By Lingen F.

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P and y – 2y = f ′ (ξ) Mε − c ε + G c − c J ε + G c c − H K H K 2 2 6 24 4 N Q 7 1 1 I 1 1 1 I F F ε –ε = ε − c ε +G c − c J ε +G H 2 3 K H 12 c c − 8 c − 2 c JK ε + ... P . 42) and simplifying, we get FG 1 c H8 IJ K 1 c2 c3 ε 04 + O (ε 50 ) . 12 Hence, the error term is proportional to ε04and the method has fourth order rate of convergence. For the equation f (x) = x4 – x – 10, we find that f (1) < 0, f (2) > 0 and a root lies in (1, 2). Let x0 = 2. 855703. 855585. 856. 33 Determine p, q and r so that the order of the iterative method xn+1 = pxn + qa + ra 2 x n2 xn5 for a1/3 becomes as high as possible.

Y = B1(x0, y0) + B2(x0, y0) + B3(x0, y0) + ... where A1 (x0, y0), B1 (x0, y0) are linear in f0, g0 ; A2 (x0, y0), B2 (x0, y0) are quadratic in f0, g0 and so on. Use Taylor series method to derive iterative methods of second and third order. Solution We have f (x0 + ∆x, y0 + ∆y) ≡ 0, g(x0 + ∆x, y0 + ∆y) ≡ 0. Expanding f and g in Taylor’s series about the point (x0, y0), we get 1 [(∆x)2 fxx + 2∆x ∆y fxy + (∆y)2 fyy] + ... ≡ 0 f (x0, y0) + [∆x fx + ∆y fy) + 2 1 g(x0, y0) + [∆x gx + ∆ygy) + [(∆x)2 gxx + 2∆x ∆y gxy + (∆y)2 gyy] + ...

FG f (x , y )IJ , n = 0, 1, ... 412524IJ . 643856 ± 10–6. 23 correct to three decimal places. 23 we have the Jacobian matrix 2 xn 2 yn J(xn, yn) = y xn n LM N and J–1 (xn, yn) = Using the method 2( xn2 OP Q 1 L x − y ) MN− y FG x IJ = FG x IJ – J Hy K Hy K n +1 n n+ 1 n 2 n –1 n n (xn , yn) OP Q − 2 yn 2 xn FG f (x H f (x 1 2 IJ , n = 0, 1, ... 23 105585 . 22185 . 22229 . 22229 . 222. 222). 46 Describe how, in general, suitable values of a, b, c and d may be estimated so that the sequence of values of x and y determined from the recurrence formula xn+1 = xn + a f (xn, yn) + bg(xn, yn) yn+1 = yn + c f (xn, yn) + dg(xn, yn) will converge to a solution of f (x, y) = 0, g(x, y) = 0.