Unit-2

Interpolation

Q1) Define interpolation and what are the conditions for interpolation.

 A1) Interpolation

Definition: Interpolation is a technique of estimating the value of a function for any intermediate value of the independent variable while the process of computing the value of the function outside the given range is called   extrapolation.

Let be a function of x.

The table given below gives corresponding values of y for different values of x.

The process of finding the values of y corresponding to any value of x which lies between   is called interpolation.

If the given function is a polynomial, it is polynomial interpolation and given function is known as interpolating polynomial.

Conditions for Interpolation

1)     The function must be   a polynomial of independent variable.

2)     The function   should be either increasing or decreasing function.

3)     The value of   the function should be increase or decrease uniformly.

Q2) Explain forward differences.

A2) Forward Difference:  Then  are called differences of y, denoted by 

The symbol  is called the forward difference operator.  Consider the forward difference table below:

Where

And third forward difference so on.

Q3) What is the Newton forward difference formula.

A3) The Newton forward difference formula is defined as-

    Where

Q4) Using Newton’s forward difference formula, find the sum

A4) Putting

It   follows that

Since is a fourth degree polynomial in n.

Further,

By Newton Forward Difference Method

Q5) Given  find   , by using Newton’s forward interpolation method.

A5) Let , then

The table of forward finite difference is given below:

By Newton’s forward difference method

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

By Formula

Q6) Find from the following table:

A6) Consider the backward difference method

Here

By Newton backward difference formula

Q7) The following table give the amount of a chemical dissolved in water:

  Compute the amount dissolve at

A7) Consider the following backward difference table:

Here

By Newton Backward difference formula

Q8) The following are the marks obtained by 492 candidates 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?

A8) 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

Q9) Give the Gauss forward and backward formula.

A9) The Gauss forward difference formula is defined as-

The Gauss backward difference formula is defined as-

Q10) By using Gauss forward difference formula obtain f (32) given that-

f (25) = 0.2707   f (35) = 0.3386

 f (30) = 0.3027   f (40) = 0.3794

A10) Here

Let us take the origin at 30, a = 30 then we get-

u = 0.4

The forward difference table-

By using Gauss forward difference formula, we get-

Q11) Using Gauss’s backward interpolation formula, find the population for the year 1936 given that-

 A11) Let origin is at 1941 and h = 10

Then-

Which gives-

The backward difference table-

By using Gauss backward formula, we get-

Hence the pop. Of the year 1936 is 32625

Q12) By using Stirling’s formula to compute from the table ( given below-

A12)

Taking the origin at

We get the following central table-

At x = 12.2, p = 0.2

Stirling’s formula-

When p = 0.2, we get-

Q13) By using Bessel’s formula to find the value of f (27.5) from the table given below-

A13) Taking the origin at

We have p = x – 27

The central table will be as follows-

At x = 27.5, p =0.5

Bessel’s formula is-

When p = 0.5, we get-

So that-

f (27.5) = 3.585

Q14) Using Everett’s formula, evaluate f (30) if

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

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

A14) Let the origin is 28, a = 28, h = 8

Then

And

The difference table is-

By using Everette’s formula-

Therefore f (30) = 4064

Q15) By means of Newton’s divided difference formula, find the values of from the following table:

A15) We construct the divided difference table is given by:

By Newton’s Divided difference formula

.

Now, putting in above we get

.

Q16) 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.

A16) 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 

Q17) What is the Lagrange’s interpolation formula.

A17)  The Lagrange’s interpolation formula is given by

Q18) 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 position6 of the independent variable?

A18) We construct the table for the given data:

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.

Q19) By means of Lagrange’s formula, prove that

A19) We construct the table:

Here x = 3, f(x)=?

  By Lagrange’s formula for   interpolation

Hence proved.

Q20) Find the cubic polynomial by using Hermite’s interpolation formula, given-

A20) We know that Hermite’s interpolation formula-

..........(1)

Now

And

Hence

From equation (1)-