Approximating square root by sexagesimal fractions

Question

I am reading The Works of Archimedes and I have found the following method for approximating the square root with sexagesimal fractions:

Ptolemy has first found the integral part of $\sqrt{4500}$ to be $67$. Now $67^2 = 4489$, so that the remainder is $11$. Suppose now that the rest of the square root is expressed by means of the usual sexagesimal fractions, and that we may therefore put $$\sqrt{4500} = \sqrt{67^2 + 11} = 67 + \frac{x}{60} + \frac{y}{60^2}$$ where $x,y$ are yet to be found. Thus $x$ must be such that $\frac{2\cdot67x}{60}$ is somewhat less than $11$, or $x$ must be somewhat less than $\frac{11\cdot60}{2\cdot67}$ or $\frac{330}{67}$, which is at the same time greater than $4$.

I am interested only in the sentence "Thus $x$ must be such that...". How are these conditions on $x$ assumed, namely the fractions $\frac{2\cdot67x}{60}$ and $\frac{11\cdot60}{2\cdot67}$?

Christian Blatter · Accepted Answer · 2020-03-24T07:32:14.077

1

Forget the $y$ for a moment. Then we want $$\sqrt{67^2+11}\approx67+{x\over60}\ .$$ This means $$67^2+11\approx 67^2+{2\cdot67\cdot x\over60}+\left({x\over60}\right)^2\ .$$ Note that $0\leq{x\over60}<1$, and the square is even smaller. Therefore we want that $${2\cdot67\cdot x\over60}<11,\quad{\rm and\ as\ large\ as\ possible.}$$ Therefore $x<{60\cdot11\over 2\cdot 67}= {660\over134}=4.925$; hence $x=4$. We now proceed with $$\sqrt{67^2+11}\approx67+{4\over60}+{y\over60^2}\ ,$$ and do a similar computation with respect to $y$.This will lead to $y=55$.

edited Mar 24 '20 at 07:32

answered Mar 23 '20 at 11:10

Christian Blatter

226,825

How do we get this fraction ${660\over134}$? – Michael Munta Mar 23 '20 at 22:16
How do you go from ${2\cdot67\cdot x\over60}<11$ to $x<{60\cdot11\over 2\cdot 67}$? – Michael Munta Sep 07 '20 at 08:16
@MichaelMunta: I multiplied with ${60\over2\cdot67}$. – Christian Blatter Sep 07 '20 at 08:22

score 0 · Answer 2 · answered Mar 23 '20 at 10:19

The first trick is to factor out $67$ :

$$\sqrt{67^2+11}=67\cdot \sqrt{1+\frac{11}{67^2}}$$

Now, the taylor series of $\sqrt{1+x'}$ gives the linear approximation $$\sqrt{1+x'}\approx 1+\frac{x'}{2}$$ for $x'\approx 0$. To avoid confusion with the $x$ here, I used $x'$.

Hence we have

$$\frac{11}{2\cdot 67}=\frac{x}{60}$$ which gives $$x=\frac{30\cdot 11}{67}=4.9254$$

This value is a bit too large which can be concluded by estimating the Lagrange residue of the taylor-series. In fact, inserting this for $x$ already gives an excellent approximation of $\sqrt{4500}$

Approximating square root by sexagesimal fractions

2 Answers2