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प्रश्न
From the following table obtain a polynomial of degree y in x.
| x | 1 | 2 | 3 | 4 | 5 |
| y | 1 | – 1 | 1 | – 1 | 1 |
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उत्तर
We will use Newton’s backward interpolation formula to find the polynomial.
`y_((x = x_"n" + "nh")) = y_"n" + "n"/(1!) ∇y_"n" + ("n"("n" + 1))/(2!) ∇^2y_"n" + ("n"("n" + 1)("n" + 2))/(3!) Delta^2y_"n" + .......`
| x | y | `Deltay` | `Delta^2y` | `Delta^3y`` | Delta^4y` |
| 1 | 1 | ||||
| – 2 | |||||
| 2 | – 1 | 4 | |||
| 2 | – 8 | ||||
| 3 | 1 | – 4 | 16 | ||
| – 2 | 8 | ||||
| 4 | – 1 | 4 | |||
| 2 | |||||
| 5 | 1 |
To find y in terms of x
xn + nh = x
5 + n(1) = x
∴ n = x – 5
`"y"_((x)) = 1 + ((x - 5))/(1!) (2) + ((x - )(x - 5 + 1))/(2!) (4) + ((x - 5)(x -5 + 1)(x - 5 + 2))/(3!) (8) + ((x - 5)(x - 5 + 1)(x - 5 + 2)(x - 5 + 3))/(4!) (16)`
= `1 + 2(x - 5) + ((x - 5)(x - 4)(4))/2 + ((x - 5)(x - 4)(x - 2)(8))/6 + ((x - 5)(x - 4)(x - 2)(16))/24`
= `1 + 2x - 10 + 2(x^2 - 9x + 20) + 4/3 (x - 5) (x^2 - 7x + 12) + 2/3(x^2 - 9x + 20)(x^2 - 5x + 6)`
= `1 + 2x - 10 + 2x^2 - 18x + 40 + 4/3 [x^3 - 7x^2 + 12x - 5x^2 + 35x - 60] + 2/3 [x^4 - 5x^3 + 6x^2 - 9x^3 + 45x^2 - 54x + 20x^2 - 100x + 120]`
= `2x^2 - 16x + 31 + 4/3 [x^3 - 12x^2 + 47x - 60] + 2/3 [x^4 - 14x^3 + 71x^2 - 154x + 120]`
= `2x^2 - 16x + 31 + 4/3 x^3 - 16x^2 + 188/3 x - 80 + 2/3 x^4 - 28/3 x^3 + 142/3 x^2 - 308/3 x + 80`
= `2/3 x^4 + (4/3 - 28/3) x^3 + (2 - 16 + 142/3) x^2 + (- 16 + 188/3 - 308/3)x + (31 - 80 + 80)`
= `2/3 x^4 + ((-24)/3)x^3+ ((6 - 48 + 142)/3) x^2 + ((-48 + 188 - 308)/3) x + 31`
`y(x) = 2/3 x^4 - 8x^3 + 100/3 x^2 - 56x + 31`
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