Question

need the MATLAB code to solve for percentage error at the 5th iteration given the following Newton-Raphson code

Using the MATLAB code finline(X13-x^2-18); dfeinline(3*x"2-2#x'); e?100 count=0; xi=0.5; while (ea>01); count = count 1 ; xil - xi - f(xi)/df(xi); ea = 100 * abs((xi 1-xi)/xi 1 ); disp([count xil.xi, ea]); xi=xi 1 ; If ea is the error defined in percentage, the value of the error (ea) at 5th iteration (i.e., count-5) is closest or equal to hoices 51% 54% | 100% 50% Submit | Attempts 1

Answer 1

Refer the output below:

f = inline (' x^3 - x62 - 18 '): df = inline (' 3 times x^2 - 2 times x '): ea = 100: count = 0: x i = .5: while (ea > 0.01) count = count + 1: xi 1 = xi - f (xi)/df (xi): ea = 100 times abs ((xi 1 - xi)/xi 1): disp ([count xi 1 xi ea]): xi = xi 1: end 1. 0000 - 72.0000 0.5000 100.6944 2. 0000 - 47.8888 - 72.0000 50.3484 3. 0000 - 31.8137 - 47.8888 50.5289 4. 0000 - 21.0945 - 31.8137 50.8152 5. 0000 - 13.9422 - 21.0945 51.2994 6. 0000 - 9.1593 - 13.9422 52.2190 7. 0000 - 5.9360 - 9.1593 54.3019 8. 0000 - 3.7043 - 5.9360 60.2440 9. 0000 - 2.0048 - 3.7043 84.7706 1.0 e + 03 times 0.0100 - 0.0001 - 0.0020 1.4085 \r\n 10. 0100 - 0.0001 - 0.0020 1.4085 11. 0000 56.3926 - 0.1329 100.2357 12. 0000 37.7094 56.3926 49.5451 13. 0000 25.2570 37.7094 49.3027 14. 0000 16.9618 25.2570 48.9053 15. 0000 11.4452 16.9618 48.1997 16. 0000 7.7968 11.4452 46.7944 17. 0000 5.4273 7.7968 43.6590 18. 0000 3.9771 5.4273 36.4639 19. 0000 3.2406 3.9771 22.7266 20. 0000 3.0196 3.2406 7.3181 21. 0000 3.0001 3.0196 0.6491 22. 0000 3.0000 3.0001 0.0048 It is seen from the output that at count = 5, ea = 51.2994 %. close to 51 % Hence choice 1 is correct

It is seen from the output that at count=5, ea= 51.2994 %. close to 51%

Hence choice 1 is correct