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 lie between

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

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

Q6) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places) near to 4.5

A6)

A7)

Correct to three decimal places in the interval  ]

The given equation is       ... (1)

Or

Or =      ... (2)

Or

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 1.524.

Q8) By iteration method, find the value of , correct to three decimal places.

A8)

Let 

Let .

Also

Therefore the root of the equation lies between 3 and 4.

Given equation can rewrite  .

Or … (2)

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 3.634.

Q9) State the Jacobi iteration method.

A9)

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.

Q10) Use Gauss –Seidel Iteration method to solve the system of equations

A10)

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

Q11) Solve the following equations by Gauss-Seidel Method

A11)

A11)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Q12) Solve the equations-

A12)

A12)

Let

 

So that-

1. 
2. 

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 get the values of x, y and z

Q13) State Cholesky method.

A13)

Consider a system of equations

AX = B … (1)

If the coefficient matrix A is symmetric and positive definite then A can be decomposed as

Where L is a lower triangular matrix.

A may also be decomposed as where U is an upper triangular matrix.

From (1) and (2),

Where

The values , 1 i n can be obtained by forward substitution and the solution ,,1 i n is obtained by back substitution.

Q14) Construct a backward difference table for y = log x, given-

A14)

X	10	20	30	40	50
y	1	1.3010	1.4771	1.6021	1.6990

A14)

The backward difference table will be-

Q15) Given  find   , by using Newton forward interpolation method.

A15)

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 of interval h = 5 and the value to be calculated at x=52.

By Formula

Q16) Find from the following table)

A16)

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

A16)

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

Q17) y means of Newton’s divided difference formula, find the values of from the following table)

A17)

x	4	5	7	10	11	13
f(x)	48	100	294	900	1210	2028

A17)

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

.

Q18) Find a polynomial satisfied by , by the use of Newton’s interpolation formula with divided difference.

A18)

F(x)

1245

33

5

9

1335

Here

We will construct the divided difference table:

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.

Q19) Deduce Lagrange’s formula for interpolation.  The observed values of a function are respectively 168,120,72 and 63 at the four position3,7,9 and 10 of the independent variables. What is the best estimate you can for the value of the function at the position of the independent variable?

A19)

We construct the table for the given data:

X	3	6	7	9	10
Y=f(x)	168	?	120	72	63

We need to calculate for x = 6, we need f (6) =?

Here

We get

By Lagrange’s interpolation formula, we have

Hence the estimated value for x=6 is 147.

Q20) Find the polynomial of fifth degree from the following data

A20)

X	0	1	3	5	6	9
Y=f(x)	-18	0	0	-248	0	13104

A20)

Here

We get 

By Lagrange’s interpolation formula