Fractal numbers, part 2
Saturday, August 28, 2010 4:05:54 PM
The whole thing with normal long division (when you get decimals as an answer - when it doesn't work out evenly) is that at each step in the process the remainder is smaller. The "real" answer may be an infinite decimal (in the case of division, a repeating decimal) but the left-to-right division algorithm produces answers that are closer and closer to this real answer. The mathematical term is "converges to", and yes there are some attempts to make this somewhat vague definition more rigorous which I don't think I need to pursue here ...
So what about one of these right-to-left algorithms? Well, to be able to use the same definitions otherwise, they define "closer" as more divisible by the base p. If they feel that they need to be formal about it, they actually define the distance between two numbers as 1/pn, where pn is the largest power of p that divides the difference between the numbers.
I don't think that helped much, did it. Okay, use the example of p = 2 again, and let's look at 1, 3 and 13. By the definitions involved, the distance between 1 and 3 is 0.5 ... (3 - 1 = 2 and 1/2 is 0.5). The distance between 3 and 13 is also 0.5 (13 - 3 = 10, the largest power of 2 that divides 10 is 2, and 1/2 is 0.5), while the distance between 1 and 13 is actually 0.25 (12 is divisible by 4 and not just 2).
Are you familiar with series? That's a mathematical addition problem where you are adding up smaller and smaller numbers to find an answer. For example,
1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2n + ... = 1
I probably don't have to tell you, that particular problem doesn't work with p-adic numbers. If p = 2, the numbers are actually getting "larger" rather than "smaller"; for other prime numbers p they aren't getting larger but they aren't smaller either.
You can guess at least one example by now, yes? Of course if p=2, you could have
1 + 2 + 4 + 8 + 16 + ... + 2n + ...
What is that answer? Well, -1 - if you know computers you already knew this, -1 is the number where all bits are set. Not too hard really...
There are some harder questions. You could have a problem like
2 + 4 + 16 + 64 + ... + 22n + ...
Obviously when p=2 this does converge, but I'm not really sure what number it converges to.
As a stranger example, how about
1 + 1 + 2 + 6 + 24 + ... + n! + ...
... where of course n! is the factorial function, n! = 1x2x3x...xn. By definition as n gets larger n! is divisible by more numbers, so in principle this series converges for every p we use as a base. But I have absolutely no idea what it converges to.
I always hate that ... math questions where we can say "Yes, there is an answer, but I have no way of knowing what it is."