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प्रश्न
From the following data find y at x = 43 and x = 84.
| x | 40 | 50 | 60 | 70 | 80 | 90 |
| y | 184 | 204 | 226 | 250 | 276 | 304 |
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उत्तर
To find y at x = 43
Since the value of y is required near the beginning of the table
We use the Newton’s forward interpolation formula.
`y_((x = 2.8)) = 34 + ((-0.2))/(1!) (23) + ((-0.2)(-0.2 + 1))/2 (14) + ((-0.2)(-0.2 + 1)(-0.2 + 2))/(3!) (6) + ......`
= `34 - 4.6 + ((-0.2)(0.8)(14))/2 + ......`
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` |
| 40 | 184 | ||||
| 20 | |||||
| 50 | 204 | 2 | |||
| 22 | 0 | ||||
| 60 | 226 | 2 | 0 | ||
| 24 | 0 | ||||
| 70 | 250 | 2 | 0 | ||
| 26 | 0 | ||||
| 80 | 276 | 2 | |||
| 28 | |||||
| 90 | 304 |
`y_((43)) = 184 + 0.3/(1!) (0) + ((0.3)(0.3 - 1))/(2!) (2) + ((0.3)(0.3 - 1)(0.3 - 2))/(3!) (0)`
= 184 + (0.3)(20) + (0.3)(– 0.7)
= 184 + 6.0 – 0.21
= 190 + 0.21
`y_((x = 43))` = 189.79
To find y at x = 84
Since the value of y is required at the end of the table, we apply backward interpolation formula.
`y_((x = x_"n" + "nh")) = y_"n" + "n"/(1!) ∇y_"n" + ("n"("n" + 1))/(2!) ∇^2y_"n" + ("n"("n" + 1)("n" + 2))/(3!) Delta^3y_"n" + .......`
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` |
| 40 | 184 | ||||
| 20 | |||||
| 50 | 204 | 2 | |||
| 22 | 0 | ||||
| 60 | 226 | 2 | 0 | ||
| 24 | 0 | ||||
| 70 | 250 | 2 | 0 | ||
| 26 | 0 | ||||
| 80 | 276 | 2 | |||
| 28 | |||||
| 90 | 304 |
xn + nh = x
90 + n(10) = 84
10n = 84 – 90
10n = – 6
∴ n = – 0.6
`y_((x = 84)) = 304 + ((0.6))/(1!) (28) + ((0.6)(-0.6 + 1))/(2!) (2) + ....`
= `304 + (0.6)(28) + ((-0.06)(0.4))/2 + 2`
= 304 – 16.8 – 0.24
= 304 – 17.04
= 286.96
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