Unit-6

Interpolation, Numerical integration, Solution of ordinary differential equations

6.1 Interpolation: Finite differences

Interpolation with an unequal interval-

Divided Difference:

In the case of interpolation, when the value of the arguments are not equispaced (unequal intervals) we use the class of differences called divided differences.

Definition:  The difference which is defined by taking into consideration the change in the value of the argument are known as divided differences.

Let be a function defined as

 Where are unequal i.e. it is a case of unequal interval.

The First-order divided differences are:

And so on.

The second-order divided difference is:

And so on.

 Similarly, the nth order divided difference is:

With the help of these we construct the divided difference table:

Newton’s Divided difference Formula:

Let be a function defined as

 Where are unequal i.e. it is a case of unequal interval.

.

Example1: Using Newton’s divided difference formula, find the values of from the following table:

We construct the divided difference table is given by:

By Newton’s Divided difference formula

.

Now, putting in above we get

.

Example2: The following values of the function f(x) for values of x are given:

Find the value of and also the value of x for which f(x) is maximum or minimum.

We construct the divide difference table:

By Newton’s divided difference formula

.

Putting in above we get

For maximum and minimum of , we have

Or

Example3: Find a polynomial satisfied by , by the use of Newton’s interpolation formula with a divided difference.

Here

We will construct the divided difference table:

By Newton’s divided difference formula

.

This is the required polynomial.

6.2 Newton’s and Lagrange interpolation formulae

Newton Forward Difference formula:

  This method is useful for interpolation near the beginning of a set of tabular values.

  Where

Example1: Using Newton’s forward difference formula, find the sum

Putting

It  follows that

Since is a fourth-degree polynomial in n.

Further,

By Newton Forward Difference Method

Example2: Given find  , by using Newton forward interpolation method.

Let , then

The table of forward finite difference is given below:

By Newton forward difference method

Here initial value = 45, difference of interval h = 5 and the value to be calculated at x=52.

By Formula

Example3: Find the missing term in the following:

Let

First, we construct the forward difference table:

Now,

Newton Backward Difference Method:

This method is useful for interpolation near the ending of a set of tabular values.

Where

Example1: Find from the following table:

 Consider the backward difference method

 Here

By Newton backward difference formula

Example2:The following table gives the amount of a chemical dissolved in water:

 Compute the amount dissolve at

Consider the following backward difference table:

Here

  By Newton Backward difference formula

Example3: The following are the marks obtained by492 candidate in a certain examination

Find out the number of candidates:

a)     Who secured more than 48 but not more than 50 marks?

b)    Who secured less than 48 but not less than 45 marks?

Consider the forward difference table given below:

        Here

 By Newton Forward Difference formula

f

a)     No. of candidate secured more than 48 but not more than 50 marks

b)    No. of candidate secured less than 48 but not less than 45 marks

Lagrange interpolation

Given a set of values of x and y, the process of finding the value of x for a certain value of y is called inverse interpolation.

Lagrange’s Inverse interpolation:

Let , be defined function we get

Where the interval is not necessarily equal. We assume f(x) is a polynomial of degree n. Then Lagrange’s inverse interpolation formula is given by

Example1: Use the inverse interpolation to find the value of x at for the following data:

Here , we have the data

Lagrange’s inverse interpolation formula is given by

.

Example2: Use the inverse  Lagrange’s method to find the root of the equation , give data

Here , we have the data

Also.

Lagrange’s inverse interpolation formula is given by

Thus the approximate root of the given equation is .

Example3: Find the value of x at for the following data:

Here , we have the data

Also.

Lagrange’s inverse interpolation formula is given by

Thus the value.

6.3 Numerical differentiation

Newton’s forward Difference formula:

This method is useful for interpolation near the beginning of a set of tabular values.

  Where

Differentiating both sides with respect to p, we get

h

This formula is applicable to compute the value of for non-tabular values of x.

For tabular values of x, we can get formula by putting

  Therefore

Similarly, we can get the formula for higher-order by differentiating the previous order formulas

  Again differentiating with respect to p, we get

Hence

Also

And so on.

Example1:Given that

Find at .

Here the first derivative is to be calculated at the beginning of the table, therefore forward difference formula will be used

Forward difference table is given below:

By Newton’s forward differentiation formula for differentiation

Here

Example2: Find the first and second derivatives of the function given below at the point :

Here the point of the calculation is at the beginning of the table,

Forward difference table is given by:

By Newton’s forward differentiation formula for differentiation

Here , 0.

Again

At

Example3: From the following table of values of x and y find for

 Here the value of the derivative is to be calculated at the beginning of the table.

Forward difference table is given by

From Newton’s forward difference formula for differentiation we get

Here

   =0.48763

Newton Backward Difference Method:

This method is useful for interpolation near the ending of a set of tabular values.

Where

Differentiating both sides with respect to p, we get

This formula is applicable to compute the value of for non-tabular values of x.

For tabular values of x, we can get formula by putting

  Therefore

Similarly, we can get the formula for higher-order by differentiating the previous order formulas

Differentiating both sides with respect to p, we get

Also

Example1:Given that

Find ?

Backward difference table:

Newton’s Backward formula for differentiation

Here

   Example2: Given that

Find the derivative of y at ?

The difference table is given below:

Since the point is at the beginning of the table therefore

From Newton’s forward difference formula for differentiation we get

Here

Since the point is at the end of the table therefore

Backward difference table is :

Newton’s Backward formula for differentiation

6.4 Numerical integration: Trapezoidal and Simpson’s Rules, Bound of truncation error

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In the case of the function of a single variable, the process is called quadrature.

Trapezoidal Method:

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n=1, we get

Or I =

The above is known as the Trapezoidal method.

Note: In this method second and higher differences are neglected and so f(x) is a polynomial of degree 1.

Geometrical Significance: The curve y=f(x), is replaced by n straight lines with the points ();() and ();…….;() and ().

The area bounded by the curve y=f(x), the ordinates, and the x-axis is approximately equivalent to the sum of the area of the n trapeziums obtained.

Example1: State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

Estimate the area bounded by the curve, the x-axis, and the extreme ordinates.

We construct the data table:

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x-axis =

Example2: Compute the value of  ?

Using the trapezoidal rule with h=0.5, 0.25, and 0.125.

Here

For h=0.5, we construct the data table:

By Trapezoidal rule

For h=0.25, we construct the data table:

By Trapezoidal rule

For h = 0.125, we construct the data table:

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Example3:Evaluate , using trapezoidal rule with five ordinates

   Here

We construct the data table:

Simpson’s Rule:

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of.

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

Note: In this rule third and higher differences are neglected a so f(x) is a polynomial of degree 2.

Example1: Estimate the value of the integral

by Simpson’s rule with 4 strips and 8 strips respectively.

For n=4, we have

Construct the data table:

By Simpson’s Rule

For n = 8, we have

By Simpson’s Rule

Example2: Evaluate

Using Simpson’s 1/3 rule with .

For , we construct the data table:

By Simpson’s Rule

Example3: Using Simpson’s 1/3 rule with h = 1, evaluate

For h = 1, we construct the data table:

By Simpson’s Rule

      = 177.3853

Simpson’s 3/8 rule

Let the interval be divided into n equal intervals such that <<….<=b.

Here.

To find the value of .

Setting n=3, we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

Note: In this rule, the fourth and higher differences are neglected and so f(x) is a polynomial of degree 3.

Example1: Evaluate

By Simpson’s 3/8 rule.

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

By Simpson’s 3/8 rule

      = 1.8278475

Example2: Evaluate

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

       =1.3571

Error in Simpson’s 3/8 Rule

The error in this rule is given by

Where is the largest value of the derivative of y(x).

Error in Trapezoidal method

The total error in the trapezoidal method is given by

Let is the largest value of the n quantities on the right-hand side of the above equation then

Error in Simpson’s Rule

The error in Simpson’s rule is given by

Where is the largest value of the fourth derivative of y(x).

Error in Simpson’s 3/8 Rule

The error in this rule is given by

Where is the largest value of the derivative of y(x).

6.5 Solution of ordinary differential equations: Euler’s method

Euler’s Method:

The general First-order differential equation

    With the initial condition

In this method, the solution is in the form of 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

Example1:Use Euler’s method to find y(0.4) from the differential equation

with h=0.1

Given equation

     Here

We break the interval into 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.

Example2:Using Euler’s method solves 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

Example3: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  

6.6 Modified Euler’s method

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 Euler’s formula

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

The 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 the third and fourth approximation is equal.

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

Example2: 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 the third and fourth approximation is 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 the first and second approximations 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 the third and fourth approximations are the same.

   Hence y = 0.25936 at x = 0.3

6.7 Runge-Kutta 4th order method, Predicator, and corrector

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 orderRunge-Kutta formula:

Where

Example1: Use Runge-Kutta method to find y when x=1.2 in the step of h=0.1 given that

Given equation

Here

Also

By Runge-Kutta formula for first interval

Again

A fourth orderRunge-Kutta formula:

To find y at

A fourth orderRunge-Kutta formula:

Example2: Apply Runge-Kutta fourth-order method to find an approximate value of y for x=0.2  in the step of 0.1, if

Given equation

Here

Also

By Runge-Kutta formula for first interval

A fourth orderRunge-Kutta formula:

Again

A fourth orderRunge-Kutta formula:

Example3: Using the Runge-Kutta method of fourth-order,  solve

Given equation

Here

Also

By Runge-Kutta formula for first interval

)

 A fourth orderRunge-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 orderRunge-Kutta formula:

Hence at x = 0.4 then y=1.37527

The solution of Second-order ODE using the 4th order Runge-Kutta method:

The Second-order differential equation

Let then the above equation reduces to a First-order simultaneous differential equation

Then

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

A fourth orderRungeKutta formula:

Where

Example1: UsingRungeKutta method of order four, solve to find

Given Second-order differential equation is

Let then above equation reduces to

   Or

(say)

Or .

By RungeKutta Method we have

A fourth orderRungeKutta formula:

Example2: Using the RungeKutta 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 RungeKutta Method we have

A fourth orderRungeKutta formula:

And  

.

Example3: Solve the differential equations

for

Using the four order RungeKutta method with initial conditions

Given differential equation are

Let

And

Also

 By RungeKutta Method we have

A fourth orderRungeKutta formula:

And  

.

Predictor and corrector-

Predicator and corrector method uses more than one previous step to calculate the next value of y.

A general form predictor and corrector method is-

Where are the constants to be determined.

Note-

When

, then the method is called the explicit method.

When

, then the method is called the implicit method.

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers