Unit - 3

Polynomial interpolation

Q1) Explain interpolation.

A1)

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.

X				….
y= f(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.

Thus interpolation is the “art of reading between the lines of a table.”

Q2) What do you understand by forward difference?

A2)

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

A3)

The backward difference table will be-

10   20   30   40   50	1   1.3010   1.4771   1.6021   1.6990	0.3010   0.1761   0.1250   0.0969	-0.1249   -0.0511   -0.0281	0.0738   0.0230	-0.0508

Q4) Explain divided difference.

A4)

Definition:   The   difference which are defined by taking into consideration   the change in the value of the argument are known as  divided differences.

Let be a function defined as

			…….
			…………

Where are unequal i.e. it is 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:

Q5) What is Newton’s divided difference formula.

A5)

Let be a function defined as

			…….
			…………

Where are unequal i.e. it is case of unequal interval.

.

Q6) By means of Newton’s divided difference formula, find the values of from the following table:

A6)

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

.

Q7) 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.

A7)

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 

Q8) Find a polynomial satisfied by , by the use of Newton’s interpolation formula with divided difference.

A8)

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.

Q9) 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 variable. What is the best estimate you can for  the value of the  function  at the position6 of  the independent variable.

A9)

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.

Q10) By means of Lagrange’s formula, prove that

A10)

We construct the table:

X	0	1	2	3	4	5	6
Y=f(x)

Here x = 3, f(x)=?

By Lagrange’s formula for   interpolation

Hence proved.

Q11) Find the polynomial of  fifth degree from  the following data

A11) Here

We get 

By Lagrange’s  interpolation  formula

Q12) Explain about the errors in interpolation.

A12)

We know that, the interpolation means to estimate a missing function value by taking a weighted average of known function value at neighbouring points.

Let y = f(x) is the known function at (n + 1) points.

A polynomial of degree n. is known as the interpolating polynomial when passes through these (n + 1) points.

can be constructed using only the numerical values and and does not need any order derivatives of f (x). The approximation is known as interpolated value when and extrapolated value when or .

Note- Error increases when the number of interpolation is increased.

The error of the polynomial is obtained by the generalized Rolle’s theorem,

It is given as-

Here contains {x, }

Note- that the error term given above is same for Lagrange polynomial and interpolating polynomial obtained from divide difference table

Q13) What are average and shift operators?

A13)

Averaging operator (:

It is defined as-

Shift operator E:

It is the operation of increasing the argument x by h so that

Ef(x) = f(x + h)

f(x) = f(x + 2h) and so on.

The inverse operator is defined by

Q14) Prove that

A14)

By taking left hand side, we get-

Q15) Prove that-

A15)
 

From (1)-

Hence proved.

Q16) 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

A16)

Here

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

u = 0.4

The forward difference table-

u	x	F(x)
-1   0   1   2	25   30   35   40	0.2707   0.3027   0.3386   0.3794	0.032   0.359   0.408	0.0039   0.0049	0.0010

By using Gauss forward difference formula, we get-

Q17) Using Gauss’s backward interpolation formula, find the population for the year 1936 given that-

A17)

Let origin is at 1941 and h = 10

Then-

Which gives-

The backward difference table-

u	F(u)
-4   -3   -2   -1   0   1	12   15   20   27   39   52	3   5   7   12   13	2   2   5   1	0   3   -4	3   -7	-10

By using Gauss backward formula, we get-

thousands

Hence the pop. Of the year 1936 is 32625

Q18) Explain spline interpolation.

A18)

The approximation of arbitrary functions on a closed interval becomes oscillatory for higher-degree polynomials. By dividing the given interval into a collection of subintervals, piecewise polynomial approximation consists of constructing a different approximating polynomial on each subinterval. Cubic spline interplation is the most common piecewise approximation using cubic polynomial, known as spline function, connecting each pair of data points. Cubic splines are third-order curves employed to connect each pair of data points.

Q19) Fit a linear spline of the data given below:

A19)

Here we have, n = 4 and there are 3 intervals.

Let us assume that-

,     ,       ,       

,     ,       ,       

And the slope,

Similarly we can find,

As we know that the

The three splines will be-

Q20) What do you understand by quadratic spline?

A20)

These are called the second order splines.

For each interval, we fit a parabola give by-

For n + 1 data points we have n intervals. So we should determine 3 × n unknown constants. The 3n condition to find the 3n unknowns are given by the following equations.

1. At the interior knots, the function values must be-

For i = 2, n giving raise to 2(n − 1) = 2n – 2 conditions.

2.     The first and last functions must pass through the two end points

Giving two conditions.

3.     At the interior knots, the first derivatives f (x) = 2ax + b must be equal.

For i = 2 to n giving n − 1 conditions

4.     Assume that the second derivative is zero at the first point. Then the first two points are connected by a straight line, so

Thus solving (2n − 2) + 2 + (n − 1) + 1 = 3n conditions above we obtain the 3n unknown constants a, b and c.