Unit-3

Numerical Integration and Differentiation

Q1) Given that

Find at .

A1) 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 

Q2) Find the first and second derivatives of the function given below at the point :

A2) 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

Q3) From the following table of values of x and y find   for 

A3) 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

Q4) Define Newton backward difference forward formula.

A4) This method is useful for interpolation near the ending of a set of tabular values.

Where 

Differentiating both side with respect to p, we get

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

Q5) Given that

Find ?

A5) Backward difference table:

Newton’s Backward formula for differentiation

Here

Q6) Given that

Find the derivative of y at ?

A6)  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

Q7) Write down the formula for Stirling, Bessel’s and Evertte’s.

A7)  Stirling’s formula-

Stirling’s formula is defined as-

Note-

The formula gives the best estimates when

Bessel’s formula-

The formula given below is called Bessel’s formula-

Note- this formula is useful when u =1/2 and gives best estimates when ¼<u<3/4

Everette’s formula-

The Everette’s formula is defined as-

Here w = 1 – u,

When u > ½, it gives best estimate.

Q8) By using Stirling formula to find , given-

A8)  Suppose the origin is at 30 and h = 5

a + hu = 28

Where h = 5 and a = 35 then-

u = -0.4

The difference table will be as follows-

By using Stirling formula, we get-

= 47691.8256

So that, we have

Q9) By using Stirling’s formula, compute from the table given below

A9)

Here let the origin is at , h = 1,

Then using Stirling’s formula-

Q10) By using Bessel’s formula to find . Given

A10)

The central difference table-

By using Bessel’s formula-

We get-

Q11) By using Everette’s formula, Evaluate f (30) if

f (20) = 2854,             f (28) = 3162

f (36) = 7088,             f (44) = 7984

A11) Let’s origin is 28,

A = 28, h = 8

A +hu = 30

28 + 8u = 30

U = 0.25

And w = 1-u = 1-0.25 = 0.75

The difference table is-

By using Everette’s formula-

So that the value of f (30) = 4064

Q12) By means of Newton’s divided difference formula, find the values of from the following table:

A12) We construct the divided difference table is given by:

By Newton’s Divided difference formula

.

Now, putting in above we get

.

Q13) 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.

A13) 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 

Q14) Write down the Lagrange’s inverse interpolation formula

A14) Lagrange’s inverse interpolation formula is given by

Q15) Use the inverse interpolation to find value of x at for the following data:

A15)  Here , we have the data

The Lagrange’s inverse interpolation formula is given by

.

Q16) Use the inverse   Lagrange’s method to find the root of the equation , give data

A16) Here , we have the data

Also.

The Lagrange’s inverse interpolation formula is given by

Thus, the approximate   root of the given equation is .

Q17) Define trapezoidal method.

A17) 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 Trapezoidal method.

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

Q18) Compute the value of   ?

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

A18) 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)]

Q19) Using Simpson’s 1/3 rule with h = 1, evaluate

A19) For h = 1, we construct the data table:

By Simpson’s Rule

= 177.3853

Q20) Define Simpson’s 3/8 rule.

A20) 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.

Q21) Evaluate

A21) 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