# How do I show that the limit is to the 600th places and is it too hasty to take the limit?

I am a little unsure about (b) and (d)

For (b), is it appropriate to just take the limit directly? (do I have to show that the sequence is bounded and monotone?) from $x_{n+1} = \frac{1}{2}(x_n+5/x_n)$? That is $\ell = \frac{1}{2}(\ell+5/\ell)$? Or is the question implying I should take the one in (a) instead (even though it is equivalent). I got it to $x_{n+1}^2-5 = \dfrac{1}{4x_n}(x_n^2-5)^2$

For (d), I was able to compute the tenth term on Mathematica as a fraction and and evaluated it using the numerical command.

So I thought that to show I would get the 600th decimal place, I would do $|x_{10}-\sqrt{5}| = 0.\underbrace{000...}_{600 \; \text{zeros}}1$. On Mathematica, I got $|x_{10}-\sqrt{5}| = 0 \times 10^{600}$. Does this show I got it to the 600th place? Because I tried 601, 602, and 605 and they seem all to be $0 \times 10^{x}$. I am terrible with counting, so should I actually take the 601th place of $x_{10}$ and subtract $\sqrt{5}$'s 600th place?

EDIT

$x_{n+1}^2-5 = \dfrac{1}{4x_n}(x_n^2-5)^2 < \dfrac{(x_n^2 - 5)^2}{20}$ for all n $\geq 2$

-
"Hence" means using result (a) to do (b)? For (d), I think you got it. –  Patrick Li Oct 8 '12 at 4:11
You are supposed to use (a), together with the easily calculated value of $x_1^2-5$, to work out the value of $x_{10}^2-5$, thereby answering (d). –  Gerry Myerson Oct 8 '12 at 4:26
Why would the expression in (a) be any easier for a computer? –  Hawk Oct 8 '12 at 4:28
"Why would the expression in (a) be any easier for a computer?" I don't understand this. Who said anything about a computer? By the way, I think you got the wrong answer for (a). –  Gerry Myerson Oct 8 '12 at 6:23
I'll underestimate $x_n$, wait for edit –  Hawk Oct 8 '12 at 18:51

Your teacher will not want to compare two 600+ digit numbers you computed with a device of finite accuracy digit by digit to have a proof of (d). Rather use (a) and $x_0^2-5$ to show that $x_{10}^2-5$ is how small? Then use $x_{10}^2-5=(x_{10}-\sqrt 5)(x_{10}+\sqrt5)$.
Where are you getting $x_0$ from? –  Hawk Oct 8 '12 at 18:17