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

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

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Answer 1 · 2015-07-11T07:16:23

Cutting the binomial series short,

${(1 + x)}^{\frac{1}{2}} ≅ 1 + \frac{x}{2} - \frac{x^{2}}{8}$ $(1+x)^(1/2) ~= 1+x/2-x^2/8$

Then $\sqrt{3} = \frac{5}{3} \sqrt{1.08} ≅ 1 + \frac{0.08}{2} - \frac{{0.08}^{2}}{8} = 1.732$ $sqrt(3) = 5/3sqrt(1.08) ~= 1+0.08/2-(0.08^2)/8 = 1.732$

Explanation:

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

$=1+x/2-x^2/8+x^3/16-...$

If $x$ is small this will converge rapidly

So with $x = 0.08$ we get:

$sqrt(1.08) = (1+0.08)^(1/2) ~= 1 + 0.08/2 - (0.08^2)/8$

$=1+0.04-0.0008 = 1.0392$

Now $1.08 = 3^3/(5^2)$

So $sqrt(1.08) = sqrt(3^3/(5^2)) = sqrt(3*(3/5)^2) = 3/5sqrt(3)$

So $sqrt(3) = 5/3sqrt(1.08) ~= 5/3*1.0392 = 1.732$

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

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