Question

Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to four decimal places.) f(x) = cos x, x1 = 1.6

Answer 1

Answer:

x₃=1.599997=1.6

Step-by-step explanation:

f(x)=cos x, x₁=1.6

$f'(x)=\frac{d}{dx}cos\ x\\=-sin\ x$

First iteration

At x₁=1.6

f(x₁)=f(1.6)

=cos(1.6)

=-0.0292

f'(x₁)=f'(1.6)

=-sin(1.6)

=-0.9996

$\frac{f(x_1)}{f'(x_1)}=\frac{-0.0292}{-0.9996}\\=0.0292\\x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1.6-(0.0292)\\\therefore x_2=1.5708$

$f(x_2)=f(1.5708)\\=cos 1.5708\\=0.000003\\f'(x_2)=-sin1.5708\\=-1\\x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.6-\frac{0.000003}{-1}\\\therefore x_3=1.599997=1.6$