Approximate using newtons method: 2sinx=x x0

Question

Approximate using newtons method: 2sinx=x x>0

anonymous · Answer

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

superdavesuper · Answer

Newton's method is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function.

So you will have to set up a table for f(x)=2sinx - x; find the function of f'(x) then calculate thro iteration using the eqn below.

anonymous · Answer

Okay I've done the that, the only problem I'm having is what to plug in for the x1

superdavesuper · Answer

0 is an obvious root but x>0 so I would try to start w 1 :)

anonymous · Answer

just guess what do you think the solution is. Say xo = 1.5, then
x1 = 1.5 - f(1.5)/f'(1.5)

whatever you get, use that value and do again

superdavesuper · Answer

actually after "cheating" w wolfram, @sourwing 's suggestion of 1.5 is better than 1....good luck :)

anonymous · Answer

Okay I got the answer thank you to both of you :)

superdavesuper · Answer

welcome :)