# How do you use the binomial theorem to approximate 1.08^(1/2) and hence find sqrt(3) to 4 significant figures?

Jul 11, 2015

Cutting the binomial series short,

${\left(1 + x\right)}^{\frac{1}{2}} \cong 1 + \frac{x}{2} - {x}^{2} / 8$

Then $\sqrt{3} = \frac{5}{3} \sqrt{1.08} \cong 1 + \frac{0.08}{2} - \frac{{0.08}^{2}}{8} = 1.732$

#### Explanation:

(1+x)^(1/2) = 1+(1/2)x+(1/(2!))(1/2)(1/2-1)x^2+(1/(3!))(1/2)(1/2-1)(1/2-2)x^3+...

$= 1 + \frac{x}{2} - {x}^{2} / 8 + {x}^{3} / 16 - \ldots$

If $x$ is small this will converge rapidly

So with $x = 0.08$ we get:

$\sqrt{1.08} = {\left(1 + 0.08\right)}^{\frac{1}{2}} \cong 1 + \frac{0.08}{2} - \frac{{0.08}^{2}}{8}$

$= 1 + 0.04 - 0.0008 = 1.0392$

Now $1.08 = {3}^{3} / \left({5}^{2}\right)$

So $\sqrt{1.08} = \sqrt{{3}^{3} / \left({5}^{2}\right)} = \sqrt{3 \cdot {\left(\frac{3}{5}\right)}^{2}} = \frac{3}{5} \sqrt{3}$

So $\sqrt{3} = \frac{5}{3} \sqrt{1.08} \cong \frac{5}{3} \cdot 1.0392 = 1.732$