Use Newton's method to approximate a root of the equation 2 sin x = x Let x1 =2. Find x2 and x3?

Question

Thanks alot!

Answer 1

f(x) = 2sin x - x

x1 = 2

x_n+1 = xn - f(xn)/f'(xn)

f'(x) = 2cos x - 1

x_n+1 = xn - (2sin xn - xn)/(2cos xn - 1)

x2 = 2 - (2sin(2) - 2)/(2cos(2) - 1)

then continue .. . .

x2 = 1.900995594

x3 = 1.895511645

btw ... the angle should be in radians

Answer 2

you can easily do this if you have a TI calculator...:do the following ; 2 enter,store x, 'ans -(2 sin ans - ans) / ( 2 cos ans -1), enter,enter,enter {x_4}.....1.89549...I have a TI 85