I am teaching a history of mathematics class this semester. We are beginning with a brief discussion of ancient number systems: Egyptian, Babylonian, Mayan, Chinese, Incan, Greek, Roman, and Hindu-Arabic. As I was writing up the first homework assignment it occured to me that I should investigate whether these numbers could be typeset using LaTeX.
It quickly became apparent that, because fonts are involved, I would have to use XeTeX rather than LaTeX. It was a fun (although time consuming) exercise. In the end I was able to typeset Egyptian hieroglyphics, Babylonian cuneiform, and Chinese rod numerals. Because the syntax was often messy, I spent a while burying the complicated TeX in the headers so that the numbers would be easy to work with in the document.
For example, to generate the Egyptian hieroglyphics for 123 I write
\Ehun\Eten\Eten\Eone\Eone\Eone.
The fraction 1/123 is
\Efrac{\Ehun\Eten\Eten\Eone\Eone\Eone}.
To express 123 in cuneiform all I have to write is
\Bnum{123}.
To create a number board with the Chinese counting rods representing 123 I type
\Cnum{|x|x|x|}{\Cvone & \Chtwo & \Cvthree}.
If you would like to give this a try, download my .tex files:
babylonian.tex and babylonian.pdf
I’d love to be able to do something similar with the Mayan numbers. I tried for a while, but couldn’t get them to work.
Disclaimer: I know my way around TeX pretty well, but I’m not a power user. It took me quite a while to get all this to work. I’m not sure I can offer much trouble-shooting advice if you can’t get this to work on your computer.
as an approximation of 
. Indeed, it is not a bad approximation: 
(although it is not within Archimedes’s bounds: 
). 
, i.e. less than 
pro mille of 
, and the second of these ways seems almost inescapable. It seems a plausible hypothesis that Plato knew of this equation, but was unable to prove whether or not it was a strict equality or only an approximation.
and 
, the 45-45-90 and 30-60-90 triangles which can be used to make the faces of the polyhedra, and the approximations to the area of a circle using these triangles.
and the area of the octagon is 
. So, it is true that the average of these areas is 
if and only if 
can be created from the integers using the operations 
, 
, 
, and 
, and by taking square roots.
(the ![{\sqrt[3]{2}}](http://s0.wp.com/latex.php?latex=%7B%5Csqrt%5B3%5D%7B2%7D%7D&bg=ffffff&fg=000000&s=0 "{\sqrt[3]{2}}")
(hence it is impossible to 
. Because 
has area 
and sector 
has area 
. Thus segment 
has area 
. The smaller circle has radius 
. So the semicircle 
has area 
. It follows that the lune has area 
. A square with the same area as the lune would have side-length 
formed by the bottom edge of the square of origami paper and a line (a fold, presumably), 
, meeting at the lower left corner of the square. Create an arbitrary horizontal fold to form the line 
, then fold the bottom edge up to 
. Let 
be the lower left corner of the square and 
be the left endpoint of 
. Open the paper and fold it to extend 
. Having accomplished this, the lines 
, which is the location of the segment 
after it is folded, 
, the fourth side of the isosceles trapezoid 
, and 
, the second diagonal of 
, where 
and 
.
and 
are parallel, 
, and because 
, 
. Thus 
. Now, 
are congruent isosceles triangles. Thus 
. It follows that 
.
-gons is related to 
that is closed under square roots, the field of origami-constructible numbers is the smallest subfield that is closed under square roots and cube roots. In fact, it is possible to solve any linear, quadratic, cubic, or quartic equation using origami!
is a critical point and 
, then the test fails.
has a critical point at 
and 
. Moreover, suppose that as we approach 
we have 
when 
and 
when 
. Is that enough to say that the critical point is a not a maximum or a minimum? His thought process was that if we look at slices 
we get curves that are concave up when 
and curves that are concave down when 
—surely that could not happen at a maximum or minimum.
. So the second derivative test fails. But observe that when 
and 
for all 
. A graph of the function is shown below. You can see the concave down cross sections for 
and 
.
as 
approaches 
is 
, or equivalently 
if for any 
there exists 
such that whenever 
, it follows that 
.
