Fractions and Addresses

As an example, suppose a1a2a3a4... = (110)infinity.

We can split this into two geometric series

1/21 + 1/24 + 1/27 + ... and 1/22 + 1/25 + 1/28 + ...

or the single series

3/22 + 3/25 + 3/28 + 3/211 + ...

and sum those, or more simply note that the point x with address (110)infinity is a fixed point of f1f1f0.

Then the equation x = f1(f1(f0(x)))gives

x = (1/2)*((1/2)*(x/2) + 1/2) + 1/2 = x/8 + 6/8

so x = 6/7.

Now suppose the address of x is (a1a2...an)infinity.

Then x is a fixed point of fa1fa2...fan.

Each fai(x) divides x by 2 and adds 1 or 0, depending on whether ai is 1 or 0.

So fa1fa2...fan(x) = x/2n + a1/2 + a2/4 + ... + an/2n = x/2n + k/2n,

where k = a12n-1 + a22n-2 + ... + an.

Then x = fa1fa2...fan(x) becomes

x = x/2n + k/2n

which gives

x = k/(2n - 1)

That is, any number whose address is a repeating block is a rational number with denominator of the form 2n - 1 for some n.

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