Examples of Newton Interpolation
Example No 1: The following supply schedule gives the quantities supplied $$\left( S \right)$$ in hundreds of a product at prices $$\left( P \right)$$ in rupees:
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$$P$$
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80
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90
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100
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110
|
120
|
|
$$S$$
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25
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30
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42
|
50
|
68
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Interpolate the quantity of the product supplied at the price dollar 85.
Solution: We construct the difference table first.
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THE DIFFERENCE TABLE
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|||||
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$$P$$
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$$S$$
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$${\Delta ^1}$$
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$${\Delta ^2}$$
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$${\Delta ^3}$$
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$${\Delta ^4}$$
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80
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25
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5
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7
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-11
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25
|
|
90
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30
|
12
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-4
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14
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|
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100
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42
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8
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10
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||
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110
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50
|
18
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|||
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120
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68
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||||
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43
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13
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3
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25
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||
Upon 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 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 the 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 using the same rule given earlier, i.e. count the number of negative signs and place a negative sign before the term if there is an odd number of negative signs, and a positive sign before the term if there is an even number, and 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 in each term, if any, 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 in alternate years:
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Year:
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1990 – 91
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1992 – 93
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1994 – 95
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1996 – 97
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1998 – 99
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Production
(000)TONS: |
35.5
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42.8
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45.8
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46.5
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50.3
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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 here 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 the 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:
\[X = \frac{{{X_o} – a}}{h} = \frac{{1997 – 1990}}{2} = \frac{7}{2} = 3.5\]
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THE DIFFERENCE TABLE
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|||||
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Years
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Production
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$${\Delta ^1}$$
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$${\Delta ^2}$$
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$${\Delta ^3}$$
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$${\Delta ^4}$$
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1990 – 91
|
355
|
73
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-43
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20
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34
|
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1992 – 93
|
428
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30
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-23
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54
|
|
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1994 – 95
|
458
|
7
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31
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||
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1996 – 97
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465
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38
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|||
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1998 – 99
|
503
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||||
|
148
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-35
|
74
|
34
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||
\[\begin{array}{*{20}{c}} {X = \frac{7}{2},\,\,f\left( a \right) = 355,\,\,\,\Delta f\left( a \right) = 73\,} \\ {\,\,\,{\Delta ^2}f\left( a \right) = – 43,\,\,\,{\Delta ^3}f\left( a \right) = 20,\,\,\,{\Delta ^4}f\left( a \right) = 34} \end{array}\]
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