Unit-6

Numerical Method

Question-1: Using Newton’s divided difference formula, find the values of from the following table:

Solution: 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

.

Question-2: Find a polynomial satisfied by , by the use of Newton’s interpolation formula with a divided difference.

Solution: 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.

Question-3: Given find, by usingNewton forwardinterpolationmethod.

Solution: 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

Question-4: Find the missing term in the following:

	0	1	2	3	4
	1	3	9	?	81

Solution: Let

First, we construct the forward difference table:

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

Now,

Question-5: The following table gives the amount of a chemical dissolved in water:

Temp.
Solubility	19.97	21.51	22.47	23.52	24.65	25.89

Solution: 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

 By Newton Backward difference formula

Question-6: Use the inverse Lagrange’s method to find the root of the equation , give data

Solution: Here , we have thedata

Also.

Lagrange’s inverse interpolation formula is given by

Thus the approximate root of the given equation is .

Question-7: Given that

Find at .

Solution: 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:

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

Question-8: 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.

Solution: 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

Question-9: Given that

Find ?

Solution: 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

Question-10: 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.

Solution: 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 =

Question-11: Evaluate, using trapezoidal rule with five ordinates

 Here

Solution: We construct the data table:

X	0
Y	0	0.3693161	1.195328	1.7926992	1.477265	0

Question-12: Evaluate

Solution: 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

Question-13: Using Euler’s method solves the differential equation for y at x=1 in five steps

Solution:  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=1in equation (i) we get

For n=2in equation (i) we get

 For n=3in equation (i) weget

 For n=4in equation (i) weget

Hence

Question-14: Using modified Euler’s method, obtain a solution of the equation

Solution: Given equation

Here

By modified Euler’s formula, the initial iteration is

The iteration formula by modified Euler’s method is

-----(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 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=0in equation(ii)we get

1814

 For n=1in equation(ii)we get

1814

 Since the first and second approximations are equal.

Hence y= 0.1814atx=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=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

Question-15: Apply Runge-Kutta fourth-order method to find an approximate value of y for x=0.2 in the step of 0.1,if

Solution: Given equation

Here

Also

By Runge-Kutta formula for first interval

A fourth order Runge-Kutta formula:

Again

A fourth order Runge-Kutta formula:

Question-16: Using Runge Kutta method of order four, solve to find

Solution: 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:

Question-17: Solve the differential equations

for

Using the four order Runge Kutta method with initial conditions

Solution: Given differential equation are

Let

And

Also

 By Runge Kutta Method we have

A fourth order Runge Kutta formula:

And

.