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I know how this works, I know what it does and what the goal of it is but I don't understand what this homework problem wants me to do and why I don't get the correct answer.

I am supposed to used newton's method with the specified initial approximation x1 to find x3, the third appromizmation to the root of the given equation

$x^5-x-1=0$, $x_1=1$ so what is this telling me to do? I approximated with $1$ and then $.75$ and then $.5$ and then $.25$ but it didn't help me at all. Do I increase values of $x$? Am I doing something wrong? Am I doing some math wrong? Is the formula not going to work?

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    When I went lower I got negative numbers and some positive numbers, it wasn't converging.2011-10-27

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@ Jordan

I agree with AMPerrine's interpretation of the question. Choose $f(x)=x^5-x-1$, which makes f'(x)=5x^4-1. The Newton's iteration take the form

x_{n+1}=x_n-\frac{\displaystyle f(x_n)}{\displaystyle f'(x_n)} = x_n-\frac{\displaystyle x_n^5-x_n-1}{\displaystyle 5x_n^4-1}

Starting with $x_1=1$ we get $x_2=1.25$, $x_3=1.1785$, $x_4=1.1675$, $x_5=1.16730$ which is the approximate root. As for your other question, the values $x_1=0.5$ or $0.25$ will not converge to right answer. Notice that $f(1)=-1$ and $f(2)=29$ So the root lies in $(1,2)$. If anything you should increase the values of $x_1$. Hope this helps.

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    @Jordan. Yes Thats what I meant. $x_1$ is the guess value. $x_2=x_1-f(x_1)/f'(x_1)$, $x_3=x_2-f(x_2)/f'(x_2)$, and so on.2011-10-27