# How do you find a one-decimal place approximation for sqrt 18?

Oct 23, 2015

Use Newton Raphson method to find

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

#### Explanation:

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

${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$

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

Then:

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{4}^{2} + 18}{2 \cdot 4} = \frac{16 + 18}{8} = \frac{34}{8} = \frac{17}{4} = 4.25$

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

${a}_{2} = \frac{{a}_{1}^{2} + n}{2 {a}_{1}} = \frac{{\left(\frac{17}{4}\right)}^{2} + 18}{2 \cdot \left(\frac{17}{4}\right)}$

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

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

That's more decimal places than we need, but at least we can say with confidence that $\sqrt{18} \approx 4.2$ 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) = [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)