I am sending another version of "Interpolation" as the previous one was not in order.
Interpolation
“Interpolation consists in reading a value which lies between two extreme points, and Extrapolation means reading a value that lies outside the two.” – W.M. Herper
We use statistical technique known as Interpolation or extrapolation when we have to estimate a value which is not known in a given series. Interpolation refers to the insertion of value in-between the given series of values and extrapolation refers to projecting a value for the future. Hence interpolation helps us to get the missing link in the series, and extrapolation helps us to forecast.
In simple interpolation, it is assumed that there is a linear relationship between x and y, where x being the independent variable and y the dependent variable.
Suppose we are given the values x0, x1, x2, xn of x and let the corresponding values of y be y0, y1, y2, yn respectively. If we want to estimate the value of yx for any value of x between the limits x0 and xn, this can be done by applying interpolation technique.
All the formulae of interpolation are based on the fundamental assumption that the given data can be expressed as a polynomial function with a fair degree of accuracy.
(Polynomial functions are functions that have this form:
f(x) = anxn + an-1xn-1 + ... + a1x + a0
The value of n must be a nonnegative integer. That is, it must be whole number; it is equal to zero or a positive integer.
Positive integers are all the whole numbers greater than zero: 1, 2, 3, 4, 5, Negative integers are all the opposites of these whole numbers: -1, -2, -3, -4, -5, We do not consider zero to be a positive or negative number.
The coefficients, as they are called, are an, an-1, a1, a0. These are real numbers.
Polynomial functions are functions that have this form:
f(x) = anxn + an-1xn-1 + ... + a1x + a0
The value of n must be a nonnegative integer. That is, it must be whole number; it is equal to zero or a positive integer.
The coefficients, as they are called, are an, an-1, a1, a0. These are real numbers.)
Finite Differences and Operators
In problems of interpolation, the independent variable x in the functional relation y=f(x) is called the argument, and the dependent variable y, the entry.
Let x0, x1, x2,…xn-1, xn be a set of equidistant values of the argument x, i.e. x1-x0=x2-x1=……=xn-xn-1 = h (constant)
and let y0, y1, y2, …….., yn-1, yn be the corresponding values of y.
Then y1-y0, y2-y1, ……, yn-yn-1 are called the fist differences (or finite differences of first order) of the function y. Denoting these differences by
∆y0 = y1-y0, ∆y1 = y2-y1…... ∆yn-1 = yn – yn-1
The differences of these first differences, i.e. the differences
∆y1 - ∆y0, ∆y2 - ∆y1, ∆yn-1 - ∆yn-2 are called the second differences. Denoting them by ∆2y0, ∆2y1, etc., we have
∆2y0 = ∆y1 - ∆y0, ∆2y1 = ∆y2 - ∆y1,….., ∆2yn-2 = ∆yn-1 - ∆yn-2
Similarly, we can define the third, fourth and higher order of differences.
Newton’s Forward Interpolation Formula
Let x0, x1, x2,…xn-1, xn be a set of equidistant values of the argument (or independent variable) x i.e.
x1-x0=x2-x1=……=xn-xn-1 = h (say)
and y0, y1, y2, …….., yn-1, yn be the corresponding values of the function y=f(x).
Then the value of y corresponding to a specified value of x lying near the beginning of the tabulated values is given by the Newton’s forward interpolation formula:
y = y0 + (u/1) y0 + {u (u-1)/1.2}2y0 + {u (u-1) (u-2)/1.2.3} 3y0 + -------------
--------------+ {u (u-1) u-2)………….. (u-n+1)/1.2.3…..n} ny0, where u = (x-x0/h.
Newton’s Backward Interpolation Formula
Let x0, x1, x2,...xn-1, xn be a set of equidistant values of the argument (or independent variable) x and y0, y1, y2, …….., yn-1, yn be the corresponding values of the function y=f(x). Then the value of y corresponding to a specified value of x lying near the end of the tabulated values is given by the Newton’s backward interpolation formula:
y = yn + (u/1) yn-1 + {u(u+1)/1.2}2yn-2 + {u(u+1)(u+2)/1.2.3}3yn-3 + --------------------------- + {u(u+1)u+2)…………..(u+n-1)/1.2.3…..n}ny0, where u = (x-x0/h and h = x1-x0=x2-x1=……=xn-xn-1
Lagrange’s Interpolation formula
Let x0, x1, x2, ….xn-1, xn be a set of values of the argument (or independent variable) x (not necessarily equidistant) and y0, y1, y2, …….., yn-1, yn be the corresponding values of the function y=f(x). Then the value of y corresponding to a specified value of x lying between the tabulated values is given by the Lagrange’s Interpolation formula:
y = {(x-x1) (x-x2)…. (x-xn)/(x0-x1) (x0-x2)…..(x1-xn)}*yo
+ {(x-x0) (x-x2)…. (x-xn)/(x1-x0) (x1-x2)…. (x1-xn)}*y1
+ ………………………………………………………..
+ {(x-x0) (x-x1)…. (x-xn-1)/ (xn-x0) (xn-x1)…. (xn-xn-1)}*yn
Inverse Interpolation
Lagrange’s Interpolation formula may also be used to find the value of argument x for a given value of the function y. This process is known as Inverse Interpolation. Lagrange’s formula for inverse interpolation is given by
X = {(y-y1) (y-y2) …. (y-yn)}/ (y0-y1) (y0-y2)…. (y0-yn)}*x0
+ {(y-y0)(y-y2) …. (y-yn)/ (y1-y0) (y1-y2)….(y1-yn)}*x1
+ {(y-y0) (y-y1)…. (y-yn-1)/ (yn-y0) (yn-y1)….(yn-yn-1)}*xn
INTERPOLATION & EXTRAPOLATION
1. Given the following logarithms, interpolate the value of log 20.5
x 20.4 20.5 20.6 20.7
Log x 1.3096 - 1.3139 1.3160
Here the arguments x0 = 20.4, x1 = 20.5, x2 = 20.6 and x3 = 20.7 are equidistant. y1 is not known (missing). This unknown entry y1 can be interpolated by binomial method.
x 20.4
x0 20.5
x1 20.6
x2 20.7
x3
Y = log x 1.3096
y0
-
y1 1.3139
y2 1.3160
y3
There are n = 3 known entries. The entry y1 is missing. Therefore ∆3y0 = 0. Thus by using Pascal’s triangle ∆3y0 = 0.
Y3 – 3y2 +3y1 – y0 = 0
1.3160 – 3*1.3139 + 3 y1 – 1.3096 = 0
3 y1 = 1.3096 – 1.3160 + 3*1.313
3 y1 = 3.9353
y1 = 1.3118
2. The following are the annual premiums for an insurance policy of Rs. 1000. Calculate the annual premium for a person aged 25 years.
Age of persons(years) 20 25 30 36
Premium(rupees) 23 - 30 35
Age of persons 20
x0 25
x1 30
x2 35
x3
Premium (rupees) 23
y0
-
y1 30
y2 35
y3
There are n = 3 known entries. The entry y1 is missing. Therefore ∆3y0 = 0. Thus by using Pascal’s triangle ∆3y0 = 0.
y3 – 3y2 +3y1 – y0 = 0
35 – 3*30 + 3* y1 – 23 = 0
3* y1 = 90 +23 – 35
3* y1 = 78 or y1 = 26
Therefore, annual premium for a person aged 25 years is y1 =Rs. 26.
3. Find the missing figure in the following series by binomial method
x 1 2 3 4 5 6
y 200 - 240 255 260 265
x 1 2 3 4 5 6
y 200
y0 -
y1 240
y2 255
y3 260
y4 265
y5
Here n = 5
Here, the arguments are equidistant and one entry, namely y1 is missing.
Using Pascal’s triangle ∆5y0 = 0.
y5 - 5y4 + 10y3 - 10y2 + 5y1 - y0 = 0
5y1 = yo + 10y2 – 10y3 + 5y4 – y5
5y1 = 200 +10*240 – 10*240 + 5*260 – 265
5y1 = 1085
y1 = 217
4. The following data gives the sales of a manufacturing firm in six successive years. Using the data estimate the sales in 2013’
Year 2007 2008 2009 2010 2011 2012
Sales
(Thousand units) 120 140 160 180 200 22
Year 2007 2008 2009 2010 2011 2012
Sales
(Thousand units)
120
y0
140
y1
160
y2
180
y3
200
y4
225
y5
Here, the arguments are equidistant But none of the enties is missing. The range of the arguments is (2007, 2011). We have to estimate the sales in 2013. The value 2013 is outside the range. Therefore, we are using binomial method for extrapolation. For 2013, y6 is the sales value.
There are n = 6 known entries. Therefore ∆6y0 = 0.
Thus by using Pascal’s triangle, ∆6y0 = 0
y6 - 6y5 + 15y4 - 20y3 + 15y2 - 6y1 + y0 = 0
y6 – 6*225 + 15*200 – 20*180 + 15*160 – 6*140 + 120
y6 = 6*225 – 15*200 +20*180 – 15*160 + 6*140 – 120
y6 = 270 (Thousand units)
5. Interpolate the missing value
x 180 190 200 210 220
y 87 83 79 - 71
x 180 190 200 210 220
y 87
y0 83
y1 79
y2 -
y3 71
y4
Here n = 4 known entries. The entry y3 is missing.
Therefore, using Pascal’s triangle ∆4y 0= 0.
Y4 – 4y3 + 6y2 – 4y1 + y0 = 0
71 - 4y3 + 6*79 – 4*83 + 87 = 0
4y3 = 71 + 6*79 – 4*83 + 87
4y3 = 300
y3 = 75
6. If e02 = 1.2214 and e04 = 1.4918, interpolate the value of e03 by binomial method.
e02 e03 e04
1.2214
y0 -
y1 1.4918
y2
Here the known entries n = 2
Thus by applying Pascal’s triangle ∆2y0 = 0
y2 – 2y1 + y0 = 0
1.4918 - 2y1 + 1.2214 = 0
2y1 = 2.7132
y1 = 1.3566
7. From the following data, estimate the population of India in census year 1981.
Census year 1991 2001 2011
Population(crore) 85 103 121
Census year 1981 1991 2001 2011
Population(crore) -
y0 85
y1 103
y2 121
y3
Here, the arguments are equidistant But none of the enties is missing. The range of the arguments is (1991, 2011). We have to estimate the population in 1981. The value 1981 is outside the range. Therefore, we are using binomial method for extrapolation. For 1981, y0 is the population value.
There are n = 3 known entries. The entry y1 is missing. Therefore ∆3y0 = 0. Thus by using Pascal’s triangle ∆3y0 = 0.
y3 – 3y2 +3y1 – y0 = 0
121 – 3*103+ 3*85 – y0 = 0
y0 = 121 – 309 + 255
y0 = 67
Hence the population in 1981 is 67 crores
8. From the missing entry in the following data regarding average yield of coconut trees.
Age(years) 1 2 3 4 5
Average yield 138 151 193 - 296
Age(years) 1 2 3 4 5
Average yield
y 138
y0 151
Y1 193
y2 -
y3 296
y4
Here n = 4 known entries. The entry y3 is missing.
Therefore, using Pascal’s triangle ∆4y 0= 0.
Y4 – 4y3 + 6y2 – 4y1 + y0 = 0
296 - 4y3 + 6*193 – 4*151 + 138 = 0
4y3 = 296 + 1158 – 604 + 138
4y3 = 988
y3 = 247