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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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I read the question as asking you to perform two iterations of Newton's method with a starting value of $1$. Is this what you tried? What result did you get and what were you expecting? – AMPerrine Oct 27 '11 at 18:40
Your rendering of the question doesn't really make sense. Did you copy it verbatim? If not, please do so. If yes, this is a badly formulated question, which I would interpret as AMPerrine did above. – joriki Oct 27 '11 at 18:51
You wrote "I approximated with 1 and then[...]", but you didn't tell us what happened when you started with 1. Why don't you tell us what you think went wrong with that before you go on to tell us you tried 0.75? – Michael Hardy Oct 27 '11 at 19:49
When I went lower I got negative numbers and some positive numbers, it wasn't converging. – user138246 Oct 27 '11 at 20:11
up vote 4 down vote accepted

@ 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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What do you mean we get? Do I take that value that x1 gives and plug that in for x2? – user138246 Oct 27 '11 at 19:45
@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. – anon Oct 27 '11 at 19:49

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