How do you approximate $\frac{2}{\sqrt{3}}$ $2/sqrt3$ ?

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How do you approximate $\frac{2}{\sqrt{3}}$ $2/sqrt3$ ?

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Answer 1 · 2015-07-06T19:42:00

Use an iterative method to get a good rational approximation for $\sqrt{3}$ $sqrt(3)$ then use that to calculate $\frac{2}{\sqrt{3}}$ $2/sqrt(3)$ to get (say)

$\frac{2}{\sqrt{3}} ≅ \frac{2}{\frac{97}{56}} = \frac{112}{97} ≅ 1.155$ $2/sqrt(3) ~= 2/(97/56) = 112/97 ~= 1.155$

Explanation:

Start with a reasonable approximation $a_{0} = 2$ $a_0 = 2$ for $\sqrt{3}$ $sqrt(3)$ .

Then iterate using the formula:

$a_{i + 1} = \frac{a_{i}^{2} + 3}{2 a_{i}}$ $a_(i+1) = (a_i^2+3)/(2a_i)$

$a_{1} = \frac{a_{0}^{2} + 3}{2 a_{0}}$ $a_1 = (a_0^2+3)/(2a_0)$

$= \frac{2^{2} + 3}{2 \cdot 2}$ $=(2^2+3)/(2*2)$

$= \frac{7}{4}$ $=7/4$

$a_{2} = \frac{a_{1}^{2} + 3}{2 a_{1}}$ $a_2 = (a_1^2+3)/(2a_1)$

$= \frac{{(\frac{7}{4})}^{2} + 3}{2 \cdot \frac{7}{4}}$ $=((7/4)^2+3)/(2*7/4)$

$= \frac{\frac{49}{16} + \frac{48}{16}}{\frac{7}{2}}$ $=(49/16+48/16)/(7/2)$

$= \frac{97}{56}$ $=97/56$

We will stop here, but if you want more accuracy, just iterate again.

In general, to find the square root of $n$ $n$ , pick a reasonable first guess as $a_{0}$ $a_0$ , then iterate using:

$a_{i + 1} = \frac{a_{i}^{2} + n}{2 a_{i}}$ $a_(i+1) = (a_i^2+n)/(2a_i)$

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