Unit-6

Numerical Methods

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 asdivided 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:

X	f(x)

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:

x	f(x)	First-order divide difference	Second-order divide difference	Third-order divide difference	Fourth-order divide difference
4     5     7     10     11     13	48     100     294     900     1210     2028				0      0

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:

x	f(x)	First-order divide difference	Second-order divide difference	Third-order divide difference
1     2     7     8	4     5     5     4			0

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 differencetable:

x	F(x)	First-order divided difference	Second-order divided difference	Third-order divided difference	Fourth-order divided difference
-4     -1     0     2     4	1245     33     5     9     1335

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 tabularvalues.

Where

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

Putting

It follows that

Since isafourth-degree polynomial inn.

Further,

By Newton Forward Difference Method

Example2:Givenfind, by usingNewton forwardinterpolationmethod.

Let, then

	0.7071	0.7660	-	0.8192	0.8660

The table of forward finite difference is given below:

45   50   55    60	0.7071    0.7660   0.8192    0.8660	0.0589    0.0532    0.0468	-0.0057    -0.0064	-0.0007

By Newton forward difference method

Here initial value = 45, difference ofintervalh = 5 andthe value to be calculated at x=52.

By Formula

Example3: Find the missing term in the following:

	0	1	2	3	4
	1	3	9	?	81

Let

First, we construct the forward difference table:

0   1    2    3    4	1   3   9     81	2   6	4

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:

	0.20	0.22	0.24	0.26	0.28	0.30
	1.6596	1.6698	1.6804	1.6912	1.7024	1.7139

 Consider the backward difference method

0.20   0.22   0.24   0.26   0.28   0.30	1.6596    1.6698    1.6804    1.6912   1.7024   1.7139	0.0102   0.0106    0.0108    0.0112   0.0115	0.0004    0.0002    0.0004    0.0003	-0.0002    0.0002   -0.0001	0.0004   -0.0003	-0.0007

Here

By Newton backward difference formula

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

Temp.
Solubility	19.97	21.51	22.47	23.52	24.65	25.89

Compute the amount dissolve at

Consider the following backward difference table:

Temp.x	Solubility y
10   15   20   25   30   35	19.97    21.51   22.47   23.52    24.65    25.89	1.54   0.96    1.05   1.13   1.24	-0.58    0.09    0.08    0.11	0.67   -0.01    0.03	-0.68   0.04	0.72

Here

 ByNewtonBackwarddifference formula

Example3:The following are the marks obtained by492candidatein a certain examination

Findout the number of candidates:

a)     Who secured more than 48butnot more than50 marks?

b)    Who secured less than 48butnotless than 45 marks?

Consider theforward difference table given below:

Marksupto x	No. of candidates y
40   45   50   55   60   65	210   210+43=253   253+54=307   307+74=381   381+32=413   413+79= 492	43   54   74   32    79	11   20   -42    47	9   -62    89	-71    151	222

 Here

 By Newton Forward Difference formula

f

a)     No. ofcandidate secured more than 48butnot more than50 marks

b)    No. ofcandidate secured less than 48 butnotlessthan 45 marks

Lagrange interpolation

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

Lagrange’s Inverse interpolation:

Let , bedefinedfunction we get

x				…..
f(Y)				……

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 thedata

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 thedata

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 thedata

Also.

Lagrange’s inverse interpolation formula is given by

Thus the value.

Key takeaways-

2.     Newton Backward Difference Method:

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

Where

3.     Then Lagrange’s inverse interpolation formula is given by

6.3 Numerical differentiation

Newton’s forward Difference formula:

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

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, thereforeforward differenceformula will be used

Forward differencetable is given below:

X	Y
1.0   1.1   1.2   1.3	0.841   0.891   0.932   0.962	0.050   0.041   0.031	-0.009   -0.010	-0.001

By Newton’s forward differentiation formula for differentiation

Here

Example2: Findthe first and secondderivatives of the function given below at thepoint :

Here the point of the calculation isat the beginning of thetable,

Forward difference table is givenby:

X	Y
1   2   3   4   5	0   1   5   6   8	1   4   1   2	3   -3   1	-6   4	-10

By Newton’s forward differentiation formula for differentiation

Here, 0.

Again

At

Example3: From the followingtable of values of x and y find for

Here thevalue of the derivative is to becalculated atthe beginning of the table.

Forward difference table is given by

X	Y
1.00   1.05   1.10   1.15   1.20   1.25   1.30	1.0000   1.02470   1.04881   1.07238   1.09544   1.11803   1.14017	0.02470   0.02411   0.02357   0.02306   0.02259   0.02214	-0.00059   -0.00054   -0.00051   -0.00047   -0.00045	0.00005   0.00003   0.00004   0.00002	-0.00002   0.00001   -0.00002	0.00003   -0.00003	-0.00006

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:

X	Y
0.1   0.2   0.3   0.4	1.10517   1.22140   1.34986   1.49182	0.11623   0.12846   0.14196	0.01223   0.01350	0.00127

Newton’s Backward formula for differentiation

Here

 Example2: Given that

Find the derivative of y at ?

Thedifference table isgiven below:

X	Y
1.0   1.2   1.4   1.6   1.8   2.0	0   0.128   0.544   1.296   2.432   4.0	0.128   0.416   0.752   0.136   1.568	0.288   0.336   0.384   0.432	0.048   0.048   0.048	0   0

Since the pointisat the beginning of the table therefore

From Newton’s forward difference formula for differentiation we get

Here

Since thepoint is at theend of thetabletherefore

Backward difference table is :

X	Y
1.0   1.2   1.4   1.6   1.8   2.0	0   0.128   0.544   1.296   2.432   4.000	0.128   0.416   0.752   0.136   1.568	0.288   0.336   0.384   0.432	0.048   0.048   0.048	0   0

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:

X	0
Y	0	0.3693161	1.195328	1.7926992	1.477265	0

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:

X	0
	0	0.50874	0.707106	0.840896	0.930604	0.98281	1

By Simpson’s Rule

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

For h = 1, we construct the data table:

X	3	4	5	6	7
	9.88751	22.108709	40.23594	64.503340	95.34959

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:

X	4.0	4.2	4.4	4.6	4.8	5.0	5.2
	1.3863	1.4351	1.4816	1.5261	1.5686	1.6094	1.6487

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:

X	0	1	2	3	4	5	6
	1	0.5	0.2	0.1	0.0588	0.0385	0.027

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

Inthismethod,the solution is in the formof tabulatedvalues.

 Integrating both side of the equation (i) we get

 Assuming thatinthisgives Euler’s formula

 In generalformula

, n=0,1,2,…..

Error estimate for the Euler’s method

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

with h=0.1

Givenequation

Here

Webreaktheinterval into four steps.

 So that

By Euler’sformula

, n=0,1,2,3 ……(i)

 For n=0 in equation (i) weget

For n=1 in equation (i) weget

.01

 For n=2 in equation (i) weget

For n=3 in equation (i) weget

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 stepsn=5 andso that

So that

 Also

By Euler’sformula

, n=0,1,2,3,4 ……(i)

 For n=0 in equation (i) weget

For n=1in equation (i) weget

For n=2in equation (i) weget

 For n=3in equation (i) weget

 For n=4in equation (i) weget

Hence

Example3:Given withthe initial conditiony=1 at x=0.Findy for x=0.1 byEuler’s method(fivesteps).

Given equation is

Here

No. of stepsn=5 and so that

So that

 Also

By Euler’sformula

, n=0,1,2,3,4 ……(i)

 For n=0 in equation (i) weget

For n=1 in equation (i) weget

 For n=2 in equation (i) weget

 For n=3 in equation (i) weget

 For n=4 in equation (i) weget

Hence

6.6 Modified Euler’s method

Modified Euler’sMethod:

Instead ofapproximating as in Euler’s method.Inthemodified Euler’smethod, wehave the iteration formula

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

Example1:Use modifiedEuler’s methodto compute yforx=0.05.Given that

The result correctto three decimal places.

Givenequation

Here

Takeh= = 0.05

By modified Euler’sformula, theinitial iteration is

)

The iterationformula bymodified Euler’s methodis

-----(i)

 For n=0in equation(i)we get

Where andasabove

For n=1in equation(i)we get

 For n=3in equation(i)we get

Sincethe third andfourth approximation is equal.

Hence y=1.0526at x =0.05correcttothree decimal places.

Example2: Using modified Euler’s method,obtain a solution ofthe equation

Given equation

Here

By modified Euler’sformula, theinitial iteration is

The iterationformula bymodified Euler’s methodis

-----(i)

 For n=0in equation(i)we get

Where andasabove

For n=1in equation(i)we get

For n=2 in equation(i)we get

For n=3in equation(i)we get

Since the third and fourth approximation is equal.

Hence y=0.0952 atx=0.1

To calculatethe valueof at x=0.2

By modified Euler’sformula, theinitial iteration is

The iterationformula bymodified Euler’s methodis

-----(ii)

For n=0in equation(ii)we get

1814

 For n=1in equation(ii)we get

1814

 Since the first and second approximations areequal.

Hencey= 0.1814atx=0.2

Tocalculate the valueof at x=0.3

By modified Euler’sformula, theinitial iteration is

The iterationformula bymodified Euler’s methodis

-----(iii)

For n=0in equation(iii)we get

For n=1in 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

Rewriteas

A fourthorderRunge-Kutta formula:

Where

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

that

Given equation

Here

Also

ByRunge-Kutta formula for firstinterval

Again

A fourthorderRunge-Kutta formula:

Tofind y at

A fourthorderRunge-Kutta formula:

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

Givenequation

Here

Also

ByRunge-Kutta formula for firstinterval

A fourthorderRunge-Kutta formula:

Again

A fourthorderRunge-Kutta formula:

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

Given equation

Here

Also

ByRunge-Kutta formula for firstinterval

)

 A fourthorderRunge-Kutta formula:

Hence at x= 0.2 then y= 1.196

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

 A fourthorderRunge-Kutta formula:

Hence atx= 0.4 theny=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 fourthorderRungeKutta 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 fourthorderRungeKutta 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 fourthorderRungeKutta 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 fourthorderRungeKutta 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