(1 point) Use Newton's method to find the second and third approximation of a root of 5sin(x)=x 5sin⁡(x)=x starting with x1=2x1=2 as the initial approximation.

This does not converge very well using an initial approximation of 1, but it does at 2.

Step-by-step explanation:

Newton's formula: formula:

 xn+1=x − f(x)/f′(x)

You have x− 5sin(x)−x/5cos(x)−1

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

1− 5sin(1)−1/5cos(1)−1 =−0.885

Now plug the -.885 in:

−.885− 5sin(−.885)−(−.885)/5cos(−.885)−1

−.885− 5sin(−.885)−(−.885)/5cos(−.885)−1=0.49264

You keep doing that until it converges to a solution.

It should converge to ±2.5957, but using an initial approximation of 1 it will 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?.

77

step-by-step explanation:

bdhfnfhjffjkfjfjfjjf