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Question
The population of a city in a censes taken once in 10 years is given below. Estimate the population in the year 1955.
| Year | 1951 | 1961 | 1971 | 1981 |
| Population in lakhs |
35 | 42 | 58 | 84 |
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Solution
Let the year be x and population be y.
To find the population for the year 1955.
i.e. The value of y at x = 1955
Since the value of y is required near the beginning of the table
We use the Newton’s forward interpolation formula.
`y_((x = x_+ "nh")) = y_0 + "n"/(1!) Deltay_0 + ("n"("n" - 1))/(2!) Delta^2y_0 + ("n"("n" - 1)("n" - 2))/(3!) Delta^3y_0 + .........`
To find y at x = 1955
∴ x0 + nh = 1955, x0 = 1951, h = 10
⇒ 1951 + n(10) = 1955
10n = 1955 – 1951
⇒ 10n = 4
n = `4/10` = 0.4
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` |
| 1951 | 35 | |||
| 7 | ||||
| 1961 | 42 | 9 | ||
| 16 | 1 | |||
| 1971 | 58 | 10 | ||
| 26 | ||||
| 1981 | 84 |
y = `35 + 0.4/(1!) (7) + ((0.4)(0.4 - 1))/(2!) (9) + (0.4(0.4 - 1)(0.4 - 2))/(3!)`
= `35 + 2.8/1 + ((0.4)(- 0.6)(9))/2 + ((0.4)(- 0.6)(- 1.6))/6`
y = 35 + 2.8 – 1.08 + 0.064
= 37.864 – 1.08
y = 36.784
∴ Population in the year 1955 is 36.784 (lakhs)
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