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Activity 1-15: Ergodic mathematics

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Activity 1-15: Ergodic 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, a5and 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 theErgodicSpreadsheet.

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, isb a.

Ergodic maths concerns itself with repeated processes.

Here we had a map from one set to itself.

This gave us the sequencea1, 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).

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).

Clearly when k = 1.

So 1 is the most common starting digit for powers of 2.

Task: find arithmetic progressions

of length3, 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, JarosławWróblewski
- found the first known case
- of 24 primes in arithmetic progression:
- 468,395,662,504,823 + 205,619223,092,870n,
- 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 RaananChermoni
- found the first known case of 25 primes:
- 6,171,054,912,832,631 + 366,384223,092,870n,
- for n = 0 to 24.

- On April 12, 2010, BenoãtPerichon
- (with software by Wróblewski and Geoff Reynolds)
- found the first known case of 26 primes:
- 43,142,746,595,714,191 + 23,681,770223,092,870n,
- 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, hello@jonny-griffiths.net