Lying in bed unable to go back to sleep this morning I worked out what is the length of a regular hexagon circumscribed about a circle of diameter 1 (it is 2*sqrt(3), about 3.46) and then started thinking about what I’ve read more than once: that Archie showed that pi is between 3 10/71 and 3 1/7 and that he did this by finding a formula for what happens to the total length of regular inscribed and circumscribed polygons when the number of sides is doubled. These turn out to be formulas involving square roots of the original ones. He then doubled the regular hexagons 4 times to get his estimates from 96-sided polygons. This left him with at least 8 square roots to extract by hand. I could do, if I had to, using the algorithm I learned in 8th grade, but, lacking decimal notation how could he have done this? The Greeks used a numbering based on alpha = 1, beta = 2, etc.
Here’s how the Babylonians did square roots:
There are ways to approximate square roots that use only ratios of integers. Take sqrt(2) as an example, for instance: We’ll get a sequence of ever-better approximations.
Our first approximation is 1 .
Now, 2/sqrt(2) = sqrt(2), so if 1 is an approximation, then so is 2/1, or 2 . One is too high and the other is two low, so a number in between them will be a better approximation. One way to get a number between two rational numbers is to add together both the numerators and denominators. So our next approximation is between 1/1 and 2/1, giving us 3/2.
By the same logic, if 3/2 is an approximation of sqrt(2), then so is 4/3 (2/(3/2)). And in between 3/2 and 4/3 is 7/5 .
So our third approximation is 7/5, or 10/7 (between 1.4 and 1.42857…, in decimal notation)
Fourth approximation is 17/12, or 24/17 (between 1.416 and 1.41176…)
Fifth approximation is 41/29, or 58/41 (between 1.41379… and 1.41463…)
We could also use this method for other numbers. Suppose, for instance, we wanted the square root of 10. We’ll start with 3/1 as our first approximation. Then our approximations are:
3/1 or 10/3
13/4 or 40/13
53/17 or 170/53
223/70 or 700/223
and so on. At that last approximation, we’re somewhere between 3.1857… and 3.1390… .
Googling “Archimedes square roots” (without the quotes) turns up some relevant links, including Archimedes and the Square Root of 3 and Archimedes’ calculations of square roots (pdf).
FWIW I just happen to currently be in the middle of Eli Maor’s book on The Pythagorean Theorem. It includes a chapter on Archimedes, but, alas, doesn’t really address your question (although it does show the method that the Babylonians supposedly used to approximate square roots, which presumably would have been available to Archimedes).
[QUOTE=Eli Maor]
It was a brilliant piece of work, all the more so if we consider that the Greeks did not have an efficient method of calculating with numbers; theirs was a hybrid between the old Babylonian base-60 system and their own system in which each letter of the alphabet had a numerical value
[/QUOTE]
He might well have come up with a method for calculating them, but my own theory is that he would have had access to a table that was created by approximation, trial and error. They understood fractions so it wasn’t hard, just tedious, to use the method described by Chronos above so that the likes of Archimedes didn’t have to.
When I was at school we had books of tables for all kind of calculations. Logarithms for example, which someone had tediously worked out for us so that we didn’t have to. When I started work in a sales office, and had to calculate commissions without a calculator, I created my own table of 16 and two thirds percent (bear in mind that we use pounds, shillings and pence in those far off days). I had tables for all the common ones, but not that.
Sixteen and two-thirds percent? As in, one-sixth? Should have been no big deal in £.s.d., surely?
Not difficult - just tedious and subject to error.
There are 20 shillings and 240 pennies in a pound. I had to do this maybe 40 or 50 times a day for various amounts which could also include halfpennies. I also had to calculate some other, simpler, ones but those I could do in my head or look up in a published table.
So, One sixth of four pounds ten shillings and ninepence halfpenny is?
Well, if the Ancients were anything like me, you ‘do it in your head’.
I can normally get down to 2 or 3 significant places when I cant sleep, much more calming than thinking of imaginary sheep.
As an aside is the contemporary mathematical system of expression for us upright apes actually the most efficient?
Just wondered if we might learn something from other, earlier civilisations.
Cube root of 2 is 1.26 near as dammit, square root is 1.414213538, both from memory. Sad, sad sad.
Peter
This is called Newton’s method and I know it well. Now explain how you can do it without good decimal notation. Imagine doing it using numerical notation even more primitive than roman numerals. Now imagine finding square roots of square roots of square roots of square roots of square roots.
Just out of interest, I once compared two programs for finding square roots. One used Newton’s method and the on-chip division on the 8088 chip (so you can imagine how long ago this was). The other used the one-bit-at-a-time version of the square root method I learned in 8th grade. It was simplified by the fact that you tried 1 and if that was too large, then the next bit had to be 0. The latter method was an order of magnitude faster.
Others have speculated on the same question, of course. Here is one such speculation.
The only similarity between this method and Newton’s, that I can see, is that both use an nth estimate to produce an n+1th estimate. And I did do it without using decimal notation (well, I used decimal notation to express the numbers, but I didn’t make use of any of the properties of decimal notation). The only operations you need to use that method are multiplication by the number that you’re taking the square root of, and addition.
Just to illustrate, here are the approximants for sqrt(2), using this method, as expressed in Roman numerals:
I/I, II/I
III/II, IV/III
VII/V, X/VII
XVII/XII, XXIV/XVII
XLI/XXIX, LVIII/XLI
I believe Archimedes used the same analytical solution technique as Fermat.
Once the general concept of fractions beyond 1 is understood -“one divided into 53 pieces, of which I have 170” - then the math is simply the question of “can you work with large numbers? (Add, subtract, multiply, divide)” The only other skill needed would be the ability to compare two fractions with different denominators.
Any decent sized civilization obviously has to do work with numbers well over 1,000 - whether taxes, bags of wheat, gold pieces, census, blocks of stone for a large building, whatever. The abacus is supposed one or two thousand years older than Archimedes. If he liked playing with big numbers, presumably he learned how to use it.
Chronos’s method would be equivalent to Newton’s method (aka, “the Babylonian method”, in this context) if, to get a value between approximations x and 2/x (or r/x, for whatever radicand r), the mean was taken.
But Chronos does not use the mean; rather, he uses the mediant. This makes the arithmetic much simpler. No longer is there any multiplication involved!
And to prove it–Newton’s method is this:
x’ = x - f(x)/f’(x)
So we set up a function where the zeroes occur when x is the square root of C:
f(x) = C - x[sup]2[/sup]
And:
f’(x) = -2x
So:
x’ = x - (C - x[sup]2[/sup])/-2x
x’ = x + C/2x - x/2
x’ = x/2 + C/2x
x’ = (x + C/x)/2
As Indistinguishable points out, that takes the mean of the two values instead of the mediant (which is a new word to me). But otherwise the same.
I know that Newton’s method converges quadratically in this instance: you double the number of correct bits with each iteration. I don’t know about the mediant version, though.
The mediant version is much slower to converge: correct bits accumulate at an asymptotically linear rate (rather than exponentially, as with the mean version).
Specifically, if we are approximating sqrt(C), calling our approximation q, then the mediant version has the property that the ratio q - sqrt(C) : q + sqrt(C) multiplies by 1 - sqrt(C) : 1 + sqrt(C) on each iteration.
My preferred mathematical sedative for such occasions is to work through a demonstration of how the differences between four successive cubes–for example 1, 8, 27, 64–when taken three times in succession, always yield the single number 6. With paper and pen it’s easy enough for me to prove the general principal with algebraic notation, but lying in the dark I prefer to take the specific case mentioned above. While not forgetting the underlying principle, I continue to find the mechanism of it fascinating.
Much more generally, the n-th repeated difference of n-th powers is always n!, because (x + 1)[sup]n[/sup] - x[sup]n[/sup] is a polynomial with leading coefficient n of degree n - 1. (This is the discrete calculus analogue of the perhaps more familiar fact that the n-th derivative of the n-th power function is n!, because the derivative of x[sup]n[/sup] is n * x[sup]n - 1[/sup].)
Whatever the method, the calculations were sure to be tedious. Archimedes didn’t necessarily do all that himself. He had students for that.
Likewise, the STEM masters of the late Middle Ages who published those massive tables of trig functions also had a workshop full of apprentices to do the bulk of the grunt work. I’m sure Archimedes did likewise.