Newtons Method for 5 sin(x) = x, w/ x1 = 1 (explain please)

[kcbrat]

Can someone help explain and solve this Newton's Method problem for me? I know the equation I just don't understand how to use it.

Use Newton's method to approximate a root of the equation 5sin(x)=x as follows: Let x1=1 be the initial approximation. What is the second (x2) and third (x3) approximations?

 

[galactus]

Re: Newtons Method?

This does not converge very well using an initial guess of 1. It does at 2, though.

Newton's formula: \(\displaystyle x_{n+1}=x-\frac{f(x)}{f'(x)}\)

You have \(\displaystyle x-\frac{5sin(x)-x}{5cos(x)-1}\)

Use x=1 first off and then plug in the successive answers you get each time around.

\(\displaystyle 1-\frac{5sin(1)-1}{5cos(1)-1}=-.885\)

Now plug the -.885 in:

\(\displaystyle -.885-\frac{5sin(-.885)-(-.885)}{5cos(-.885)-1}=0.49264\)

You keep doing that until it converges to a solution.

It should converge to \(\displaystyle \pm{2.5957}\), but using an initial guess of 1 it does not. Use 2 and it will.

Here is an animated graph showing what it does using x=1. The first graph is using x=2. See how it converges compared to the second one, which uses x=1?.