How do you use the binomial theorem to approximate 1.0812 and hence find 3 to 4 significant figures?

Cutting the binomial series short, (1+x)12=∼1+x2−x28 Then 3=531.08=∼1+0.082−0.0828=1.732 Explanation: (1+x)12=1+(12)x+(12!)(12)(12−1)x2+(13!)(12)(12−1)(12−2)x3+⋯ =1+x2−x28+x316−⋯ If x is small this will converge rapidly So with x=0.08 we get: 1.08=(1+0.08)12=∼1+0.082−0.0828 =1+0.04−0.0008=1.0392 Now 1.08=3352 So 1.08=3352=3⋅(35)2=353 3=531.08=∼53⋅1.0392=1.732

Expert answers to "How do you use the binomial theorem to..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Terrance Khan

Answered question

2022-02-05

How do you use the binomial theorem to approximate

{1.08}^{\frac{1}{2}}

and hence find

\sqrt{3}

to 4 significant figures?

Answer & Explanation

ljmolerovae

Beginner2022-02-06Added 15 answers

Cutting the binomial series short,

{(1 + x)}^{\frac{1}{2}} \tilde{=} 1 + \frac{x}{2} - \frac{x^{2}}{8}

Then

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

Explanation:

{(1 + x)}^{\frac{1}{2}} = 1 + (\frac{1}{2}) x + (\frac{1}{2!}) (\frac{1}{2}) (\frac{1}{2} - 1) x^{2} + (\frac{1}{3!}) (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) x^{3} + \dots

= 1 + \frac{x}{2} - \frac{x^{2}}{8} + \frac{x^{3}}{16} - \dots

If x is small this will converge rapidly
So with

x = 0.08

we get:

\sqrt{1.08} = {(1 + 0.08)}^{\frac{1}{2}} \tilde{=} 1 + \frac{0.08}{2} - \frac{{0.08}^{2}}{8}

= 1 + 0.04 - 0.0008 = 1.0392

Now

1.08 = \frac{3^{3}}{5^{2}}

\sqrt{1.08} = \sqrt{\frac{3^{3}}{5^{2}}} = \sqrt{3 \cdot {(\frac{3}{5})}^{2}} = \frac{3}{5} \sqrt{3}

\sqrt{3} = \frac{5}{3} \sqrt{1.08} \tilde{=} \frac{5}{3} \cdot 1.0392 = 1.732

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