Algebra involving errors and root 2

**mitch_nufc** · March 2nd 2009, 01:48 AM

I can do the first bit, which is just manipulation of the LHS, but I'm not sure how to start on the second and third bit.

Say we wish to estimate the square rot of 2. We begin with a good first estimate $x_1$ , say $x_1 = 1.5$ . In fact, it is a little too big. We note that 2/1.5 is a little too small, so we form a new guess by averaging them, $x_2 = \frac{1}{2}(1.5+2/1.5)$ . We continue this method to give a sequence:

$x_{n+1} = \frac{1}{2}\bigg(\frac{2}{x_n} + x_n\bigg)$

To ease notation, we set $x_n = \sqrt{2} + \epsilon_n$ , so, $\epsilon_1 < 1/2^3$

i) Show that

$\bigg(\frac{2}{\sqrt{2} + \epsilon_n} - \big(\sqrt{2} - \epsilon_n\big)\bigg) = \frac{\epsilon_n^2}{\sqrt{2} + \epsilon_n}$ ;

ii) show that

$\epsilon_{n+1} = \frac{1}{2} \frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n} < \frac{1}{2}\epsilon_n^2$

iii) How big do we require $n$ so that $x_n$ is within $2^{-15}$ of $\sqrt{2}$ . Find the fraction $x_n$

**Grandad** · March 2nd 2009, 06:04 AM

Hello mitch_nufc

Originally Posted by mitch_nufc

I can do the first bit, which is just manipulation of the LHS, but I'm not sure how to start on the second and third bit.

Say we wish to estimate the square rot of 2. We begin with a good first estimate $x_1$ , say $x_1 = 1.5$ . In fact, it is a little too big. We note that 2/1.5 is a little too small, so we form a new guess by averaging them, $x_2 = \frac{1}{2}(1.5+2/1.5)$ . We continue this method to give a sequence:

$x_{n+1} = \frac{1}{2}\bigg(\frac{2}{x_n} + x_n\bigg)$

To ease notation, we set $x_n = \sqrt{2} + \epsilon_n$ , so, $\epsilon_1 < 1/2^3$

i) Show that

$\bigg(\frac{2}{\sqrt{2} + \epsilon_n} - \big(\sqrt{2} - \epsilon_n\big)\bigg) = \frac{\epsilon_n^2}{\sqrt{2} + \epsilon_n}$ ;

ii) show that

$\epsilon_{n+1} = \frac{1}{2} \frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n} < \frac{1}{2}\epsilon_n^2$

iii) How big do we require $n$ so that $x_n$ is within $2^{-15}$ of $\sqrt{2}$ . Find the fraction $x_n$

As you say, part (i) is straightforward. So:

(ii) $x_{n+1} = \sqrt{2}+ \epsilon_{n+1}$

$\Rightarrow \epsilon_{n+1} = x_{n+1} - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{x_n}+ x_n\right) - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}+ \sqrt{2} +\epsilon_n\right) - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}- (\sqrt{2} -\epsilon_n)\right) - \sqrt{2}$

$= \frac{1}{2}\frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n }$

$= \frac{1}{2}\frac{\epsilon_n^2}{x_n}$

$< \frac{1}{2}\epsilon_n^2$ , since $x_n > 1$

(iii) $\epsilon_1 < 2^{-3} \Rightarrow \epsilon_2 < \tfrac{1}{2}\cdot 2^{-6}=2^{-7}$

$\Rightarrow \epsilon_3 < \tfrac{1}{2}\cdot 2^{-14}= 2^{-15}$

$x_2 = \frac{1}{2}\left(\frac{3}{2}+ \frac{4}{3}\right) = \frac{17}{12}$

$\Rightarrow x_3 = ... = \frac{577}{408}$

Grandad

**Grandad** · March 2nd 2009, 06:35 AM

Hello mitch_nufc

Sorry, I left an extra $\sqrt{2}$ term in part (ii), so here's the corrected version:

(ii) $x_{n+1} = \sqrt{2}+ \epsilon_{n+1}$

$\Rightarrow \epsilon_{n+1} = x_{n+1} - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{x_n}+ x_n\right) - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}+ \sqrt{2} +\epsilon_n\right) - \sqrt{2}$

$= \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}- (\sqrt{2} -\epsilon_n)\right)$

$= \frac{1}{2}\frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n }$

$= \frac{1}{2}\frac{\epsilon_n^2}{x_n}$

$< \frac{1}{2}\epsilon_n^2$ , since $x_n > 1$

(iii) $\epsilon_1 < 2^{-3} \Rightarrow \epsilon_2 < \tfrac{1}{2}\cdot 2^{-6}=2^{-7}$

$\Rightarrow \epsilon_3 < \tfrac{1}{2}\cdot 2^{-14}= 2^{-15}$

$x_2 = \frac{1}{2}\left(\frac{3}{2}+ \frac{4}{3}\right) = \frac{17}{12}$

$\Rightarrow x_3 = ... = \frac{577}{408}$

Grandad

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Correction to part (ii)

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