Question

Using the Bisection method, find an approximate root of the equation sin(x)=1/x that lies between x=1 and x=1.5 (in radians). Compute upto 5 iterations. Determine the approximate error in each iteration. Give the final answer in a tabular form.

Answer 1

An f(x) = x sinx-1 f(1) = sin() - 1 --O. 15849 20 f(1.5) = 1.5 sin (1.5) -1 = 0.49 625 > 0 The root lies between 1 and 1.5. First appropriate root x= 14 lis = 2.5 = 1.25 f(xi) = f(1.25) = 1.25 Sin (1.25)-1 = 0.1862730 Root lies between , and 1.25. x = 1+1.25_ 2.25 1.125

f(x2) = f(1.125) = 1.125 Sin (1.125)-1 = 0. 01509 so Root lies between 's and 2 = 1.125. So approximate error Fa - Present approximate - Previous approximal = 0.1509-0. 18627 = – 0.17 118 .. Third approximate root is z = 1+ 1.125 _ 2.125 1.8625 f(1) = 1.0625 Sin(1.0625) - =-0.0718 20 in Root lies between &z and xy. Approximate essor is Eaz = f(az) - f(x2) = -0.0718- 0.01509 -0.089689. 1. 062 .. Fourth approximate to the noot is + 1.125 74= 2 = 1.09375 flan) = 1.09375 sin (1.0937) - = -0.02836 <0 Root lies between 4 and 2

Approximate eroor is Eaz = f(x4) - f(az) = -0.02836 - (-0.8-768) = 0.04344. .. Fifth approximate to the groot is 2 = 1.09375 + 1.125 2 = 1-10937 flag) - 1. 10937 Sin (1.10937) - = – 0. 00421 60 Rost lies between x = 1. 18937 and 2 = 1.125 Approximate error is Eag = f(ar) – fren) - ~ 0.80421 - (-0.828 36) = 0.02415 . Appoonimate goots upto 5th iterration. {x,,,, 1, 2+, *} = {125, 1125, 1.0625, 1.09375, 0.109373 . Apprsnimate eroor is {Ea, Eaz, Egg, Ead} = {- 0.1718, - 0.08689, 0.04344, 0.029415}.