By C. B. Gupta

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This formula is called the Stirling’s difference formula. yu = y0 + u SOLVED EXAMPLES Example 1. Given u0 = 580, u1 = 556, u2 = 520, u3 = —, u4 = 384, find u3. Solution. Let the missing term u3 = X 26 ADVANCED MATHEMATICS ∴ The forward difference table is x ux 0 580 1 556 ∆u x ∆2u x ∆ 3u x ∆ 4 ux – 24 – 12 – 36 2 X – 472 520 X – 484 1860 – 4X X – 520 3 1388 – X X 904 – 2X 384 – X 4 384 Here four values of ux are given. Therefore, we can assume ux to be a polynomial of degree 3 in x ∴ ∆ 4 ux = 0 or or 1860 – 4X = 0 X = 465.

X 10 11 12 13 14 105ux 23967 28060 31788 35209 38368 Solution. Here h = 1. 2 1 The difference table is as under: x u yu 10 –2 23967 11 –1 28060 ∆yu ∆2yu ∆3yu ∆4yu 4093 – 365 3728 12 0 58 31788 – 307 3421 13 1 – 13 45 35209 – 262 3159 14 2 38368 Stirling formula is yu = y0 + u u(u 2 − 1) 3 u 2 (u 2 − 1) 4 ( ∆y0 + ∆y−1 ) u 2 2 + ∆ y–1 + (∆ y–1 + ∆3y–2) + ∆ y–2 + ..... 4! 2! 3! 2) 2 + (– 307) + (45 + 58) 3! 2 2! 2) 2 − 1) (– 13) 4! 4848. Example 16. 5 from the following data. 5563025 Solution. Here h = 10.

Thus f (x) is supposed to be polynomial in x of degree n. Then y = f (x) = a0 (x – x1) (x – x2) ..... (x – xn) + a2(x – x0) (x – x1) ..... (x – xn) ..... + an (x – x0) (x – x1) ..... , an are constants. 7), we get y0 = a0(x0 – x1) (x0 – x2) ...... (x0 – xn) ⇒ a0 = y0 ( x 0 − x1 ) ( x 0 − x 2 ) ...... 7), we get y1 = a1 (x1 – x0) (x1 – x2) ...... (x1 – xn) ⇒ a1 = y1 ( x1 − x 0 ) ( x1 − x 2 ) ...... ( x1 − x n ) Proceeding in this way, we get an = yn ( x n − x 0 ) ( x n − x1 ) ...... 7), we get y = f (x) = ( x − x1 ) ( x − x 2 ) ......

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