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By Gupta, C.B.,Malik, A.K. , Kumar, Vipin

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A3 n x n ] a33 . : xn = 1 [bn − a n1 x1 − a n 2 x 2 ...... S. 6), we get x1( 2 ) = 1 [b1 − a12 x 2(1) − a13 x3(1) ...... a1n x n(1) ] a11 45 SOLUTION OF LINEAR SIMULTANEOUS EQUATIONS Now put x1( 2 ) x 2(1) x 3(1) ...... S. 6), we get x 2( 2 ) = 1 [b2 − a21 x1( 2 ) − a23 x 3(1) ...... 6), we get x 3( 2 ) = 1 [ b3 − a31 x1( 2 ) − a32 x 2( 2 ) ...... 6), we get 1 [ bn − an1 x1( 2 ) − an 2 x 2( 2 ) ...... , xn are obtained to desired accuracy Gauss-Seidel method is also known as a method of successive displacement.

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. Aliter: 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 (E – I)4 ux = 0 or (E4 – 4C1 E3 + 4C2E2 – 4C3 E + I) ux = 0 or E4 ux – 4E3 ux + 6E2 ux – 4Eux + ux = 0 or ux + 4h – 4ux + 3h + 6ux + 2h – 4ux + h + ux = 0 Putting x = 0 and h = 1, we get u4 – 4u3 + 6u2 – 6u1 + u0 = 0 or 384 – 4X + 6 × 520 – 6 × 556 + 580 = 0 1860 – 4X = 0 ⇒ X = 465.

3! + (u + 1) u (u − 1) (u − 2) 4 ∆ y–2 + ...... 4! 8) and Gauss’s backward difference formula is yu = y0 + u∆y– 1 + (u + 2) (u + 1) u (u − 1) u (u + 1) 2 (u + 1) u (u − 1) 3 ∆ y–1 + ∆ y–2 + ∆4y–2 + ...... 4! 2! 3! 9) is 3 3 u(u 2 − 1) ( ∆ y− 1 + ∆ y−2 ) u 2 (u 2 − 1) 4 ( ∆y0 + ∆y−1 ) u 2 2 ∆ y + + ∆ y–2 + ..... + –1 3! 2 4! 2 2! 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.

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