4.0 INTRODUCTION

Let be a function of x. Suppose we are given a set of value of y, say corresponding to given values of x, say x₀, x₁, x₂, …, x_i, …, x_n. The values of x are called arguments and values of y are called entries.

Interpolation may be defined as the process of finding the values of a function f(x) for any intermediate value of the argument x between x₀ and x_n. This process has been described by Thiele as “the art of reading between the lines of a table”.

In the broader sense interpolation is the process of replacing the unknown function or complicated function f(x) by a simpler function ϕ(x) which assumes the same values of y for the given values of x. The function ϕ(x) is called the interpolating function or interpolating formula. In many engineering applications this function is called a smoothing function.

A desirable characteristic of interpolating function is that it must be simple. Since polynomials are the simplest of the functions, we usually choose ϕ(x) to be a polynomial and so it is called interpolating polynomial. Nearly all the standard formulae of interpolation are polynomials. In case the given values indicate that the function is periodic, we represent it by a finite trigonometric series. But we consider here only polynomial interpolation.

The process of representing by a polynomial is justified by the Weierstrass theorem: “Every function f(x), which is continuous in an interval (a,b) can be represented in that interval, to any desired degree of accuracy by a polynomial ϕ(x)”.

Some of the important interpolating polynomial formulae are

1. Newton’s forward and backward formulae
2. Newton’s divided difference formula
3. Lagrange’s formula
4. The central difference formulae
   1. Gauss forward and backward formulae
   2. Stirling’s formula
   3. Everrett’s formula
   4. Bessel’s formula etc

All these formulae are expressed interms of difference operators of various types. So we shall first introduce the different difference operators.

Extrapolation: The process of obtaining the value of a tabulated function outside the interval of the given values of the arguments is called extrapolation or prediction.

4.1 FINITE DIFFERENCE OPERATORS

When the changes in the independent variable or argument of a function is discrete, the infinitesimal calculus cannot be applied to study such functions. But the calculus of finite differences enable us to study such functions. The calculus of finite differences form the basis of many processes and is used in the derivation of many formulae in numerical analysis.

4.1.1 Forward Difference Operator Δ

Let y₀, y₁, y₂,…, y_n be the values of the function at equally spaced arguments x₀, x₁, x₂,…, x_n respectively.

Let the equal space or interval be h.

Then

Consider the differences

We denote these as

and are called first order forward differences or first differences and Δ is called the forward difference operator.

The difference of first order differences are called second order differences.

They are

Similarly we can find 3^rd, 4^th, …, n^th order differences.

Thus

If , then

Forward difference table

The first entry y₀ is called the leading term and the differences in the diagonal through y₀ are called the leading differences.

4.1.2 Backward Difference Operator ∇ (Read as Del or Nebla)

The differences are denoted by and are called first order backward differences. The operator ∇ is called backward difference operator.

∴

The differences

are called second order backward differences and so on.

The backward difference table is

Prove that

Prove:

We have

Similarly

4.1.3 Shift Operator or Displacement Operator E

The operator E is defined by where h is the interval of differencing. That is E shifts to next higher value

In other words , where and

The operator is defined as , where h is the interval of differencing.

That is shifts to preceeding lower value .

In other words, where and

Note: and where 1 is the identity operator

4.1.4 Relation Between the Operators E, Δ, ∇

1. Prove that where 1 is the identity operator.

   Proof:

   We have

   

   and

   

   

   ∴ for any

   ∴

   Similarly

   ,

   

   Note: The method of writing the operator equality is known as the method of separation of symbols. However it should be remembered that operators cannot really stand alone and the operand is always understood.

2. Prove that

   Proof:

   We have

   

   and

   

   

   

   ∴ for any

   ∴

4.1.5 Properties of Δ and E

1. Δ is linear operator

   ie.

   and , where C is a constant.

2. E is a linear operator

   ie.

   and

3. Index laws

   If m and n are positive integers, then

   

   Similarly

   

   is the inverse operator of

   is the inverse operator of

4. Prove that Δ∇ = ∇Δ

   Proof:

   We have

   

   

   ⇒

   

   

5. Prove that

   Proof:

   We have

   

   and

   

   

   

   ∴ for any

   ∴

6. Prove that

   Proof:

   We have

   

   

   

   ∴ for any

   ⇒

   Because of this property, we shall write each side

   Theorem 4.1

   The n^th differences of a polynomial of n^th degree are constant.

   ie. If is an n^th degree polynomial, then

   constant, where h is the interval of differencing.

   Note:

   1. More generally,
   2. The converse of this theorem is also true.

   That is, if the n^th differences of a function tabulated at equally spaced arguments are constants, then the function is a polynomial of n^th degree.

   The Converse is of great use in practical situations. If in a difference table the n^th differences are constant or nearly so (since rounding off errors may prevent them being exactly equal), then the function may be represented by a polynomial of n^th degree by a suitable interpolation formula.

7. The Central difference operator

   The Central difference operator δ is defined by .

   

8. The averaging operator or the mean value operator μ is defined by

   

   

9. Prove that

   Proof:

   

   

   

10. Prove that

    Proof:

    We have

    (1)

    and

    

    

11. Relation between differentical operator and difference operator

    We know ,

    

    We have Taylor’s series for as

    

    But

    

    ∴

WORKED EXAMPLES

Example 1

If then show that where k is a constant.

Solution

Given

We have

Example 2

Prove that

Solution

Example 3

Test whether and are equal.

Solution

(2)

∴ from (1) and (2), we get

Example 4

Find the value of taking .

Solution

Let

Example 5

Prove that .

Solution

(1)

(2)

From (1) and (2),

Example 6

Evaluate if the interval of differencing is 2.

Solution

Let

Given

We know that for a polynomial of n^th degree , where h is the interval of differencing and is coefficient of .

When is expanded as a polynomial in x, the leading term is

Example 7

Prove the following with usual notation.

1. 
2. (1-—)
3. 

Solution

1. We know that

   

2. we have

   

   

3. We have

   and

   

Example 8

Prove that .

Solution

We know that

and

From (1) and (2), we get

Example 9

Using the method of separation of symbols, prove that

Solution

We know

Example 10

Using the method of separation of symbols, prove that

Solution

Example 11

Using the method of separation of symbols prove that

Solution

[ = ΔE^r]

Example 12

Show that

Solution

Example 13

If and . Obtain the values of x, assuming the second differences are constant.

Solution

Given

and

.

∴ the arguments are 1, 2, 3, 4. Since the second differences are constants, third differences are zero.

Example 14

If find the value of assuming the second differences are constant.

Solution

Given the arguments are 0, 1, 2, 3, 4, 5 and and the second differences are constant.

We form the difference table.

Since the second differences are constant, say k, we have

(1)

(2)

(3)

(4)

(5)

(6)

Substituting in (3), we get

(7)

(8)

Given

Example 15

Estimate the missing value in the table.

x	0	1	2	3	4
	1	3	9	–	81

Solution

Since four values are given, the fourth differences are zero.

Let a be the missing value

We from the difference table

Since , we get

⇒

∴ the missing value is 31

Aliter:

Since four values are given, the fourth differences are zero.

Put

Example 16

Obtain the missing values in the table.

x	1	2	3	4	5	6	7	8
	1	8	–	64	–	216	343	512

Solution

Since six values are given, the sixth differences are zero.

(1)

Put in (1), we get

(2)

Put in (1), we get

(3)

Subtracting, we get

Exercises 4.1

1. Evaluate (a) (b) (c) taking h as the interval of differencing.
2. Taking interval of differencing as h, show that
   1. (a) (b)
   2. (c) (d)
3. Show that
4. Prove the following with usual notation
   1. (a) (b)
   2. (c) (d) and
5. Using method of separation of symbols, prove the following.
   1. (a)
   2. (b)

      

   3. (c)
   4. (d)
6. Find the missing values from the table.

   x	0	1	2	3	4
   yx	1	2	4	–	16

7. Find the missing values from the table.

   x	0	1	2	3	4	5	6
   f(x)	–4	–2	–	–	220	546	1148

8. Assuming as a polynomial of degree 4, compute the missing values from the following table.

   x	0	1	2	3	4	5	6	7
   yx	0	1	2	1	0	–	–	–

Answers 4.1

(6) (7) (8)

4.1.6 Factorial Polynomial

The product of the form starting with x and the successive factors decrease by a constant is called factorial polynomial of degree r and it is denoted by , where r is a positive integer

If

we shall use this factorial polynomial in our discussions.

Differences of

similarly

Note:

1. Since

   it is analogous to differentiation

   When r = 0, [x]⁰ = 1

2. 

So inverse operator behaves like integration for factorial polynomial.

Thus

Because of this special property, it is convenient to represent a polynomial in terms of factorial polynomial.

Example 1

Represent as a factorial polynomial.

Solution

Let

where a₁, a₂, a₃ are constants.

Put x = 0, then a₃ = 12.

Put x = 1, then a₂ + a₃ = 1 + 3 + 12 + 12

⇒ a₂ + 12 = 16 + 1 a₂ = 16

Put x = 2, then

a₁ 2 (2 - 1) + a₂ 2 + a₃ = 8 + 12 + 24 + 12

⇒ 2a₁ = 56 - 2a₂ - a₃

= 56 - 2 × 16 - 12 ⇒ a₁ = 6

∴ x³ + 3x² + 12x + 12 = [x]³ + 6[x]² + 16[x] + 12

Aliter:

By Synthetic division, we can find the coefficients a₁, a₂, a₃.

∴ x³ + 3x² + 12x + 12 = [x]³ + 6[x]² + 16[x] + 12

Example 2

Represent and its successive differences in factorial notation.

Solution

Let

We shall use synthetic division method to express f(x) in factorial notation.

and higher differences are zero

4.2 INTERPOLATION WITH EQUALLY SPACED ARGUMENTS OR INTERPOLATION WITH EQUAL INTERVALS

Let the function y = f(x) take values for equidistant values of the arguments Let the equal interval be h ie. Then We have to find the interpolating polynomial which represents f(x) in the interval

Since (n + 1) values of the function are known, we can assume to be a polynomial of n^th degree and it is determined uniquely. at and approximately equal in intermediate points. Hence we write f(x) itself in the place of

4.2.1 Newton’s Forward Formula for Interpolation

Newton’s forward formula is

where

Proof: Let y = f(x) be a function which takes values at equally spaced arguments Let h be the interval so that Let be the interpolating polynomial of the n^th degree which may be written in the form

(1)

where are constants to be determined such that

Put successively in (1)

we get

Similarly, and

∴ (1) becomes

(2)

This is Newton’s forward formula interms of x.

Now

Now

Remark:

1. Newton forward formula is also known as Newton-Gregory formula for forward interpolation.
2. The formula involves y₀ and its leading differences. ie. it involves y₀ and the values of the function to the right of y₀. Hence it is called forward interpolation formula.
3. The formula is used for interpolating the values of y near the beginning of a set of tabular values.
4. It can also be used for extrapolating values of y at a short distance on the left side of x₀ (ie. x < x₀ and x is close to x₀).

4.2.2 Newton’s Backward Formula for Interpolation

Newton’s backward formula is

where

Proof: Let y = f(x) be a function which takes values for equally spaced arguments

where

∴

Let be the interpolating polynomial of n^th degree which may be written as

(1)

where are constants to be determined in such a way that

Substituting in succession in (1) we get

Similarly,

Substituting in (1), we get

Remark:

1. This formula is also known as Newton-Gregory’s backward formula.
2. Since the formula involves y_n and the values of the function to the left of it, it is called backward formula.
3. It is used for interpolating values of y near the end of a set of tabular values.
4. It can also be used for extrapolating values of y at a short distance on the right of x_n (ie.x > x_n and x is close to x_n).

Note: If the tabulated function is a polynomial then for any value of x, both the forward and backward formula of Newton will give the exact value of the function whether it is interpolation or extrapolation.

WORKED EXAMPLES

Example 1

Using Newton’s forward interpolation formula find the cubic polynomial which takes the following values.

x	0	1	2	3
f(x)	1	2	1	10

Evaluate f(4).

Solution

We use Newton’s forward formula to find the polynomial in x.

Newton’s forward formula is

where

Here ∴

Now we form the forward difference table.

∴

When x = 4,

Example 2

A third degree polynomial passes through the points (0, −1), (1, 1), (2, 1) and (3, –2). Using Newton’s forward formula, find the polynomial. Hence find the value at 1.5.

Solution

We use Newton’s forward formula to find the polynomial passing through (0, –1) (1, 1), (2, 1) and (3, –2).

Newton’s forward formula is

where

Here

(1)

Now we form the forward difference table.

Substituting in (1), we get

Since u = x, the polynomial is

Example 3

The population of a city in Census taken once in 10 years is given below. Estimate the population in the year 1955.

Year	1951	1961	1971	1981
Population in thousands	35	42	58	84

Solution

Let us denote year as x, and population as y.

Let y = f(x)

x = 1955 is near the beginning of the table. So we use Newton’s forward difference formula to find y.

Newton’s forward formula is

(1)

where . Here

When

Now we form the forward difference table.

∴

and when x = 1955, u = 0.4

Substituting in (1), we get

Example 4

From the data given below find the number of students whose weight is between 60 to 70.

Weight in lbs:	0–40	40–60	60–80	80–100	100–120
No. of students	250	120	100	70	50

Solution

Let weight be denoted by x and number of students be denoted by y

Let y = f(x)

We use Newton’s forward formula to find y when x lies between 60–70.

We rewrite the table as cumulative table showing the number of students less than x lbs.

Newton’s forward formula is

(1)

where .

Here ∴

We shall find y when x = 70

∴

Now we form the forward table.

Hence

and when

Substituting in (1), we get

No. of students whose weight is below 70 is 424

∴ no. of students whose weight is between 60 – 70 is = 424 – 370 = 54

Example 5

If

x	0	1	2	3	4	5
y	27	32	25	36	32	41

then find approximately the value of y when x = −0.5.

Solution

Let y =f(x). To find y when x = –0.5.

x = –0.5 is outside the table value and is near the beginning of the table.

∴ It is extrapolation. So, we use Newton’s forward formula to find y when x = –0.5

Newton’s forward formula is

(1)

where . Here

When

Now we form the difference table.

Hence

and when

Substituting in (1), we get

Example 6

The following table gives melting point of an alloy of zinc and lead, θ is the temperature and x is the percentage of lead. Using Newton’s interpolation formula find θ when x = 84.

x	40	50	60	70	80	90
θ	184	204	226	250	276	304

Solution

Let y = f(x), where y = θ,

x = 84 is near the end of the table.

∴ we use Newton’s backward formula to find θ when x = 84.

Newton’s backward formula is

(1)

where

When

,

Now we form the table.

Hence

and when

Substituting in (1) we get

Example 7

If l_x represents the number of people living at age x in a life table, find l₄₇, given l₂₀ = 512, l₃₀ = 439, l₄₀ = 346 and l₅₀ = 243.

Solution

Let y = l_x.

x = 47 is near the end of the table values. So, we use Newton’s backward formula to find l₄₇.

Newton’s backward formula is

(1)

where . Here ∴

When

Now we form the difference table.

∴

and when x = 47, v = –0.3

Substituting in (1), we get

Hence the number of people expected to live at the age of 47 is 274.

Example 8

A function y is given by the following table. Estimate the value of y when x = 5.

x	0	1	2	3	4
y	79	91	105	116	127

Solution

Let y = f(x).

We want to find y when x = 5, which is outside the end of the table and hence extrapolation. So, we use Newton’s backward formula to find y when x = 5.

Newton’s backward formula is

(1)

where . Here ∴

When

Now we form the difference table.

Here

and when

Substituting in (1), we get

Example 9

The following are data from the steam table.

Temperature °C	140	150	160	170	180
Pressure kg/cm2	3.685	4.854	6.302	8.076	10.225

Find the pressure at temperature 142°C and 175°C.

Solution

Let temperature be x°C and pressure be y kg/cm²

Let y = f(x)

Since x = 142°C is near the beginning of the table, we use Newton’s forward formula to find y.

Newton’s forward formula is

(1)

where . Here ∴

When

We now form the forward difference table.

Here

When temperature is x = 142°C, then pressure

(ii) Since x = 175°C is near the end of the table, we use Newton’s backward formula.

Newton’s backward formula is

(1)

where

When

From the difference table, we have

Example 10

Estimate the value of f(22) and f(42) from the following data:

x	20	25	30	35	40	45
f (x)	354	332	291	260	231	204

Solution

We find f(22) using Newton’s forward formula.

Newton’s forward formula is

(1)

where . Here ∴

When

From forward difference table

To find y when

Given y = f(x)

x = 42 is near the end of the table. So, we use Newton’s backward formula to find f(42).

Newton’s backward formula is

(1)

where . Here

When

and when x = 42, v = –0.6

Substituting in (1) we get

f(42) = 218.66 = 219 approximately

Exercises 4.2

1. Use Newton’s backward formula to find a polynomial of degree 3 for the data f(–0.75) = –0.0781250, f(–0.5) = –0.024750, f(–0.25) = 0.33493750, f(0) = 1.10100. Hence find
2. From the following table, find the value of tan 45° 15′

   x0	45	46	47	48	49	50
   tan x0	1.0	1.03553	1.07237	1.11061	1.15037	1.19175

3. Find f(2.5) using Newton’s forward difference formula for the given data.

   x	1	2	3	4	5
   	0	1	8	27	64

4. Using Newton’s backward difference formula find the area of a circle of diameter 98 from the given table of diameter and area of a circle.

   Diameter	80	85	90	95	100
   Area	5026	5674	6362	7088	7854

5. Find given that
6. (6) A function f(x) is given by the following table. Estimate f(0.2) by any appropriate formula.

   x	0	1	2	3	4	5	6
   f(x)	176	185	194	203	212	220	229

7. Find the value of f(x) when x = 32, given the following table:

   x	30	35	40	45	50
   f(x)	15.9	14.9	14.1	13.3	12.5

8. Find the number of students from the following data who secured marks not more than 45.

   Marks	30–40	40–50	50–60	60–70	70–80
   No of students	35	48	70	40	22

9. The following are the annual premiums charge by the L.I.C of India for a policy of Rs. 1000. Calculate the premium payable at the age of 22.

   Age in year	20	25	30	35	40
   Premium Rs.	23	26	30	35	42

10. Calculate the value of sin 33° 13′ 30″ from the following table of sines.

    x°	30	31	32	33	34
    sin x°	0.5000	0.5150	0.5290	0.5446	0.5592

11. The table gives the distance in nautical miles of the visible horizon for the given height in feet above the earths surface. Find the values of y when x = 218′ and x = 410′.

    x	100	150	200	250	300	350	400
    y	10.63	13.03	15.04	16.81	18.42	19.90	21.27

12. Find u₆ given that

    u0	u1	u2	u3	u4	u5
    25	25	22	18	15	5

13. By appropriate formula estimate the population for the year 2006, given

    Year:	2001	2002	2003	2004	2005
    Population in thousands:	251	279	319	383	483

14. A function y is given by the following table. Estimate the value of y when x = 5.

    x	0	1	2	3	4
    y	79	91	105	116	127

15. Two variables x and y have the following related values

    x	0	10	20	30	40	50	60
    y	2501	2795	2838	3030	3050	3381	3888

    Find y when x = 54.

16. Given

    x	0.5	0.6	0.7	0.8	0.9	1.0	1.1	1.2
    y = f(x)	0.34375	0.87616	1.47697	2.17408	3.00139	4	5.21941	6.71872

    Find f(1.18).

Answers 4.2

1. (1) ,
2. (2) 1.00876 (3) 3.4 (4) 7543 (5) –161
3. (6) 177.67 (7) 15.45 (8) 51 (9) 24.04
4. 0.54788 (11) 15.697, 21.53 (12) 20 (13) 631
5. 149 (15) 3674.97 (16) 6.39328

4.3 CENTRAL DIFFERENCE INTERPOLATION FORMULAE

We have seen that Newton’s formward formula is suitable to interpolate near the beginning of a table of values and Newton’s backward formula is suitable to interpolate near the end of the table with equally spaced arguments.

For interpolating near the middle of the table the central difference formulae are better suited than others. The central difference formulae employ the difference lying as nearly as possible on a horizontal line through y₀ near the centre.

We will discuss the following central difference formulae.

1. Gauss’s forward formula
2. Gauss’s backward formula
3. Stirling’s formula
4. Bessel’s formula
5. Everett’s formula.

Of these Bessel’s and Stirlings are the most important ones.

Central difference table

Let be the function. Let be the values of y corresponding to and be the values corresponding to

Then, the difference table with these values on either side of is given by the table below.

Put , where is the origin.

The values of u corresponding to

are respectively

4.3.1 Gauss’s Forward Formula for Interpolation

Gauss’s forward Interpolation formula is

Proof: We derive Gauss’s forward difference interpolation formula from Newton’s forward formula and using even differences on the horizontal line through and odd differences on the horizontal line between and from the central difference table.

Newton’s forward formula is

(1)

where

From the Central difference table, we have

Similarly,

Similarly, and so on.

Substituting for in (1) we get

If the symbol , then the formula can be written as

Note:

1. Gauss’s forward formula involves even differences on the central line and odd differences below the central line as
   
2. Gauss’s forward formula is mainly used to interpolate if

   .

3. Since it is measured from the origin forwardly, this formula is called forward formula.

WORKED EXAMPLES

Example 1

Use Gauss’s forward formula to find the value of y when from the following table.

x	1.5	2	2.5	3	3.5	4
y	37.9	246.2	409.3	537.2	636.3	715.9

Solution

We use Gauss’s forward formula to find y when since it is near the middle of the given table.

Gauss’s forward formula is

where and is the origin. Here

We take , since lies is between and

When x = 2.7,

We shall form the forward difference table.

Substituting in (1), we get

When

When x = 2.7, y = 464.31

Example 2

Using Gauss’s forward formula find , given that

Solution

The given values of x are 21, 25, 29, 33, 37 and the corresponding y values are

is near the middle of the table.

So, we use Gauss’s forward formula to find y when and formula is

(1)

where and is the origin.

Here and take since lies between and .

When x = 30,

Now we shall form the central difference table.

Substituting in (1), we get

When

∴

Example 3

Use Gauss’s forward interpolation formula to find log sin 0°16′8.5″ from the given table.

x	0°16′7″	0°16′8″	0°16′9″	0°16′10″
	7.67100	7.67145	7.67190	7.67235

Solution

We use Gauss’s forward formula to find , since is near the middle of the table.

Gauss’s forward formula is

(1)

where and is the origin

Here and take , since lies between and .

∴

When

We shall form the central difference table.

Substituting in (1), we get

When

∴ log_e sin 0° 16¢8.5′ = 7.671675

Example 4

Estimate the value of from the following data.

x	0	1	2	3	4	5	6
	176	185	194	203	212	220	229

Solution

We use Gauss’s forward formula to find since is near the middle of the table.

Gauss’s forward formula is

(2)

where and is the origin.

Here and take , since lies between and .

∴ .

When ,

We shall form the central difference table.

Substituting in (1), we get

When

Example 5

Interpolate by means of Gauss’s forward formula the value of given that

Solution

The given values of the arguments x are 25, 30, 35, 40 and the corresponding values of y are is near the middle of the table.

So, we use Gauss’s forward formula to find .

Gauss’s forward formula is

(1)

where and is the origin.

Here and take , since lies between and .

∴

When

,

We shall form the central difference table.

Substituting in (1), we get

When

4.3.2 Gauss’s Backward Formula for Interpolation

Gauss’s Backward formula for Interpolation is

Proof: We derive Gauss’s backward difference formula from Newton’s forward formula by using even differences along the central line and odd differences above the central line.

Newton’s forward formula is

(1)

where and x₀ is the origin.

We have Δy₀ – Δy_–1 = Δ²y_–1

⇒ Δy₀ = Δy_–1 + Δ²y_–1

Similarly

Δ²y₀ = Δ²y_–1 + Δ³y_–1

Δ³y₀ = Δ³y_–1 + Δ⁴y_–1 and so on.

Also

Δ³y_–1 -Δ³y_–2 = Δ⁴y_–2

⇒ Δ³y_–1 = Δ³y_–2 + Δ⁴y_–2

Δ⁴y_–1 = Δ⁴y_–2 + Δ⁵y_–2 and so on.

Substituting in (1) we get

⇒

where

Note:

1. Gauss’s backward formula uses the differences on the indicated path
   
2. Gauss’s backward formula is used for interpolation if .

   Because of this reason, Gauss’s backward formula is some times known as Gauss’s formula for negative interpolation.

WORKED EXAMPLES

Example 1

Given that Show by Gauss’s backward formula

Solution

Given values of x are 12500, 12510, 12520, 12530.

Required the value of by Gauss’s backward formula.

Let Then to find

Gauss’s backward formula is

(1)

where and is the origin.

Here and take , since lies between and

∴

When ,

We shall form the central difference table.

Substituting in (1), we get

When

Example 2

Given the following data

x	1.72	1.73	1.74	1.75	1.76
	0.17907	0.17728	0.17552	0.17377	0.17204

Find the value of using Gauss’s backward formula.

Solution

Required the value of when by Gauss’s backward formula

(1)

Where and is the origin.

Here and take since lies between and

∴

When

We shall form the central difference table.

Substituting in (1), we get

When we get

∴

Example 3

Using Gauss’s backward formula estimate the number of persons earnings wages between Rs. 60 and 70 from the following data.

Wages (Rs.)	Below 40	40–60	60–80	80–100	100–120
No of person’s (in thousands)	250	120	100	70	50

Solution

We take the given intervals as 20–40, 40–60, 60–80, 80–100, 100–120.

Let the middle points of the intervals be the x-values and numbers of persons as the corrsponding y values.

Values of x	30	50	70	90	110
y	250	120	100	70	50

To find the number of persons earning wages between Rs. 60 and 70, we use Gauss’s backward formula.

Gauss’s backward formula is

(1)

where and is the origin.

Here

For the interval 60–70, middle value is

∴ take , since lies between and

∴

When ,

We shall form the central difference table.

Substituting in (1), we get

When , we get

∴ the number of persons in the wage range 60 to 70 is equal to 104.

Exercises 4.3

1. Using Gauss’s forward formula find given that

   x	2	3	4	5
   	2.426	3.454	4.784	6.986

2. Given that

   Year	1930	1932	1934	1936	1938	1940
   Population (in thousand)	12	16	21	27	32	40

   Using Newton-Gauss’s forward formula find the population of the town in 1935.

3. Using Gauss forward formula find the value of y_28.3 given that y₂₆ = 0.038462,
4. Given sin 25° 41′ 40″ = 0.433572, sin 25° 42′ 0′ = 0433659, sin 25° 42′ 20″ = 0.433746, sin 25° 42′ 40″ = 0.433834. Find the value of sin 25° 42′ 10″ by Gauss’s forward formula.
5. If find approximately.
6. Given , find the value of by Gauss’s backward formula.
7. The specific gravities of Zinc sulphate solutions of different concentrations at are given below

   Concentration (%):	10	12	14	16	18	20	22
   Specific gravity:	1.059	1.073	1.085	1.097	1.110	1.124	1.137

   Find the specific gravity of 15.8% solution at , using Gauss’s Backward formula.

8. Interpolate by means of Gauss’s backward formula the sales of a concern for the year 1997 given that

   Year:	1962	1972	1982	1992	2002	2012
   Sales (in cores of Rs.)	12	15	20	27	39	52

9. Find the value of by Gauss’s backward formula, given that

   x:
   cos x:	0.6428	0.6293	0.6152	0.6018	0.5878

Answers 4.3

1. 4.034 (2) 24.046875 (3) 0.0353355
2. 0.43 3703 (5) 7.6 (6) 3165.36
3. 1.0958 (8) 32.625 crores of Rs. (9) 0.6198

4.3.3. Stirling’s Formula for Interpolation

Stirling’s formula is

Proof: Stirling’s formula is derived by taking the average Gauss’s forward and backward formulae.

Gauss’s forward formula is

(1)

Gauss’s backward formula is

(2)

Adding (1) and (2), we get

This is called Stirling’s formula or Newton-Stirling’s formula.

Note:

1. Stirling’s formula involves average of odd difference above and below the central line and even differences on the central line as shown.
   
2. Stirling’s formula gives fairly accurate results when , though this formula can be used if .

WORKED EXAMPLES

Example 1

Given

	0	0.0875	0.1763	0.2679	0.3640	0.4663	0.5774

Find the value of using Stirling’s formula.

Solution

Let

Given

To find the value of .

We use Stirling’s formula to find y when x = 16

Stirling’s formula is

where and x₀ is the origin.

Here and take , since lies between and

When ,

We shall form the central difference table.

Substituting in (1), we get

When , we get

Example 2

The following table gives the values of the probability integral Find the value o the integral when .

x	0.51	0.52	0.53	0.54	0.55	0.56
	0.52924	0.53789	0.54646	0.55494	0.56332	0.57162

Solution

Required to find when by using Stirling’s formula.

Stirling’s formula is

(1)

where and is the origin.

Here and take , since lies between and

∴

When

We shall form the central difference table.

Substituting in (1), we get

When , we get

when

Example 3

Using Stirling’s formula find the annual net premium at the age of 25 from the table of annual net premium given below.

Age	20	24	28	32
Premium	0.01427	0.01581	0.01772	0.01996

Solution

Let us denote age by x and premium as y.

Required, the net premium at the age of 25 using Stirling’s formula.

That is to find y when x = 25.

Stirling’s formula is

(1)

where and is the origin.

Here and take since lies between and

∴

When

We shall form the central difference table.

Here h = 4 and take x₀ = 24, since x = 25 lies between x = 24 and x = 28

Substituting in (1), we get

When

∴ net Premium when is 0.01625

Example 4

Find the value of and from the following table using Stirling’s formula.

x	1.50	1.60	1.70	1.80	1.90
	17.609	20.412	23.045	25.527	27.875

Solution

Let

Required the values of y when and by using Stirling’s formula.

Stirling’s formula is

(1)

where and is the origin.

Here and take the orgin as since x = 1.63 lies between 1.60 and 1.70

∴

When

We shall form the central difference table when

Substituting in (1), we get

When

We use Stirlings formula if only. So we cannot take the origin as x = 1.60 to find y when x = 1.67, since

So, we change the origin to x = 1.70, to find y when x = 1.67

∴

When

[ lies is the interval ]

When the difference table is

Substituting in (1), we get

4.3.4 Bessel’s Formula for Interpolation

Bessel’s formula is

Proof: We derive the Bessel’s formula using Gauss’s forward and backward formula.

Gauss’s forward formula is

(1)

Gauss’s backward formula is

(2)

Shifting the origin of u from 0 to 1, that is replace u by in backward formula, we get

(3)

Adding (1) and (3) we get

This is Bessel’s formula.

If we put then and Bessel’s formula reduces to a more symmetrical form.

Note:

1. Bessel’s formula employs odd differences below the central line and the means of even differences on and below the central line.
2. If Bessel’s formula is preferable. However it will give more accurate result when interpolating near the middle of the interval

   When Bessel’s formula gives best results with minimum number of terms.

3. When we get the special case of Bessel’s formula.

   

   This is called formula for interpolating to halves.

WORKED EXAMPLES

Example 1

Apply Bessel’s formula to find log₁₀3375, given log₁₀310 = 2.49137, log₁₀320 = 2.250515, log₁₀330 = 2.51851, log₁₀340 = 2.53148, log₁₀350 = 2.54407, log₁₀360 = 2.55630.

Solution

Required the value of using Bessel’s formula,

Bessel’s formula is

(1)

where and is the origin.

The given function is and the x values are 310, 320, 330, 340, 350, 360.

Here and take the origin as , since lies between and

∴

When

We shall form the difference table.

Substituting in (1), we get

Example 2

The area A of a circle and diameter d is given by the following table.

d	80	85	90	95	100
Area A	5026	5674	6362	7088	7854

Find the area when the diameter is 91.

Solution

Let us denote the diameter d as x and area A as y.

Required, the area y when x = 91.

We use Bessel’s formula.

Bessel’s formula is

(1)

where and is the origin.

Here and take the origin as since lies between and

∴

When

We shall form the central difference table.

Substituting in (1), we get

∴ area = 6504.1248 sq. unit

Example 3

Given and fifth differences are constant.

Prove that

Where

Solution

Given the values are such that the fifth differences are constant.

∴ 6^th, 7^th … differences are zero.

(1)

Bessel’s formula for is

Now shift the origin to 2, then and so on.

∴

But

Similarly,

Example 4

Using Bessel’s interpolation formula show that assuming suitable level of approximation.

Solution

Bessels formula is

(1)

Replacing u by x in (1) we get

Assuming the fourth order and higher differences are very small, neglecting them and putting we get

Now shifting the origin to x,

4.3.5 Laplace-Everett Formula for Interpolation

Laplace-Everett’s formula for interpolation is

where v = 1−u.

Proof: Laplace-Everett’s formula is obtained from Gauss’s forward formula by replacing the odd differences interms of even differences

Gauss forward formula is

(1)

Now

Substituting in (1), we get

This formula is usually written in a more convenient form by putting in the terms with negative sign.

where

Note:

1. This formula involves the even differences on and below the central line.
2. This formula is used when u lies between 0 and 1, but more accurate results will be obtained when or when or vice-versa.

WORKED EXAMPLES

Example 1

From the following table find using Everett’s formula.

x	20	25	30	35	40
	11.4699	12.7834	13.7648	14.4982	15.0463

Solution

Everett’s formula is

(1)

where and and is the origin.

Here and take , since lies between and .

∴

When x = 34,

We shall form the difference table.

Substituting in (1), we get

∴ when

Example 2

Apply Everett’s formula to obtain given

Solution

Everett’s formula is

where , and is the origin.

The given values of x are 20, 24, 28, 32.

Required the value of y when x = 25

Here and take since lies between and

We shall form the difference table.

Substituting in (1), we get

∴ when

Example 3

If the third differences of are constant, show that

where

Solution

We shall prove the result using Everett’s formula.

Since the third differences are constants the fourth and higher differences are zero.

Everett’s formula is

where , and .

Here

∴ the formula becomes

Exercises 4.4

1. Using Stirlings formula find given

   x	1	1.1	1.2	1.3	1.4
   	0.841	0.891	0.932	0.963	0.985

2. Using Sitrling’s formula find given

   x	1	6	11	16	21
   	831	723	592	430	392

3. Using Stirling’s formula find given , where represents the number of persons at age x years in a life table.
4. Using Stirling’s formula find the value given that

   x	1.01	1.015	1.02	1.025	1.03
   	1.64463	2.10524	2.69159	3.43711	4.38391

5. (5) Apply Stirling’s formula to find the value of given that

   x	3.4	4.4	5.4	6.4	7.4	8.4
   	1156	1936	2916	4096	5476	7056

6. (6) Using Bessel’s formula to find the value of y when x = 15, given that

   x	10	12	14	16	18	20
   y	51.21	60.24	75.32	96.02	119.78	151.45

7. (7) Using Bessel’s formula find given

   x	15	20	25	30	35	40
   	10.3797	12.4622	14.0939	15.3725	16.3742	17.1591

8. (8) Use Bessel’s formula to obtain given
9. (9) Given

   x	20	24	28	32
   	24	32	35	40

   Find using Bessel’s formula.

10. If

    x	30	35	40	45	50	55	60
    	771	862	1001	1224	1572	2123	2983

    then find by Everett’s method.

11. Use Everett’s formula to obtain given that

Answers 4.4

1. 0.934 (2) 673.58 (3) 395 (4) 2.86155
2. 2959.36 (6) 85.189 (7) 15.0818 (8) 3250.8715
3. 32.945 (10) 1431.5 (11) 1.073

4.4 INTERPolatiON WITH UNEQUAL INTERVALS

Newton’s forward and backward formulae can be applied when the arguments are equally spaced. When the arguments are unequally spaced, we use Lagrange’s interpolation formula.

4.4.1 Lagrange’s Interpolation Formula

Theorem 4.2

Let be the entries corresponding to the arguments which are not necessarily equally spaced, then Lagrange’s interpolation formula is

Proof: Let be the entries corresponding to the arguments not necessarily equally spaced, then the interpolating polynomial for is of degree n and let

where are constants to be determined such that

When

When

Similarly, when we get

(1)

In the place of we can write

WORKED EXAMPLES

Example 1

Find f(x) as a polynomial in x from the given data and find f(8).

x	3	7	9	10
f(x)	168	120	172	63

Solution

Given the values of x and y are

and

The values of x are not equally spaced. So we use Lagrange’s formula to find y = f(x)

Lagrange’s formula for a set of four pairs of values is

Note:

Example 2

Find the polynomial f(x) by using Lagrange’s formula and hence find f(3) for the following values of x and y.

x	0	1	2	5
y	2	3	12	147

and hence find f(3).

Solution

The values of x are not equally spaced so we use Lagrange’s formula to find y = f(x)

Lagrange’s formula for a set of four pairs of values is

Example 3

Using Lagrange’s formula prove that

Solution

From the given equation, the values of x involved are and the corresponding y values are

The values of x are not equally spaced.

So, we use Lagrange’s formula to find y = f(x)

Lagrange’s formula for a set of four pairs of values is

When x = 1, we get

Example 4

Given the values

x	5	7	11	13	17
y = x f(x)	150	392	1452	2366	5202

Evaluate f(9) using Lagrange’s formula.

Solution

Given the values of x and y are

The values of x are not equally spaced, so we use Lagrange’s formula to find y = f(x).

Lagrange’s formula for a set of five pairs of values is

When x = 9,

Exercises 4.5

1. Using Lagrange’s formula fit a polynomial to the data and find f(5).

   x	–1	1	2
   f(x)	7	5	15

2. Using Lagrange’s formula find y (10) given that y (5) = 12, y (6) = 13, y (9) = 14, y (11) = 16.
3. Using Lagrange’s interpolation formula calculate the profit in the year 2000 from the following data:

   Year	1997	1999	2001	2002
   Profit in Lakhs Rs.	43	65	159	248

4. Apply Lagrange’s formula to find f(5) and f(6) given that f(l) = 2, f(2) = 4, f(3) = 8, f(4) = 16, and f(7) = 128
5. Find the form of the function u_x given that
6. Use Lagrange’s formula to find f(5) from the following data:

   x	2	3	4	6	7
   f(x)	1	5	13	61	125

7. Find the form of u_x, give that
8. Find u₅ given that u₁ = 4, u₂ = 7, u₄ = 13 and u₇ = 30
9. The following table gives the premium payable at the ages in years completed. Interpolate the premium payable at the age of 35 completed.

   Age completed in years:	25	30	40	60
   Premium in Rs.	50	55	70	95

10. From the following table find the value of y when x = 10

    x	5	6	9	11
    y	12	13	14	16

Answers 4.5

1. (2)
2. (4)
3. (6)
4. (8)
5. (10)

4.4.2 Divided Differences

In forward and backward differences for equally spaced arguments we considered only differences of the entries. But in divided differences, we divide this difference by difference of the corresponding arguments.

Let a function y = f(x) take values corresponding to the arguments , not necessarily equally spaced.

The first divided difference for the arguments x₀, x₁ is defined as .

It is denoted by

We shall denote,

Similarly,

The second divided difference for the arguments is defined as

The second difference for is

The third divided difference for is

The third difference for is

and so on.

Properties of divided differences

1. 1. The divided differences are symmetric functions of their arguments.

   For example

   

2. 2. The n^th divided differences of a polynomial of degree n are constants.
3. 3.

   and is constant.

   This shows that Δ is a linear operator.

4. 4. The n^th divided difference is a quotient of two determinants, each of order n + 1.

   We have

   

   is a quotient of two determinant of same order 2.

   Similarly we can write the other divided differences as a quotient of determinants of the same order.

WORKED EXAMPLES

Example 1

For the function , prove that the third divided difference with arguments a, b, c, d is equal to

Solution

Given

Similarly

Now

Similarly

Example 2

Prove that

Solution

Given the function f(x) = x³ and the arguments are x, y, z

∴

Similarly,

Now

Divided difference table

A divided difference table can be constructed using same principle as for ordinary difference tables. It is a diagonal difference table as illustrated by the following example.

4.4.3 Newton’s General Interpolation Formula or Newton’s Divided Difference Formula for Interpolation

Let be the given values of the function f(x) corresponding to unequally spaced arguments .

Newton’s general interpolation formula is

Proof: Corresponding to the given unequally arguments the values of the function f(x) are

Let x be the argument for which the value is required.

Then for the arguments x₀, x

(1)

Again for the arguments x₀, x₁, x

Substituting in (1), we get

(2)

Now for the arguments x₀, x₁, x₂, x

Substituting in (2) we get,

(3)

Proceeding in this way we obtain

where is the remainder term.

Let the interpolating polynomial be

If we put

then

We see that because

Hence the interpolation formula is or

Note: Using the relation between divided differences and ordinary differences from the general formula, we can deduce Newton’s forward and backward difference formulae for interpolation.

WORKED EXAMPLES

Example 1

Construct the divided difference table for the following data and find the value of f(2).

x	4	5	7	10	11	12
f(x)	50	102	296	800	1010	1224

Solution

Given value of x are

The values of x are unequally spaced, so we use Newton’s divided difference formula to find f(x) when x = 2.

Newton’s divided difference formula is

(1)

We form the divided difference table.

(1) becomes

when x = 2,

Example 2

By using Newton’s divided difference formula find f(8), given

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

Also find f(6), f(9), f(15).

Solution

Given the values of x are

The values of x are unequally spaced, so, we use Newton’s divided difference formula to find f(x) when x = 8.

Newton’s divided difference formula is

(1)

We form the divided difference table.

(1) becomes

When x = 15

Example 3

Given the data

x	0	1	2	5
f(x)	2	3	12	147

Find the cubic function of x, using Newton’s divided difference formula, and hence find f(2).

Solution

Given

The values of x are unequally spaced. We use Newton’s divided difference formula to find f(x)

Newton’s divided difference formula is

(1)

We form the divided difference table

∴ (1) becomes

When

Example 4

Find the third divided difference with arguments 2, 4, 9, 10 of the function f(x) = x³ – 2x.

Solution

Given the values of x are

and

We form the divided difference table to find the divided differences.

The third divided difference is Δ³f(x) = 1

Example 5

Given the following data, find f(x ) as a polynomial in x.

x	0	2	3	4	7	9
f (x)	4	26	58	112	466	922

Solution

Given

The values of x are unequally spaced. So, we find f(x) using Newton’s divided difference formula.

(1)

We form the divided difference table

∴ (1) becomes

[∴ all higher powers are zero ]

Example 6

If f(0) = 0, f(l) = 0, f(2) = –12, f(4) = 0, f(5) = 600, f(7) = 7308, find a polynomial that satisfies this data using Newton’s divided difference formula. Hence find f(6).

Solution

Given the values of f(x) are

The values of x are unequally spaced. We use Newton’s divided difference formula to find f(x).

We form the divided difference table

∴ Newton’s divided difference formula is

(1)

when

Aliter:

Since 6 values of f(x) are given, f(x) is polynomial of degree 5.

Since f(0) = 0, f(1) = 0, f(4) = 9,

f(x) = 0 when x = 1 and x = 4.

So, by factor theorem are factors of f(x).

, where g(x) is a polynomial of degree 2.

We shall find this polynomial g(x) by using the other three values

.

When

When

When

We shall find g(x) by using Newton’s divided difference formula

By Newton’s divided difference formula

∴ the polynomial

Now

Exercises 4.6

1. Given the following data:

   x	3	7	9	10
   f(x )	168	120	72	63

   Calculate f(6).

2. Using Newton’s divided difference formula, find u(3) given
3. Find f(10) using Newton’s divided difference formula given that

   x	11	17	21	23
   f(x )	14646	83526	194486	279845

4. Find the function f(x) from the following table:

   x	0	1	4	5
   f (x )	8	11	78	123

5. Given and log₁₀ 661 = 2.8202, find by divided difference the value of
6. Find the equation of the curve passing through the points (–1, –21), (1, 15), (2, 12), (3, 3). Find also f(0).
7. Find the cubic function from the following table:

   x	0	1	3	4
   f (x )	1	4	40	85

8. Given , find y₂ using Newton’s divided difference formula.

Answers 4.6

1. 147 (2) 100 (3) 130198
2. (5) 2.8168 (6)
3. f(x) = x³ + x² + x + 1 (8) y₂ = 4

4.5 ERRORS IN INTERPOLATION FORMULAE

We state some results without proof

1. 1. Rolle’s Theorem: If the real function f is (i) Continuous on the closed interval [a, b] (ii) differentiable in the open interval (a, b) and (iii) then there exists at least one point c in (a, b) such that f’

   Corollary: If is a polynomial of degree n and if then there is at least one such that

   From this it follows that between two roots of there is a root of

   In other words, the real roots of separate the real roots of the equation

   More generally, if is a polynomial of degree n with a and b as the smallest and largest roots of then

   has (n –1) roots in (a, b)

   has (n –2) roots in (a, b)

   

   has at least one root in (a, b).

4.5.1 Remainder Term in Interpolation Formulae

Let be a function defined at points and let be continuous and it has continuous derivatives of all orders.

Let be a polynomial of degree not exceeding n such that i = 0, 1, 2, … , n, be an approximation for

Let (1)

Since for i = 0, 1, 2, … , n, we get

⇒ g(x_i) = 0, i = 0, 1, 2, …, n

∴ are roots of

Hence

⇒ where is to be determined.

∴ (2)

Now consider the function

(3)

∴ for

Since and

for i = 0, 1, 2, …,n and for t = x by (2).

Hence we find at n + 2 points

Also is continuous and differentiable n times in where x₀ is the least root and is the largest root.

So, by Rolle’s theorem,

has at least (n + 1) roots in

has at least n roots in

has at least one root, say in

ie

But

Since is a polynomial of degree n,

Since we get

⇒

∴

⇒

Since is difference between the given function and the polynomial at any point x, it represents the error comitted in approximating the given function by the polynomial p(x)

Hence the error = R_n

This error is called the truncation error for

Remainder term in Newton’s forward formula

Let be equally spaced arguments with interval h.

Then

∴

and

∴

Suppose the analytical form of is not known, then we cannot find the derivative

∴ we replace by difference,

we have the formula

⇒

Put

∴

Take then

∴

∴

Similarly we can find the error or remainder term in all the interpolation formulae.

We shall list them.

1. Newton’s forward formula
   1. (a)
   2. (b) where
2. Newton’s backward formula
   1. (a)
   2. (b) where
3. Stirling’s formula
   1. (a)

      

4. Bessells formula in terms of v
   1. (a)
   2. (b)
5. Lagrange’s formula

   

WORKED EXAMPLES

Example 1

Find the error in the computation of sin 52° by using Newton’s forward formula, given that

x	45°	50°	55°	60°
y = sin x	0.7071	0.7660	0.8192	0.8660

Solution

The error in Newton’s forward formula is

where

We form the difference table.

Here (highest difference is third order)

∴

When x = 52,

∴

⇒

Example 2

The values of for certain equidistant values of x are given below

x	1.72	1.73	1.74	1.75	1.76
	0.1791	0.1773	0.1755	0.1738	0.1720

Find the error in computing by Stirling formula.

Solution

The error in Stirling’s formula is

where x₀ is the origin. Here x₀ = 1.73, h = 0.01

When x = 1.735,

Here

∴

Example 3

The following table gives certain values of Using Lagrange’s interpolation formula find the value of log 323.5. Also find the error.

x	321	322.8	324.2	325
	2.50651	2.50893	2.51081	2.51188

Solution

The given values of x are x₀ = 321, x₁ = 322.8, x₂ = 324.2, x₃ = 325

Using Lagrange’s formula, it can be seen that log 323.5 = 2.50987

We shall now compute the error.

Here

∴

But

(Taking t₀ = 321)

This shows that the error is less than 1 in the 11^th place.

4.6 INTERPOLATION WITH A CUBIC SPLINE

4.6.0 Introduction

The interpolation formulae we have seen so far represent a single polynomial passing through the points (or nodes) in a given interval. If the number of points is more, then the interpolating polynomial for many functions will be of higher degree, which tend to oscillate more and more between nodes as the degree increases. Hence the values computed using these interpolating polynomials will be very rough, except the case where the given set of points is for a polynomial function. This drawback is overcome by the method of splines which was introduced by I.J. Schoenberg in 1946.

Instead of using a single high-degree interpolating polynomial in an interval we subdivide the interval into a number of subintervals and in each subinterval we use a lower degree polynomial and join them together to get an interpolating function, called spline. Thus spline interpolation is a piece wise polynomial interpolation. Though splines can be of any degree, cubic splines are the most popular ones. The name ‘spline is borrowed from the draftman’s spline, a device which is an elastic rod used for drawing a smooth curve through a set of points.

4.6.1 Cubic Spline Interpolation

Let y = f(x) be the given function on the interval

Divide [a, b] into n subintervals by the points

where

Let S (x) be the cubic spline that approximates f(x) such that

1. (i) for
2. is a third degree polynomial in each subinterval
3. and are continuous on (a, b)

ie.

ie. at the joining point x_i, the successive cubics have the same slope and same curvature.

In each interval is a cubic polynomial, and so is linear. By Lagrange’s formula, for the arguments we have

Since

and and

Let

and

∴

Integrating twice w.r.to x, we get

(2)

where c₁ and c₂ are arbitrary constants.

To find c₁ and c₂ use the hypothesis and

Put in (2)

(3)

Put in (2)

(4)

Subtracting, we get

where and are unknown quantities to be determined by using the continuity of at the point

⇒

ie. the left and right limits at are equal.

Now differentiating (5) w.r.to x, we get

(6)

Replacing i by i + 1, we get

(7)

From (6),

(8)

From (7),

(9)

Since

we get from (8) and (9), (putting in the RHS)

(10)

This is true for all

Equations (10) give a system of (n – 1) linear equations in n + 1 unknowns

To solve for these unknowns we need two more equations. These two conditions may be taken in different ways.

We usually assume that outside the interval is flat or a straight line, then

∴

Thus we can solve for and these values are substituted in (5) to get the cubic spline

Note:

1. The end conditions are called natural conditions because they yield a natural spline. It is called a natural spline because the draftman’s spline behaves like this.
2. Functions with abrupt local changes can be better approximated by cubic splines.

Working Rule: Given the table of values for y = f(x)

x	x0	x1	…	xn
y	y0	y1	…	yn

The cubic spline approximation S(x) for these points is in the interval is

where and

WORKED EXAMPLES

Example 1

Obtain the cubic spline approximation for the function y = f(x) from the following data, given that

x	–1	0	1	2
y	–1	1	3	35

Solution

The given values of x and y are

x₀ = -1, x₁ = 0, x₂ = 1, x₃ = 2

and

y₀ = -1, y₁ = 1, y₂ = 3, y₃ = 35

The values of x are equally spaced with h = 1. Here n = 3.

The cubic spline for the interval

(1)

where (2)

Given and

Put i = 1 in (2), we get,

Put i = 2 in (2), we get,

The cubic spline for is

(3)

Put i = 1 in (1), then the interval is

Then the cubic spline for is

Put i = 2 in (1), then the interval is

Then the cubic spline for is

Put i = 3 in (1), then the interval is

Then the cubic spline for is

∴ the cubic spline approximation for the given function is

Example 2

Find the cubic spline approximation for the function f(x) given by the data:

x	0	1	2	3
f(x)	1	2	33	244

With . Hence estimate the value of f(2.5), f(1.5).

Solution

Let y = f(x)

The given the values of x and y are

x₀ = 0 , x₁ = 1 , x₂ = 2 , x₃ = 3

y₀ = 1 , y₁ = 2 , y₂ = 33 , y₃ = 244

The values of x are equally spaced with h = 1, Here n = 3.

The cubic spline for the interval is

(1)

where (2)

and

Put i = 1 in (2),

Put i = 2 in (2),

Solving we get

Put i = 1 in (1), then the interval is

Then the cubic spline for the interval is

Put i = 2 in (1), then the interval is

Then the cubic spline for the interval is

Put i = 3 in (1), then the interval is

Then the cubic spline for is

∴ the cubic spline is

When x = 2.5,

When x = 1.5,

Example 3

The following values of x an y are given

x	1	2	3	4
y	1	2	5	11

Find the cubic splines and evaluate y(1.5) and y′(3).

Solution

Given the values of x and y are

x₀ = 1, x₁ = 2, x₂ = 3, x₃ = 4

y₀ = 1, y₁ = 2, y₃ = 5, y₃ = 11

The values of x are equally spaced with h = 1. Here n = 3

The cubic spline for the interval is

(1)

where

(2)

We assume the conditions of natural spline

Putting i = 1 in (2) we get,

(3)

Putting i = 2 in (2) we get,

(4)

Solving (3) and (4), we get

Now Putting i = 1 in (1), then the interval is .

Then the cubic spline for the interval is

Putting i = 2 in (1), then the interval is

Then the cubic spline for the interval is

Putting i = 3 in (1), then the interval is

Then the cubic spline for is

∴ the cubic spline is

When x = 1.5,

Differentiating S(x) w.r.to x, we get

The derivates at the joining points should exist for the spline.

can be obtained from [2, 3] or [3, 4] and they should be equal.

So we can find from [2, 3],

Note:

If we find from [3, 4],

Exercises 4.7

1. Given the following table, find f(2.5) using cubic spline functions.

   i	0	1	2	3
   xi	1	2	3	4
   f(xi)

2. Find a natural cubic spline to the data, given

   x:	1	2	3	4
   y:	1	5	11	8

   Hence find y(1.5) and y′(2).

3. Find the natural cubic spline valid for the interval [3, 4] for the data

   x:	1	2	3	4
   y:	3	10	29	65

   and hence find the value of y(3.2).

4. Find the natural cubic spline for the data

   x	0	2	4	6
   f(x)	4	0	4	80

5. Find the cubic spline interpolation for the data

   x:	0	2	4	6
   f(x):	1	9	41	41

   Given

Answers 4.7

1. 
2. 
3. 
4. 
5. 

Short Answer Questions

1. When Newton’s forward interpolation formula is used ?
2. When Newton’s backward formula for interpolation is used ?
3. State Newton’s backward formula for interpolation.
4. State Newton’s forward formula for interpolation.
5. What do you mean by interpolation?
6. Given , find the value of u₂.
7. A third degree polynomial passes through (0, –1), (1, 1), (2, 1) and (3, –2). Find its value at x = 4.
8. Given f(0) = –1, f(1) = 1 and f(2) = 4, find the root of the Newton’s interpolating polynomial equation f(x) = 0.
9. Form the divided difference table for the following data:

   x	2	5	10
   y	5	29	109

10. Find the second degree polynomial through the points (0, 2), (2, 1), (1, 0).
11. What is the Lagrange’s formula to find y, if are given?
12. What are the advantages of Lagrange’s formula over Newton’s formula ?
13. Write down the Lagrange’s interpolating polynomial for the data (xi, f(xi)), i = 0,1,2,3.
14. Find the second degree polynomial fitting the data:

    x	1	2	4
    y	4	5	13

15. Show that the second divided difference [x₀, x₁, x₂] is independent of the arguments.
16. Find the second divided differences with arguments a, b, c of the function .
17. Form the divided difference table for the data (0, 1), (1, 4) (3, 40) and (4, 85).
18. Form the divided difference table

    x	–1	1	2	4
    y	–1	5	23	119

19. Find y(l), given

    x	0	3	4
    y	12	6	8

20. Define a cubic spline S(x) which is commonly used for interpolation.
21. State the properties of cubic spline.
22. For cubic splines what are the 4n conditions required to evaluate the unknowns.
23. Find the divided differences of f(x)= x3 + x + 2 for the arguments 1, 3, 6.