Examples of Newton Interpolation

Example No 1: The following supply schedule gives the quantity supplied \left( S \right) in hundreds of a product at prices \left( P \right) in rupees:

P
80
90
100
110
120
S
25
30
42
50
68

 
Interpolate, what will be the quantity supplied of the product at a price Dollor 85.

Solution: Constructing the difference table first.

THE DIFFERENCE TABLE
P
S
{\Delta ^1}
{\Delta ^2}
{\Delta ^3}
{\Delta ^4}
80
25
5
7
-11
25
90
30
12
-4
14
100
42
8
10
110
50
18
120
68
43
13
3
25

On checking, we found that the table is correctly prepared, now, we have

\begin{array}{*{20}{c}} {a = 80,\,\,\,h = 10,\,\,\,{X_o} = 85} \\ {f\left( a \right) = 25,\,\,\,\Delta f\left( a \right) = 5,\,\,\,{\Delta ^2}f\left( a \right) = 7,\,\,\,{\Delta ^3}f\left( a \right) = - 11,\,\,\,{\Delta ^4}f\left( a \right) = 25} \end{array}

Also X = \frac{{{X_o} - a}}{h} = \frac{{85 - 80}}{{10}} = \frac{5}{{10}} = \frac{1}{2} = 0.5

We may either put X = \frac{1}{2} = 0.5 in the formula, but the latter is subject to errors due to misleading of the decimal point in multiplication.

\begin{gathered} f\left( {{X_o}} \right) = f\left( {a + Xh} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,f\left( a \right) + X\Delta f\left( a \right) + \frac{{X\left( {X - 1} \right)}}{{2 \times 1}}{\Delta ^2}f\left( a \right) + \frac{{X\left( {X - 1} \right)\left( {X - 2} \right)}}{{3 \times 2 \times 1}}{\Delta ^3}f\left( a \right) + \cdots \, \\ \end{gathered}

Now, using formula and substituting the corresponding values, we have

\begin{gathered} S\left( {85} \right) = 25 + \left( {\frac{1}{2}} \right)\left( 5 \right) + \frac{{\frac{1}{2}\left( {\frac{1}{2} - 1} \right)}}{{2 \times 1}}\left( 7 \right) + \frac{{\frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right)}}{{3 \times 2 \times 1}}\left( { - 11} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{{\frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right)\left( {\frac{1}{2} - 3} \right)}}{{4 \times 3 \times 2 \times 1}}\left( {25} \right) \\ \end{gathered}

\begin{gathered} S\left( {85} \right) = 25 + \frac{5}{2} + \left( {\frac{1}{2}} \right)\left( { - \frac{1}{2}} \right)\left( {\frac{7}{2}} \right) + \left( {\frac{1}{2}} \right)\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)\left( { - \frac{{11}}{6}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {\frac{1}{2}} \right)\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)\left( { - \frac{5}{2}} \right)\left( {\frac{{25}}{{24}}} \right) \\ \end{gathered}

Now determine the sign of each term by the same rule given earlier, i.e. count the number of negative signs and place a negative sign before the term, if the number of negative signs are odd, and a positive sign before the term if the number of negative signs are even, then forget the individual signs.

S\left( {85} \right) = 25 + \left( {\frac{5}{2}} \right) - \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{7}{2}} \right) - \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{3}{2}} \right)\left( {\frac{{11}}{6}} \right) + \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{3}{2}} \right)\left( {\frac{5}{2}} \right)\left( {\frac{{25}}{{24}}} \right)

Cancel the common factors, if any, in each term and simplify
S\left( {85} \right) = 25 + \frac{5}{2} - \frac{7}{8} - \frac{{11}}{{16}} - \frac{{125}}{{128}} = 24.961 \approx 25

Example No 2: The production of vegetable oil is recorded on a fiscal – year basis alternate years:

Year:
1990 – 91
1992 – 93
1994 – 95
1996 – 97
1998 – 99
Production
(000)TONS:
35.5
42.8
45.8
46.5
50.3

Estimate the production during 1997 – 98.

Solution: In this situation again we can multiply the values in the second column by 10, and the adjustment may be made in the final answer by dividing the result by 10.

Another problem will arise hare as to which should be regarded as a and X. The equal class interval is obvious and that is 2. The value of X may be determined by either considering the lower limits of the years or by considering the upper limits of the years. Thus, in order to calculate the value of X from Newton formula of interpolation, we can either take {X_o} = 1997 and a = 1990 or we can take {X_o} = 1998 and a = 1991, both will provide the same value of X. Thus

X = \frac{{{X_o} - a}}{h} = \frac{{1997 - 1990}}{2} = \frac{7}{2} = 3.5

THE DIFFERENCE TABLE
Years
Production
{\Delta ^1}
{\Delta ^2}
{\Delta ^3}
{\Delta ^4}
1990 – 91
355
73
-43
20
34
1992 – 93
428
30
-23
54
1994 – 95
458
7
31
1996 – 97
465
38
1998 – 99
503
148
-35
74
34

\begin{array}{*{20}{c}} {X = \frac{7}{2},\,\,f\left( a \right) = 355,\,\,\,\Delta f\left( a \right) = 73\,} \\ {\,\,\,{\Delta ^2}f\left( a \right) = - 43,\,\,\,{\Delta ^3}f\left( a \right) = 20,\,\,\,{\Delta ^4}f\left( a \right) = 34} \end{array}

Using the formula for Newton Interpolation, we have

\begin{gathered} f\left( {{X_o}} \right) = 355 + \left( {\frac{7}{2}} \right)\left( {73} \right) + \frac{{\frac{7}{2}\left( {\frac{7}{2} - 1} \right)}}{{2 \times 1}}\left( { - 43} \right) + \frac{{\frac{7}{2}\left( {\frac{7}{2} - 1} \right)\left( {\frac{7}{2} - 2} \right)}}{{3 \times 2 \times 1}}\left( {20} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{{\frac{7}{2}\left( {\frac{7}{2} - 1} \right)\left( {\frac{7}{2} - 2} \right)\left( {\frac{7}{2} - 3} \right)}}{{4 \times 3 \times 2 \times 1}}\left( {34} \right) \\ \end{gathered}

\begin{gathered} f\left( {{X_o}} \right) = 355 + \frac{{511}}{2} + \left( {\frac{7}{2}} \right)\left( {\frac{5}{2}} \right)\left( { - \frac{{43}}{2}} \right) + \left( {\frac{7}{2}} \right)\left( {\frac{5}{2}} \right)\left( {\frac{3}{2}} \right)\left( {\frac{{20}}{6}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {\frac{7}{2}} \right)\left( {\frac{5}{2}} \right)\left( {\frac{3}{2}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{{34}}{{24}}} \right) \\ \end{gathered}

f\left( {{X_o}} \right) = 355 + 255.5 - \frac{{1505}}{8} + \frac{{175}}{4} + \frac{{595}}{{64}} = 475.42

For readjustment divide this result by 10, therefore,
Production (1997 – 98) = 47.542