Question

Determine the lowest positive root of f (x) = 8sin(x)e–x – 1:

(a) Graphically.

(b) Using the Newton-Raphson method (three iterations, xi = 0.3).

(c) Using the secant method (three iterations, xi–1 = 0.5 and xi = 0.4.

(d) Using the modified secant method (five iterations, xi = 0.3, δ = 0.01).

Answer 1

a. Write the following code on the console.

Code:

>> x = linspace(0,4);

>> y = 8*sin(x).*exp(-x)-1;

>> plot(x,y); grid

Output:

The lowest positive root seems to be at approximately 0.15.

b. Newton- Raphson

i	x i	f ( x ) f ' ( x ) εa
0 1 2 3	0.3 0.107 8 44 0.143 4 87 0.145 0 12	0.751414 3.910431 -0.22695 6.36737 178.180% -0.00895 5.868388 24.841% -1.6E-05 5.847478 1.052%

c. Secant

i x i −1 f(xi−1)	x i	f ( x i ) εa
0 0.5 1.32629 1 0.4 1.088279 2 -0.05724 -1.48462 3 0.206598 0.334745 4 0.158055 0.075093 5 0.144016 -0.00585	0.4 -0.05 7 24 0.206 5 98 0.158 0 55 0.144 0 16 0.145 0 3	1.088279 -1.48462 798.821% 0.334745 127.706% 0.075093 30.713% -0.00585 9.748% 9E-05 0.699%

d. Modified Secant

i	x	x + dx f ( x ) f ( x + d x ) f ' ( x ) εa
0 1 2 3 4 5	0.3 0.107 0 07 0.143 4 63 0.145 0 15 0.145 0 15 0.145 0 15	0.303 0.751414 0.763094 3.893468 0.108077 -0.23229 -0.22547 6.371687 180.357% 0.144897 -0.00909 -0.00069 5.858881 25.412% 0.146465 -1.2E-06 0.008464 5.83752 1.070% 0.146465 2.12E-09 0.008465 5.837517 0.000% 0.146465 -3.6E-12 0.008465 5.837517 0.000%