pdf23ds comments on Bizarre Illusions - Less Wrong

11 Post author: MrHen 27 January 2010 06:25PM

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Comment author: MrHen 28 January 2010 06:40:17PM *  1 point [-]

You may be a good person to ask this question:

I was wondering if there was a function f(x, y, z) so that x and z represent the left and right sides of common mathematic operators and y represents the level of operation. So f(1, 2, 4) would be 1 + 4 and f(2, 2, 4) would be 2 * 4. Better versions of f(x, y, z) would have fewer end cases hardcoded into it.

The reason behind this is to handle operator levels greater than addition, multiplication, and exponents. The casual analysis from my grade school and undergrad level math shows the pattern that multiplication is repeated addition and exponents are repeated multiplication.

My quick attempts at coming up with such a function are spiraling into greater and greater complexities. I figured someone else has to have thought about this. Do you know of a place I can start reading up on ideas similar to this? Is what I am doing even plausible?

Quick thoughts based on me playing around:

  • Addition may be level 0, not level 1
  • The sequences never really look exactly like multiplication tables, but the patterns are similar enough to appease me
  • Ideally, everything can be reduced to the simple concept of X + 1 so as to walk along the number line
  • In practical terms, I have no idea how to express "negative" levels. Division and roots are unapproachable at this point in my playing around.
Comment author: pdf23ds 28 January 2010 06:55:50PM *  1 point [-]

Hyper operators. You can represent even bigger numbers with Conway chained arrow notation. Eliezer's 3^^^^3 is a form of hyper operator notation, where ^ is exponentiation, ^^ is tetration, ^^^ is pentation, etc.

If you've ever looked into really big numbers, you'll find info about Ackermann's function, which is trivially convertable to hyper notation. There's also Busy Beaver numbers, which grow faster than any computable function.

Comment author: MrHen 28 January 2010 07:07:33PM 0 points [-]

Yes, this is exactly what I was looking for. Thank you.