# Pi in the Bi

Binary is beautiful — both simple and subtle. What could be simpler than using only two digits to count with?

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000...

But the simple patterns in the two digits of binary involve two of the most important numbers in mathematics: π and e (aka Euler’s number):

π = 3.141592653589793238462643383...
e = 2.718281828459045235360287471...

It’s easy to write π and e in binary:

π = 11.00100 10000 11111 10110 10101 00010...
e = 10.10110 11111 10000 10101 00010 11000...

But how do π and e appear in the patterns of binary 1 and 0? Well, suppose you use the digits of binary to generate the sums of distinct integers. For example, here are the sums of distinct integers you can generate with three digits of binary, if you count the digits from right to left (so the rightmost digit is 1, the the next-to-rightmost digit is 2, the next-to-leftmost digit is 3, and the leftmost digit is 4):

0000 → 0*4 + 0*3 + 0*2 + 0*1 = 0
0001 → 0*4 + 0*3 + 0*2 + 1*1 = 1*1 = 1
0010 → 0*4 + 0*3 + 1*2 + 0*1 = 1*2 = 2
0011 → 0*4 + 0*3 + 1*2 + 1*1 = 1*2 + 1*1 = 3
0100 → 1*3 = 3
0101 → 1*3 + 1*1 = 4
0110 → 3 + 2 = 5
0111 → 3 + 2 + 1 = 6
1000 → 4
1001 → 4 + 1 = 5
1010 → 4 + 2 = 6
1011 → 4 + 2 + 1 = 7
1100 → 4 + 3 = 7
1101 → 4 + 3 + 1 = 8
1110 → 4 + 3 + 2 = 9
1111 → 4 + 3 + 2 + 1 = 10

There are 16 sums (16 = 2^4) generating 11 integers, 0 to 10. But some integers involve more than one sum:

3 = 2 + 1 ← 0011
3 = 3 ← 0100

4 = 3 + 1 ← 0101
4 = 4 ← 1000

5 = 3 + 2 ← 0110
5 = 4 + 1 ← 1001

6 = 3 + 2 + 1 ← 0111
6 = 4 + 2 ← 1010

7 = 4 + 2 + 1 ← 1011
7 = 4 + 3 ← 1100

Note the symmetry of the sums: the binary number 0011, yielding 3, is the mirror of 1100, yielding 7; the binary number 0100, yielding 3 again, is the mirror of 1011, yielding 7 again. In each pair of mirror-sums, the two numbers, 3 and 7, are related by the formula 10-3 = 7 and 10-7 = 3. This also applies to 4 and 6, where 10-4 = 6 and 10-6 = 4, and to 5, which is its own mirror (because 10-5 = 5). Now, try mapping the number of distinct sums for 0 to 10 as a graph:

Graph for distinct sums of the integers 0 to 4

The graph show how 0, 1 and 2 have one sum each, 3, 4, 5, 6 and 7 have two sums each, and 8, 9 and 10 have one sum each. Now look at the graph for sums derived from three digits of binary:

Graph for distinct sums of the integers 0 to 3

The single taller line of the seven lines represents the two sums of 3, because three digits of binary yield only one sum for 0, 1, 2, 4, 5 and 6:

000 → 0
001 → 1
010 → 2
011 → 2 + 1 = 3
100 → 3
101 → 3 + 1 = 4
110 → 3 + 2 = 5
111 → 3 + 2 + 1 = 6

Next, look at graphs for sums derived from one to sixteen binary digits and note how the symmetry of the lines begins to create a beautiful curve (the y axis is normalized, so that the highest number of sums reaches the same height in each graph):

Graph for sums from 1 binary digit

Graph for sums from 2 binary digits

Graph for sums from 3 binary digits

Graph for sums from 4 binary digits

Graph for sums from 5 binary digits

Graph for sums from 6 binary digits

Graph for sums from 7 binary digits

Graph for sums from 8 binary digits

Graph for sums from 9 binary digits

Graph for sums from 10 binary digits

Graph for sums from 11 binary digits

Graph for sums from 12 binary digits

Graph for sums from 13 binary digits

Graph for sums from 14 binary digits

Graph for sums from 15 binary digits

Graph for sums from 16 binary digits

Graphs for 1 to 16 binary digits (animated)

You may recognize the shape emerging above as the bell curve, whose formula is this:

Formula for the normal distribution or bell curve (image from ThoughtCo)

And that’s how you can find pi in the bi, or π in the binary digits of 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101…

Post-Performative Post-Scriptum

I asked this question above: What could be simpler than using only two digits? Well, using only one digit is simpler still:

1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111...

But I don’t see an easy way to find π and e in numbers like that.