# Finding roots of equation using Newtons method

1. Hi really struggling with this question any help would be great.
Use Newtons method to find the root of 4sin^2x - x = 0 which lies closest to x=2, correct to 3sf.

## The Attempt at a Solution

HallsofIvy
Homework Helper
1. Hi really struggling with this question any help would be great.
Use Newtons method to find the root of 4sin^2x - x = 0 which lies closest to x=2, correct to 3sf.

## The Attempt at a Solution

The Newton-Raphson method starts with an initial value $x_0$ and then calculates a sequence $x_1$, $x_2$, ... using the formula
$$x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}$$

Here $f(x)= 4sin^2(x)- x$ so $f'(x)= 8 sin(x)cos(x)- 1$

What initial value did you use and what results did you get?

Well I am at the same point now, however say i choose initial value of 1 when substituting into formula xn -(4sin^2x -x/ 8sinxcosx-1) my calculator is just saying error! is this because of the sin^2 part of the equation?

D H
Staff Emeritus
Newton's method does not always converge. For example suppose somewhere along the line you get a value of xn that is close to a zero of the derivative function. This will send xn+1 into never-never land. This particular function, $$f(x)=4\sin^2(x)-x$$, has a very narrow interval of convergence wrt Newton's method. An initial value of 1 is outside that interval.