of h=0.1 given that

Given equation

Here

Also

By  Runge Kutta formula for first  interval

Again

A fourth  order Runge Kutta formula:

To  find y at 

A fourth  order Runge Kutta formula:

Example 2: Apply Runge Kutta fourth order method to find an approximate value of  y for  x=0.2   in step of 0.1,  if

Given equation 

Here 

Also

By  Runge Kutta formula for first  interval

A fourth  order Runge Kutta formula:

Again

A fourth  order Runge Kutta formula:

Example 3: Using Runge Kutta method of fourth order,   solve

Given equation 

Here 

Also

By  Runge Kutta formula for first  interval

)

A fourth  order Runge Kutta formula:

Hence at x  = 0.2 then y  = 1.196

To find the  value of y at x=0.4. In this case

A fourth  order Runge Kutta formula:

Hence at  x  = 0.4 then  y=1.37527

Simultaneous equation using Runge Kutta method of 2 orders:

The second order differential equation

Let then the above equation reduces to first order simultaneous differential equation

Then

This can be solved as we discuss above by Runge Kutta Method. Here for and for .               

A fourth  order Runge Kutta formula:

Where 

Example 1: Using Runge Kutta method of order four , solve   to find

Given second order differential equation is

Let   then above equation reduces to

Or

(say)

Or .

By Runge Kutta Method we have

A fourth  order Runge Kutta formula:

Example 2: Using Runge Kutta method, solve

for correct to four decimal places with initial condition .

Given second order differential equation is

Let   then above equation reduces to

Or

(say)

Or .

By Runge Kutta Method we have

A fourth  order Runge Kutta formula:

And

.

Example 3: Solve the differential equations

    for

Using four order Runge Kutta method with initial conditions

Given differential equation are

Let

And

Also

By Runge Kutta Method we have

A fourth  order Runge Kutta formula:

And  

.

Key takeaways-

Runge-kutta methods-

• A second order Runge Kutta formula

Where

• A fourth order Runge Kutta formula:

Where 

1.7 Milne’s predictor- corrector method

For given dy/dx = f(x,y) and y = and x = , to find the value of y for x = , by using Milne’s method,

We follow the steps given below-

The value being given, here we calculate-

By Taylor’s series or Picard’s method.

Now we calculate-

Then to find

We substitute Newton’s forward interpolation formula-

In the relation-

By putting x = ,  dx = h dn

Neglecting fourth and higher order differences and expressing in terms of the function values, we get-

This is called a predictor.

Now having found we obtain a first approaximation to

Then the better value of is found by simpson’s rule as-

Which is called corrector.

Then an improved value of is computed and again corrector is applied to find a better value of .

We continues this step until remains unchanged.

Once and are obtained to desired degree of accuracy,

is found from the predictor as-

And

is calculated.

Then the better approximation to the value of we get from the corrector as-

We repeat until becomes stationary and we proceed to calculate .

This is called the Milne’s predictor-corrector method.

Adams - Bashforth predictor and corrector formula-

This is called Adams - Bashforth predictor formula.

And

This is called Adams - Bashforth corrector formula.

Example: Find the solution of the differential equation   in the range for the boundary conditions y = 0 and x = 0 by using Milne’s method.

Sol.

By using Picards method-

Where

To get the first approximation-

We put y = 0 in f(x, y),

Giving-

In order to find the second approximation, we put y  = in f(x,y)

Giving-

And the third approximation-

Now determine the starting  values of the Milne’s method from equation  (1), by choosing h = 0.2

Now using the predictor-

X = 0.8

,      

And the corrector-

,                 ................(2)

Now again using corrector-

Using predictor-

X = 1.0,  

,      

And the corrector-

,      

Again using corrector-

,       which is same as before

Hence

Example: Solve the initial value problem ,  y(0) = 1 to find y(0.4) by using Adams-Bashforth method.

Starting solutions required are to be obtained using Runge-Kutta method of order 4 using step value h = 0.1

Sol.

Here we have-

Here

So that-

Thus

To find y(0.2)-

Here

Thus,

Y(0.2) = 

To find y(0.3)-

Here

Thus,

Y(0.3) = 

Now the starting values of Adam’s method with h = 0.1-

x = 0.0                   y-3 = 1.0000                    f-3 = 0.0 – (1.0)2                                 = - 1.0000

x = 0.1                   y-2 = 0.9117                    f-2 = 0.1 – (0.9117)2                          = - 1.7312

x = 0.2                   y-1 = 0.8494                    f-1 = 0.2 – (0.8494)2                          = - 0.5215

x = 0.3                   y0 = 0.8061                     f0 = 0.3 – (0.8061)2                           = - 0.3498

Using predictor-

= 0.7789                                                                             f1 = - 0.2067

Using corrector-

Hence

Key takeaways-

1. Predictor-
    
2. Corrector-
    
3. Adams - Bashforth predictor formula-

4.     Adams - Bashforth corrector formula.

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.