Here is a beautiful and astonishingly simple formula for π created by the French mathematician François Viète (1540-1603):
• 2 / π = √2/2 * √(2 + √2)/2 * √(2 + √(2 + √2))/2…
I can remember testing the formula on a scientific calculator that allowed simple programming. As I pressed the = key and the results began to home in on π, I felt as though I was watching a tall and elegant temple emerge through swirling mist.
The factorial function, n!, is easy to understand. You simply take an integer and multiply it by all integers smaller than it (by convention, 0! = 1):
0! = 1
1! = 1
2! = 2 = 2*1
3! = 6 = 3*2*1
4! = 24 = 4*3*2*1
5! = 120 = 5*4*3*2*1
6! = 720 = 6*120 = 6*5!
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
The gamma function, Γ(n), isn’t so easy to understand. It allows you to find the factorials of not just the integers, but everything between the integers, like fractions, square roots, and transcendental numbers like π. Don’t ask me how! And don’t ask me how you get this very beautiful and unexpected result:
Γ(1/2) = √π = 1.77245385091...
But a blog called Mathematical Enchantments can tell you more:
Post-Performative Post-Scriptum
glamour | glamor, n. Originally Scots, introduced into the literary language by Scott. A corrupt form of grammar n.; for the sense compare gramarye n. (and French grimoire ), and for the form glomery n. 1. Magic, enchantment, spell; esp. in the phrase to cast the glamour over one. 2. a. A magical or fictitious beauty attaching to any person or object; a delusive or alluring charm. b. Charm; attractiveness; physical allure, esp. feminine beauty; frequently attributive colloquial (originally U.S.). — Oxford English Dictionary
How radians work (from Wikipedia)
Here’s √2 in base 2:
√2 = 1.01101010000010011110... (base=2)
And in base 3:
√2 = 1.10201122122200121221... (base=3)
And in bases 4, 5, 6, 7, 8, 9 and 10:
√2 = 1.12220021321212133303... (b=4)
√2 = 1.20134202041300003420... (b=5)
√2 = 1.22524531420552332143... (b=6)
√2 = 1.26203454521123261061... (b=7)
√2 = 1.32404746317716746220... (b=8)
√2 = 1.36485805578615303608... (b=9)
√2 = 1.41421356237309504880... (b=10)
And here’s π in the same bases:
π = 11.00100100001111110110... (b=2)
π = 10.01021101222201021100... (b=3)
π = 03.02100333122220202011... (b=4)
π = 03.03232214303343241124... (b=5)
π = 03.05033005141512410523... (b=6)
π = 03.06636514320361341102... (b=7)
π = 03.11037552421026430215... (b=8)
π = 03.12418812407442788645... (b=9)
π = 03.14159265358979323846... (b=10)
Mathematicians know that in all standard bases, the digits of √2 and π go on for ever, without falling into any regular pattern. These numbers aren’t merely irrational but transcedental. But are they also normal? That is, in each base b, do the digits 0 to [b-1] occur with the same frequency 1/b? (In general, a sequence of length l will occur in a normal number with frequency 1/(b^l).) In base 2, are there as many 1s as 0s in the digits of √2 and π? In base 3, are there as many 2s as 1s and 0s? And so on.
It’s a simple question, but so far it’s proved impossible to answer. Another question starts very simple but quickly gets very difficult. Here are the answers so far at the Online Encyclopedia of Integer Sequences (OEIS):
2, 572, 8410815, 59609420837337474 – A049364
The sequence is defined as the “Smallest number that is digitally balanced in all bases 2, 3, … n”. In base 2, the number 2 is 10, which has one 1 and one 0. In bases 2 and 3, 572 = 1000111100 and 210012, respectively. 1000111100 has five 1s and five 0s; 210012 has two 2s, two 1s and two 0s. Here are the numbers of A049364 in the necessary bases:
10 (n=2)
1000111100, 210012 (n=572)
100000000101011010111111, 120211022110200, 200011122333 (n=8410815)
11010011110001100111001111010010010001101011100110000010, 101201112000102222102011202221201100, 3103301213033102101223212002, 1000001111222333324244344 (n=59609420837337474)
But what number, a(6), satisfies the definition for bases 2, 3, 4, 5 and 6? According to the notes at the OEIS, a(6) > 5^434. That means finding a(6) is way beyond the power of present-day computers. But I assume a quantum computer could crack it. And maybe someone will come up with a short-cut or even an algorithm that supplies a(b) for any base b. Either way, I think we’ll get there, π and by.
A roulette is a little wheel or little roller, but it’s much more than a game in a casino. It can also be one of a family of curves created by tracing the path of a point on a rotating circle. Suppose a circle rolls around another circle of the same size. This is the resultant roulette:
The shape is called a cardioid, because it looks like a heart (kardia in Greek). Now here’s a circle with radius r rolling around a circle with radius 2r:
That shape is a nephroid, because it looks like a kidney (nephros in Greek).
This is a circle with radius r rolling around a circle with radius 3r:
And this is r and 4r:
The shapes above might be called outer roulettes. But what if a circle rolls inside another circle? Here’s an inner roulette whose radius is three-fifths (0.6) x the radius of its rollee:
The same roulette appears inverted when the inner circle has a radius two-fifths (0.4) x the radius of the rollee:
But what happens when the circle rolling “inside” is larger than the rollee? That is, when the rolling circle is effectively swinging around the rollee, like a bunch of keys being twirled on an index finger? If the rolling radius is 1.5 times larger, the roulette looks like this:
If the rolling radius is 2 times larger, the roulette looks like this:
Here are more outer, inner and over-sized roulettes:
And you can have circles rolling inside circles inside circles:
And here’s another circle-in-a-circle in a circle:
What’s special about the binary number 10011 and the ternary number 1001120221? To answer the question, you have to see double. 10011 contains all possible pairs of numbers created from 0 and 1, just as 1001120221 contains all possible pairs created from 0, 1 and 2. And each pair appears exactly once. Now try the quaternary number 10011202130322331. That contains exactly one example of all possible pairs created from 0, 1, 2 and 3.
But there’s something more: in each case, the number is the smallest possible number with that property. As the bases get higher, that gets less obvious. In quinary, or base 5, the smallest number containing all possible pairs is 10011202130314042232433441. The digits look increasingly random. And what about base 10? There are 100 possible pairs of numbers created from the digits 0 to 9, starting with 00, 01, 02… and ending with …97, 98, 99. To accommodate 100 pairs, the all-pair number in base 10 has to be 101 digits long. It’s a string of digits, so let’s call it a d-string:
1, 0, 0, 1, 1, 2, 0, 2, 1, 3, 0, 3, 1, 4, 0, 4, 1, 5, 0, 5, 1, 6, 0, 6, 1, 7, 0, 7, 1, 8, 0, 8, 1, 9, 0, 9, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 6, 7, 6, 8, 6, 9, 7, 7, 8, 7, 9, 8, 8, 9, 9, 1
Again, the digits look increasingly random. They aren’t: they’re strictly determined. The d-string is in harmony. As the digits are generated from the left, they impose restrictions on the digits that appear later. It might appear that you could shift larger digits to the right and make the number smaller, but if you do that you no longer meet the conditions and the d-string collapses into dischord.
Now examine d-strings containing all possible triplets created from the digits of bases 2, 3 and 4:
1, 0, 0, 0, 1, 0, 1, 1, 1, 0 in base 2 = 558 in base 10
1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 1, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 2, 2, 1, 0 in base 3 = 23203495920756 in base 10
1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 1, 0 in base 4 = 1366872334420014346556556812432766057460 in base 10
Note that there are 8 possible triplets in base 2, so the all-triplet number has to be 10 digits long. In base 10, there are 1000 possible triplets, so the all-triplet number has to be 1002 digits long. Here it is:
1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 0, 4, 0, 0, 4, 1, 0, 5, 0, 0, 5, 1, 0, 6, 0, 0, 6, 1, 0, 7, 0, 0, 7, 1, 0, 8, 0, 0, 8, 1, 0, 9, 0, 0, 9, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 1, 4, 0, 1, 4, 1, 1, 5, 0, 1, 5, 1, 1, 6, 0, 1, 6, 1, 1, 7, 0, 1, 7, 1, 1, 8, 0, 1, 8, 1, 1, 9, 0, 1, 9, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 2, 4, 0, 2, 4, 1, 2, 5, 0, 2, 5, 1, 2, 6, 0, 2, 6, 1, 2, 7, 0, 2, 7, 1, 2, 8, 0, 2, 8, 1, 2, 9, 0, 2, 9, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 1, 3, 4, 0, 3, 4, 1, 3, 5, 0, 3, 5, 1, 3, 6, 0, 3, 6, 1, 3, 7, 0, 3, 7, 1, 3, 8, 0, 3, 8, 1, 3, 9, 0, 3, 9, 1, 4, 2, 0, 4, 2, 1, 4, 3, 0, 4, 3, 1, 4, 4, 0, 4, 4, 1, 4, 5, 0, 4, 5, 1, 4, 6, 0, 4, 6, 1, 4, 7, 0, 4, 7, 1, 4, 8, 0, 4, 8, 1, 4, 9, 0, 4, 9, 1, 5, 2, 0, 5, 2, 1, 5, 3, 0, 5, 3, 1, 5, 4, 0, 5, 4, 1, 5, 5, 0, 5, 5, 1, 5, 6, 0, 5, 6, 1, 5, 7, 0, 5, 7, 1, 5, 8, 0, 5, 8, 1, 5, 9, 0, 5, 9, 1, 6, 2, 0, 6, 2, 1, 6, 3, 0, 6, 3, 1, 6, 4, 0, 6, 4, 1, 6, 5, 0, 6, 5, 1, 6, 6, 0, 6, 6, 1, 6, 7, 0, 6, 7, 1, 6, 8, 0, 6, 8, 1, 6, 9, 0, 6, 9, 1, 7, 2, 0, 7, 2, 1, 7, 3, 0, 7, 3, 1, 7, 4, 0, 7, 4, 1, 7, 5, 0, 7, 5, 1, 7, 6, 0, 7, 6, 1, 7, 7, 0, 7, 7, 1, 7, 8, 0, 7, 8, 1, 7, 9, 0, 7, 9, 1, 8, 2, 0, 8, 2, 1, 8, 3, 0, 8, 3, 1, 8, 4, 0, 8, 4, 1, 8, 5, 0, 8, 5, 1, 8, 6, 0, 8, 6, 1, 8, 7, 0, 8, 7, 1, 8, 8, 0, 8, 8, 1, 8, 9, 0, 8, 9, 1, 9, 2, 0, 9, 2, 1, 9, 3, 0, 9, 3, 1, 9, 4, 0, 9, 4, 1, 9, 5, 0, 9, 5, 1, 9, 6, 0, 9, 6, 1, 9, 7, 0, 9, 7, 1, 9, 8, 0, 9, 8, 1, 9, 9, 0, 9, 9, 2, 2, 2, 3, 2, 2, 4, 2, 2, 5, 2, 2, 6, 2, 2, 7, 2, 2, 8, 2, 2, 9, 2, 3, 3, 2, 3, 4, 2, 3, 5, 2, 3, 6, 2, 3, 7, 2, 3, 8, 2, 3, 9, 2, 4, 3, 2, 4, 4, 2, 4, 5, 2, 4, 6, 2, 4, 7, 2, 4, 8, 2, 4, 9, 2, 5, 3, 2, 5, 4, 2, 5, 5, 2, 5, 6, 2, 5, 7, 2, 5, 8, 2, 5, 9, 2, 6, 3, 2, 6, 4, 2, 6, 5, 2, 6, 6, 2, 6, 7, 2, 6, 8, 2, 6, 9, 2, 7, 3, 2, 7, 4, 2, 7, 5, 2, 7, 6, 2, 7, 7, 2, 7, 8, 2, 7, 9, 2, 8, 3, 2, 8, 4, 2, 8, 5, 2, 8, 6, 2, 8, 7, 2, 8, 8, 2, 8, 9, 2, 9, 3, 2, 9, 4, 2, 9, 5, 2, 9, 6, 2, 9, 7, 2, 9, 8, 2, 9, 9, 3, 3, 3, 4, 3, 3, 5, 3, 3, 6, 3, 3, 7, 3, 3, 8, 3, 3, 9, 3, 4, 4, 3, 4, 5, 3, 4, 6, 3, 4, 7, 3, 4, 8, 3, 4, 9, 3, 5, 4, 3, 5, 5, 3, 5, 6, 3, 5, 7, 3, 5, 8, 3, 5, 9, 3, 6, 4, 3, 6, 5, 3, 6, 6, 3, 6, 7, 3, 6, 8, 3, 6, 9, 3, 7, 4, 3, 7, 5, 3, 7, 6, 3, 7, 7, 3, 7, 8, 3, 7, 9, 3, 8, 4, 3, 8, 5, 3, 8, 6, 3, 8, 7, 3, 8, 8, 3, 8, 9, 3, 9, 4, 3, 9, 5, 3, 9, 6, 3, 9, 7, 3, 9, 8, 3, 9, 9, 4, 4, 4, 5, 4, 4, 6, 4, 4, 7, 4, 4, 8, 4, 4, 9, 4, 5, 5, 4, 5, 6, 4, 5, 7, 4, 5, 8, 4, 5, 9, 4, 6, 5, 4, 6, 6, 4, 6, 7, 4, 6, 8, 4, 6, 9, 4, 7, 5, 4, 7, 6, 4, 7, 7, 4, 7, 8, 4, 7, 9, 4, 8, 5, 4, 8, 6, 4, 8, 7, 4, 8, 8, 4, 8, 9, 4, 9, 5, 4, 9, 6, 4, 9, 7, 4, 9, 8, 4, 9, 9, 5, 5, 5, 6, 5, 5, 7, 5, 5, 8, 5, 5, 9, 5, 6, 6, 5, 6, 7, 5, 6, 8, 5, 6, 9, 5, 7, 6, 5, 7, 7, 5, 7, 8, 5, 7, 9, 5, 8, 6, 5, 8, 7, 5, 8, 8, 5, 8, 9, 5, 9, 6, 5, 9, 7, 5, 9, 8, 5, 9, 9, 6, 6, 6, 7, 6, 6, 8, 6, 6, 9, 6, 7, 7, 6, 7, 8, 6, 7, 9, 6, 8, 7, 6, 8, 8, 6, 8, 9, 6, 9, 7, 6, 9, 8, 6, 9, 9, 7, 7, 7, 8, 7, 7, 9, 7, 8, 8, 7, 8, 9, 7, 9, 8, 7, 9, 9, 8, 8, 8, 9, 8, 9, 9, 9, 1, 0
Consider the quadruplet number in base 10. There are 10000 possible quadruplets, so the all-quadruplet number is 10003 digits long. And so on. In general, the “all n-tuplet” number in base b contains b^n n-tuplets and is (b^n + n-1) digits long. If b = 10 and n = 4, the d-string starts like this:
1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 3, 1, 0, 0, 4, 0, 0, 0, 4, 1, 0, 0, 5, 0, 0, 0, 5, 1, 0, 0, 6, 0, 0, 0, 6, 1, 0, 0, 7, 0, 0, 0, 7, 1, 0, 0, 8, 0, 0, 0, 8, 1, 0, 0, 9, 0, 0, 0, 9, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 0, 0, 1, 2, 1, 0, 1, 3, 0, 0, 1, 3, 1, 0, 1, 4, 0, 0, 1, 4, 1, 0, 1, 5, 0, 0, 1, 5, 1, 0, 1, 6, 0, 0, 1, 6, 1, 0, 1, 7, 0, 0, 1, 7, 1, 0, 1, 8, 0, 0, 1, 8, 1, 0, 1, 9, 0, 0, 1, 9, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 2, 2, 0, 0, 2, 2, 1, 0, 2, 3, 0, 0, 2, 3, 1, 0, 2, 4, 0, 0, 2, 4, 1, 0, 2, 5, 0, 0, 2, 5, 1, 0, 2, 6…
What about when n = 100? Now the d-string is ungraspably huge – too big to fit in the known universe. But it starts with 1 followed by a hundred 0s and every digit after that is entirely determined. Perhaps there’s a simple way to calculate any given digit, given its position in the d-string. Either way, what is the ontological status of the d-string for n=100? Does it exist in some Platonic realm of number, independent of physical reality?
Some would say that it does, just like √2 or π or e. I disagree. I don’t believe in a Platonic realm. If the universe or multiverse ceased to exist, numbers and mathematics in general would also cease to exist. But this isn’t to say that mathematics depends on physical reality. It doesn’t. Nor does physical reality depend on mathematics. Rather, physical reality necessarily embodies mathematics, which might be defined as “entity in interrelation”. Humans have invented small-m mathematics, a symbolic way of expressing the physical embodiment of big-m mathematics.
But small-m mathematics is actually more powerful and far-ranging, because it increases the number, range and power of entities and their interaction. Where are √2 and π in physical reality? Nowhere. You could say that early mathematicians saw their shadows, cast from a Platonic realm, and deduced their existence in that realm, but that’s a metaphor. Do all events, like avalanches or thunderstorms, exist in some Platonic realm before they are realized? No, they arise as physical entities interact according to laws of physics. In a more abstract way, √2 and π arise as entities of another kind interact according to laws of logic: the concepts of a square and its diagonal, of a circle and its diameter.
The d-strings discussed above arise from the interaction of simpler concepts: the finite set of digits in a base and the ways in which they can be combined. Platonism is unnecessary: the arc and spray of a fountain are explained by the pressure of the water, the design of the pipes, the arrangement of the nozzles, not by reference to an eternal archetype of water and spray. In small-m mathematics, there are an infinite number of fountains, because small-m mathematics opens a door to a big-U universe, infinitely larger and richer than the small-u universe of physical reality.
“In 1997, Fabrice Bellard announced that the trillionth digit of π, in binary notation, is 1.” — Ian Stewart, The Great Mathematical Problems (2013).
Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral 