06/30/2009, 11:03 PM
(This post was last modified: 06/30/2009, 11:16 PM by Base-Acid Tetration.)
(06/30/2009, 09:09 PM)robo37 Wrote: Now let’s look at exponentiation (hyper 3). 4^5=1024, 4^2=16 and 4^3=64. 3^6=726, 3^5=243 and 3^1=3. 1024÷16=64, so 4^5÷4^2=4^3. 726÷243=3 so 3^6÷3^5=3^1. (x^y) ÷ (x^z)=x^(y-z).
One would expect then that with tetration (hyper 3) (x^^z)√(x^^y)=x^^(y-z), and if (x^^z)√(x^^y)≠x^^(y-z) something’s wrong. If 256=2^^4 (and 4=2^^2) then the 4th root of 256 should be 4, which it is, but including to this article 256 doesn't equal 2^^4 at all, which to me makes no sense whatsoever.
There is more than one tetration. in fact, there are infinitely many "tetrations":
*The standard, right-to-left tetration (we talk about it the most) has: a^^(b-1) = log_a(a^^b) adding 1 to tetrant is doing a to power of a^^b. We say this by saying "the right-to-left-tetration is a "superfunction" of a^x."
*the left-to-right tetration: a^^(b-1) = a√(a^^b) (superfunction of x^a)
*balanced tetration (i would call that 1:1 tetration), you could think of 2:3 tetration or 3:2 tetration, 4:1 etc. I think any ratios between two real numbers might work in this infinite set of "center" tetrations)
etc. etc.
There are different tetrations because exponentiation is not associative like addition and multiplication. Whereas we have simple identities for exponentiation, no more so for any of the tetrations. like: a^(x+y) = a^x*a^y, and a^(xy) = (a^x)^y. These were allowed because multiplication was associative (grouping didn't matter, so we could remove parentheses to multiply the two expressions). But (a^^3)^(a^^4) = (a^a^a)^(a^a^a^a) =/= a^a^a^a^a^a^a. We can not just remove the parentheses to make it a^^7 because grouping does matter for exponentiation. The identity a^(xy) = (a^x)^y allowed us to define nth roots as the 1/n-th power, but the "nth superroot" is not tetration to the 1/n.
... Yeah, that's basically why we're here.
As for your left to right tetration, if you set x^^0 = 1, then x^^y can be simply written as x^(x^(y-1)). basically as bo198214 explained.
Extending THAT tetration to the reals is trivial (very easy). just set b as any base in b^(b^(x-1)), and you automatically get b^^x that can be used for any x.
But right-to-left tetration is not easy to extend to the reals. Let me summarize the situation for you: there are several methods (and they are all really complicated to me! I don't understand any of them) of real extensions of right-to-left tetrations, and we don't know if they are actually the one and the same right-to-left tetration. we are extending tetration to the complex numbers and trying to look for something that would make tetration unique.
Enough broad, off-topic summaries here.