829058 967799 2 35 4 375 967799 2829058 527746 3 375 4 3875 527746 2829058

829058 967799 2 35 4 375 967799 2829058 527746 3 375

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2.829058 -9.67799 2 3.5 4 3.75 -9.67799 2.829058 -5.27746 3 3.75 4 3.875 -5.27746 2.829058 -1.75589 4 3.875 4 3.9375 -1.75589 2.829058 0.394922 5 3.875 3.9375 3.90625 -1.75589 0.394922 -0.7148 6 3.90625 3.9375 3.921875 -0.7148 0.394922 -0.16866 7 3.921875 3.9375 3.929688 -0.16866 0.394922 0.110936 8 3.921875 3.929688 3.925782 -0.16866 0.110954 -0.0294 9 3.921578 3.929688 3.925633 -0.17919 0.110954 -0.03471 10 3.925633 3.929688 3.927661 -0.03471 0.110954 0.037973 11 3.925633 3.927611 3.926622 -0.03471 0.036195 0.000706 12 3.925633 3.926622 3.926128 -0.03471 0.000706 -0.01701 Therefore, maximum is at 3.926
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6. Determine the smallest positive root between (0,2) of the given equation using Bisection method with an error less than 0.02. cosx = 1 x 1 n x Solution: f(x) = cosx - 1 x 1n x using incremental search we get x 0 1 2 f(x) 0.540302 -0.76272 Therefore root lies between 1 and 2 a = 1 and b =2 f(a)=0.5403 and f(b) = -0.763 p = 1.5 and f(p)= -0.1996 Therefore p is the new b. n A b p fa fb fp er 1 1 2 1.5 0.540302 -0.76272 -0.19957 2 1 1.5 1.25 0.540302 -0.19957 0.136808 0.2 3 1.25 1.5 1.375 0.136808 -0.19957 -0.03706 0.090909 4 1.25 1.375 1.3125 0.136808 -0.03706 0.048246 0.047619 5 1.3125 1.375 1.34375 0.048246 -0.03706 0.00522 0.023256 6 1.34375 1.375 1.359375 0.00522 -0.03706 -0.01601 0.011494
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15. Use Newton-Raphson method to determine the positive root of the equation sinx 4 x 2 + 1 = 0 Correct to 3 decimal places. Solutuion: sin(x) – 4x 2 + 1 = 0 f’(x) = cos(x) – 8x from incremental we get (x) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 f(x) 1 1.0598 1.038669 0.9355 0.7494 0.4794 0.1246 -0.316 Therefore, x 0 = 0.6 f(x 0 ) = f(0.6) = sin(0.6) – 4(0.6) 2 + 1 = 0.124642 f’(x 0 ) = f’(0.6) = cos(0.6) – 8(0.6) = -3.97466 x 1 = x 0 ( f ( x 0 ) f ' ( x 0 ) ) = 0.6 - ( 0.124642 3.97466 ) = 0.63136 Continuing in table form i x i f(x i ) f’(x i ) x i + 1 0 0.6 0.124642 -3.79466 0.63136 0.04967 1 0.63136 -0.0042 -4.24365 0.630366 0.001576 2 0.630366 -4.2 * 10 -6 -4.23512 0.630365 1.59 * 10 -6 x = 0.63036
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25. Find cubic root of 8 using Secant method accurate to 4 digits. (Hint: solve x 3 – 8 = 0). Solution: x 3 – 8 = 0 Cubic root of 8 is 2. Therefore we start at 1.9; x n = 1.9 and x n – 1 = 1.8 f’(x n ) = f ( x n ) f ( x n 1 ) x n x n 1 = ( 1.141 ) −(− 2.168 ) 1.9 1.8 = 10.27 x n + 1 = x n ( f ( x n ) f ' ( x n ) ) = 1.9 - ( 1.141 10.27 ) = 2.0111 Continuing in table form n x n – 1 x n f(x n ) f(x n – 1 ) f’(x n ) x n + 1 0 1.8 1.9 -1.141 -2.168 10.27 2.0111 1 1.9 2.0111 0.133944 -1.141 11.4756 1.999428 2 2.0111 1.999428 -0.00686 0.133944 12.0633 1.999997 Therefore, x
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