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

x₃=1.599997=1.6

Explanation:

f(x)=cos x, x₁=1.6

`f'(x)=(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

`(f(x_1))/(f'(x_1))=(-0.0292)/(-0.9996)\\=0.0292\\x_2=x_1-(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-(f(x_2))/(f'(x_2))=1.6-(0.000003)/(-1)\\\therefore x_3=1.599997=1.6`