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, …, xi, …, xn. 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 xn. 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
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.
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.
Let y0, y1, y2,…, yn be the values of the function at equally spaced arguments x0, x1, x2,…, xn 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 3rd, 4th, …, nth 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.
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
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
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.
Proof:
We have
and
∴ for any
∴
ie.
and , where C is a constant.
ie.
and
If m and n are positive integers, then
Similarly
is the inverse operator of
is the inverse operator of
Proof:
We have
⇒
Proof:
We have
and
∴ for any
∴
Proof:
We have
∴ for any
⇒
Because of this property, we shall write each side
Theorem 4.1
The nth differences of a polynomial of nth degree are constant.
ie. If is an nth degree polynomial, then
constant, where h is the interval of differencing.
Note:
That is, if the nth differences of a function tabulated at equally spaced arguments are constants, then the function is a polynomial of nth degree.
The Converse is of great use in practical situations. If in a difference table the nth 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 nth degree by a suitable interpolation formula.
The Central difference operator δ is defined by .
Proof:
Proof:
We have
(1)
and
We know ,
We have Taylor’s series for as
But
∴
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 nth 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.
Solution
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
[ = ΔEr]
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
x | 0 | 1 | 2 | 3 | 4 |
yx | 1 | 2 | 4 | – | 16 |
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
f(x) | –4 | –2 | – | – | 220 | 546 | 1148 |
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
yx | 0 | 1 | 2 | 1 | 0 | – | – | – |
Answers 4.1
(6) (7) (8)
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:
it is analogous to differentiation
When r = 0, [x]0 = 1
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
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 nth degree and it is determined uniquely. at and approximately equal in intermediate points. Hence we write f(x) itself in the place of
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 nth 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:
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 nth 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:
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.
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 lx 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 = lx.
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
x0 | 45 | 46 | 47 | 48 | 49 | 50 |
tan x0 | 1.0 | 1.03553 | 1.07237 | 1.11061 | 1.15037 | 1.19175 |
x | 1 | 2 | 3 | 4 | 5 |
0 | 1 | 8 | 27 | 64 |
Diameter | 80 | 85 | 90 | 95 | 100 |
Area | 5026 | 5674 | 6362 | 7088 | 7854 |
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
f(x) | 176 | 185 | 194 | 203 | 212 | 220 | 229 |
x | 30 | 35 | 40 | 45 | 50 |
f(x) | 15.9 | 14.9 | 14.1 | 13.3 | 12.5 |
Marks | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 |
No of students | 35 | 48 | 70 | 40 | 22 |
Age in year | 20 | 25 | 30 | 35 | 40 |
Premium Rs. | 23 | 26 | 30 | 35 | 42 |
x° | 30 | 31 | 32 | 33 | 34 |
sin x° | 0.5000 | 0.5150 | 0.5290 | 0.5446 | 0.5592 |
x | 100 | 150 | 200 | 250 | 300 | 350 | 400 |
y | 10.63 | 13.03 | 15.04 | 16.81 | 18.42 | 19.90 | 21.27 |
u0 | u1 | u2 | u3 | u4 | u5 |
25 | 25 | 22 | 18 | 15 | 5 |
Year: | 2001 | 2002 | 2003 | 2004 | 2005 |
Population in thousands: | 251 | 279 | 319 | 383 | 483 |
x | 0 | 1 | 2 | 3 | 4 |
y | 79 | 91 | 105 | 116 | 127 |
x | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
y | 2501 | 2795 | 2838 | 3030 | 3050 | 3381 | 3888 |
Find y when x = 54.
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
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.
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
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:
.
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
∴ loge 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
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:
Because of this reason, Gauss’s backward formula is some times known as Gauss’s formula for negative interpolation.
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
x | 2 | 3 | 4 | 5 |
2.426 | 3.454 | 4.784 | 6.986 |
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.
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.
Year: | 1962 | 1972 | 1982 | 1992 | 2002 | 2012 |
Sales (in cores of Rs.) | 12 | 15 | 20 | 27 | 39 | 52 |
x: | |||||
cos x: | 0.6428 | 0.6293 | 0.6152 | 0.6018 | 0.5878 |
Answers 4.3
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:
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
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:
When Bessel’s formula gives best results with minimum number of terms.
This is called formula for interpolating to halves.
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.
∴ 6th, 7th … 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,
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:
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
x | 1 | 1.1 | 1.2 | 1.3 | 1.4 |
0.841 | 0.891 | 0.932 | 0.963 | 0.985 |
x | 1 | 6 | 11 | 16 | 21 |
831 | 723 | 592 | 430 | 392 |
x | 1.01 | 1.015 | 1.02 | 1.025 | 1.03 |
1.64463 | 2.10524 | 2.69159 | 3.43711 | 4.38391 |
x | 3.4 | 4.4 | 5.4 | 6.4 | 7.4 | 8.4 |
1156 | 1936 | 2916 | 4096 | 5476 | 7056 |
x | 10 | 12 | 14 | 16 | 18 | 20 |
y | 51.21 | 60.24 | 75.32 | 96.02 | 119.78 | 151.45 |
x | 15 | 20 | 25 | 30 | 35 | 40 |
10.3797 | 12.4622 | 14.0939 | 15.3725 | 16.3742 | 17.1591 |
x | 20 | 24 | 28 | 32 |
24 | 32 | 35 | 40 |
Find using Bessel’s formula.
x | 30 | 35 | 40 | 45 | 50 | 55 | 60 |
771 | 862 | 1001 | 1224 | 1572 | 2123 | 2983 |
then find by Everett’s method.
Answers 4.4
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.
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
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
x | –1 | 1 | 2 |
f(x) | 7 | 5 | 15 |
Year | 1997 | 1999 | 2001 | 2002 |
Profit in Lakhs Rs. | 43 | 65 | 159 | 248 |
x | 2 | 3 | 4 | 6 | 7 |
f(x) | 1 | 5 | 13 | 61 | 125 |
Age completed in years: | 25 | 30 | 40 | 60 |
Premium in Rs. | 50 | 55 | 70 | 95 |
x | 5 | 6 | 9 | 11 |
y | 12 | 13 | 14 | 16 |
Answers 4.5
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
For example
and is constant.
This shows that Δ is a linear operator.
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.
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.
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.
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
x | 3 | 7 | 9 | 10 |
f(x ) | 168 | 120 | 72 | 63 |
Calculate f(6).
x | 11 | 17 | 21 | 23 |
f(x ) | 14646 | 83526 | 194486 | 279845 |
x | 0 | 1 | 4 | 5 |
f (x ) | 8 | 11 | 78 | 123 |
x | 0 | 1 | 3 | 4 |
f (x ) | 1 | 4 | 40 | 85 |
Answers 4.6
We state some results without proof
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).
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(xi) = 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 = Rn
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.
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 11th place.
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.
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
ie.
ie. at the joining point xi, 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:
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
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
i | 0 | 1 | 2 | 3 |
xi | 1 | 2 | 3 | 4 |
f(xi) |
x: | 1 | 2 | 3 | 4 |
y: | 1 | 5 | 11 | 8 |
Hence find y(1.5) and y′(2).
x: | 1 | 2 | 3 | 4 |
y: | 3 | 10 | 29 | 65 |
and hence find the value of y(3.2).
x | 0 | 2 | 4 | 6 |
f(x) | 4 | 0 | 4 | 80 |
x: | 0 | 2 | 4 | 6 |
f(x): | 1 | 9 | 41 | 41 |
Given
Answers 4.7
x | 2 | 5 | 10 |
y | 5 | 29 | 109 |
x | 1 | 2 | 4 |
y | 4 | 5 | 13 |
x | –1 | 1 | 2 | 4 |
y | –1 | 5 | 23 | 119 |
x | 0 | 3 | 4 |
y | 12 | 6 | 8 |