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प्रश्न
Find the value of f(x) when x = 32 from the following table:
| x | 30 | 5 | 40 | 45 | 50 |
| f(x) | 15.9 | 14.9 | 14.1 | 13.3 | 12.5 |
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उत्तर
Since the value of f(x) is required near the beginning of the table
We use the Newton’s forward interpolation formula.
`y_((x = x_0 + "nh")) = y_0 + "n"/(1!) Deltay_0 + ("n"("n" - 1))/(2!) Delta^2y_0 + ("n"("n" - 1)("n" - 3))/(3!) Delta^3y_0 + .........`
To find y at x = 32
∴ x0 + nh = 32;
30 + n(5) = 32
5n = 32 – 30
⇒ 5n = 2
n = `2/5`
∴ n = 0.4
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` |
| 30 | 15.9 | ||||
| – 1 | |||||
| 35 | 14.9 | 0.2 | |||
| – 0.8 | – 0.2 | ||||
| 40 | 14.1 | 0 | 0.2 | ||
| – 0.8 | 0 | ||||
| 45 | 13.3 | 0 | |||
| – 0.8 | |||||
| 50 | 12.5 |
`y_((x = 32)) = 15.9 + ((0.4))/(1!) (1) + ((0.40.4 - 1))/(2!) (0.2) + ((0.4)(0.4 - 1)(0.4 - 2))/(3) (- 0.2) + ((0.4)(.4 - 1)(0.4 - 2)(0.4 - 3))/(4!) (0.2) +`
= 15.9 – 0.4 – 0.024 – 0.0128 – 0.00832
= 15.9 – 0.44512
= 15.45488
= 15.45
∴ When x = 32, f(x) = 15.45
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