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Activity 1-15: Ergodic mathematics
What follows is an introduction
to a newish branch of mathematicsERGODIC mathematics.
Firstly Task: how many irrational numbers do you know?
Write down five.
Now pick any number between 0 and 1, let’s call it t.
Now pick another number between 0 and 1,
let’s call it α,
with the condition that α is irrational.
Work out t + α, and then throw away everything
except the decimal part (call this a1).
Now add α again to this, throwing away everything
except the decimal part once more (call this a2.)
Find a1, a2, a3, a4, a5 and a6.
Can we say anything about the numbers a1, a2, a3...?
They will all be between 0 and 1:
are some parts of (0, 1) more likely to be hit than others?
It is a fundamental theorem of ergodic maths
that all parts of (0, 1) will be hit equally often.
This can be explored using an Excel spreadsheet.
Task: explore the Ergodic Spreadsheet.
This fundamental ergodic theorem says:
in the long run, the probability that this process ends up
giving us a value in the interval (a, b), where 0 < a < b < 1, is
b  a.
Ergodic maths concerns itself with repeated processes.
Here we had a map from one set to itself.
This gave us the sequence a1, a2, a3, a4, a5, a6.....
This sequence of points is called the orbit of a1.
The nature of the orbit tells us a lot about α.
If α is rational, the orbit is finite.
If α is irrational, the orbit is infinite,
and ‘equally spread’ on [0,1).
Another problem that classes as ergodic we all know this sequence:
1, 2, 4, 8, 16, 32, 64, 128...
Consider this sequence,
found by taking the starting digits from the above:
1, 2, 3, 8, 1, 3, 6, 1...
Which digit occurs most often in this sequence?
Define {t} as ‘the fractional part of t’.
Consider 500  512 < 600.
 5  102  28 < (5 + 1)  102.
Taking logs to base 10, log (5) + 2  8  log(2) < log (5 + 1) + 2.
 log(5)  {8 log (2)} < log (5 + 1).
Can we generalise this?
Consider k  10j  2n < (k + 1)  10j.
Taking logs to base 10, log (k) + j  n  log(2) < log (k + 1) + j.
 log(k)  {n log (2)} < log (k + 1).
Now log(2) is irrational, and 0  log(k) < log (k + 1)  1.
So by our fundamental ergodic theorem,
P(starting digit of 2n = k) = log(k+1)  log (k)
= log((k+1)/k).
So considering the values of k from {1, 2 ... 9},
when is log(k + 1)/log(k) greatest?
Clearly when k = 1.
So 1 is the most common starting digit for powers of 2.
Here is another theorem that deserves the adjective ergodic:
Task: find arithmetic progressions
of length 3, 4, 5…
by looking at the smallest prime numbers.
Prime Arithmetic
Progression
Spreadsheet
http://www.s253053503.websitehome.co.uk/
carom/carom-files/carom-1-15-2.xlsm
Below we have
the arithmetic sequences using the smallest possible numbers
of the given length made up purely of prime numbers.
What are
the longest known arithmetic sequences
made up of prime numbers?
On January 18, 2007, Jarosław Wróblewski
found the first known case
of 24 primes in arithmetic progression:
468,395,662,504,823 + 205,619223,092,870n,
for n = 0 to 23.
The constant 223 092 870 here is the product of
all the prime numbers up to 23.
On May 17, 2008, Wróblewski and Raanan Chermoni
found the first known case of 25 primes:
6,171,054,912,832,631 + 366,384223,092,870n,
for n = 0 to 24.
On April 12, 2010, Benoãt Perichon
(with software by Wróblewski and Geoff Reynolds)
found the first known case of 26 primes:
43,142,746,595,714,191 + 23,681,770223,092,870n,
for n = 0 to 25.
With thanks to:
Manfred Einsiedler and Tom Ward,
and their book, Introduction to Ergodic Theory.
Wikipedia, for another excellent article.
Graeme McRae, for his helpful site.
Carom is written by Jonny Griffiths, [email protected]