Finding roots of equation using Newtons method

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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.



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  • #2
HallsofIvy
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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.



Homework Equations





The Attempt at a Solution


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

Here [itex]f(x)= 4sin^2(x)- x[/itex] so [itex]f'(x)= 8 sin(x)cos(x)- 1[/itex]

What initial value did you use and what results did you get?
 
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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?
 
  • #4
D H
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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, [tex]f(x)=4\sin^2(x)-x[/tex], has a very narrow interval of convergence wrt Newton's method. An initial value of 1 is outside that interval.

That said, you should be able to solve this equation without resorting to Newton's method. The convergence interval does contain this known point. Choose an initial value that is close to this known solution.
 

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