How do you find a one-decimal place approximation for $\sqrt{18}$ $sqrt 18$ ?

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How do you find a one-decimal place approximation for $\sqrt{18}$ $sqrt 18$ ?

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Answer 1 · 2015-10-23T20:55:51

Use Newton Raphson method to find

$\sqrt{18} \approx 4.2$ $sqrt(18) ~~ 4.2$ to one decimal place.

Explanation:

To find an approximation for the square root of a number $n$ $n$ , start with a reasonable approximation $a_{0}$ $a_0$ and apply the following formula to make a better approximation. Repeat to get better approximations.

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

In our case $n = 18$ $n = 18$ and it makes sense to choose $a_{0} = 4$ $a_0 = 4$ since $4^{2} = 16$ $4^2 = 16$ is fairly close.

Then:

$a_{1} = \frac{a_{0}^{2} + n}{2 a_{0}} = \frac{4^{2} + 18}{2 \cdot 4} = \frac{16 + 18}{8} = \frac{34}{8} = \frac{17}{4} = 4.25$ $a_1 = (a_0^2+n)/(2a_0) = (4^2+18)/(2*4) = (16+18)/8 = 34/8 = 17/4 = 4.25$

Unfortunately $a_{1}$ $a_1$ is right in the middle between $4.2$ $4.2$ and $4.3$ $4.3$ , so let's try another iteration...

$a_{2} = \frac{a_{1}^{2} + n}{2 a_{1}} = \frac{{(\frac{17}{4})}^{2} + 18}{2 \cdot (\frac{17}{4})}$ $a_2 = (a_1^2+n)/(2a_1) = ((17/4)^2+18)/(2*(17/4))$

$= \frac{\frac{289}{16} + 18}{\frac{17}{2}} = \frac{289 + 18 \cdot 16}{17 \cdot 8} = \frac{289 + 288}{136}$ $=(289/16+18)/(17/2) =(289+18*16)/(17*8)=(289+288)/136$

$= \frac{577}{136} \approx 4.24265$ $=577/136 ~~ 4.24265$

That's more decimal places than we need, but at least we can say with confidence that $\sqrt{18} \approx 4.2$ $sqrt(18) ~~ 4.2$ to one decimal place.

Alternatively, you can use the method described in http://socratic.org/questions/given-an-integer-n-is-there-an-efficient-way-to-find-integers-p-q-such-that-abs-
to find a continued fraction expansion for $\sqrt{18}$ $sqrt(18)$ ...

$sqrt(18) = [4;bar(4,8)] = 4+1/(4+1/(8+1/(4+1/(8+1/(4+...)))))$

Then you can approximate $sqrt(18)$ by truncating the continued fraction.

For example:

$sqrt(18) ~~ [4;4,8] = 4+1/(4+1/8) = 4+8/33 = 140/33 = 4.dot(2)dot(4)$

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How do you find a one-decimal place approximation for $\sqrt{18}$ $sqrt 18$ ?

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How do you find a one-decimal place approximation for $\sqrt{18}$ $sqrt 18$ ?