Working from right to left.  Printable Version + Tetration Forum (https://math.eretrandre.org/tetrationforum) + Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) + Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) + Thread: Working from right to left. (/showthread.php?tid=311) Pages:
1
2

Working from right to left.  robo37  06/30/2009 To me this doesn't seem to make much sense. Let’s look at addition (hyper 1). 9+4=13, 9+3=12 and 9+1=10. 7+5=12, 7+2=9 and 7+3=10. 1312=1 and 129=3, so (x+y)(x+z)=yz. Now let’s look at multiplication (hyper 2). 8 × 6=48 right, and 8×4=32. Also, 8×2=16. I'll make up some more; say 5×7=35, 5×6=30 and 5×1=5. 4836=16, so 8×68×4=8×2. 3530=5, so 5×75×6=5×1. From this it's clear that (x×y)(x×z)=x×(yz). 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^(yz). One would expect then that with tetration (hyper 3) (x^^z)√(x^^y)=x^^(yz), and if (x^^z)√(x^^y)≠x^^(yz) 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. Wouldn't the logical thing to do be to do what every calculator in the world does already, and do tetration from left to right? You do everything else from left to right even when it does make a difference like with division and subtraction for example. Doing it from left to right would also make it easier as going from x^^y to x^^(y+1) would only involve putting x^^y to the power of x just like going from x^y to x^(y+1) only involves multiplying x^y by x and going from x×y to x×(y+1) only involves adding x to x×y. RE: Working from right to left.  bo198214  06/30/2009 (06/30/2009, 09:09 PM)robo37 Wrote: Doing it from right to left would also make it easier as going from x^^y to x^^(y+1) would only involve putting x^^y to the power of x just like going from x^y to x^(y+1) only involves multiplying x^y by x and going from x×y to x×(y+1) only involves adding x to x×y. Well if you do it from left to right, there does not come up something new. If we have x^^0=x x^^(y+1)=(x^^y)^x then must x^^y = x^(x^y). RE: Working from right to left.  BaseAcid Tetration  06/30/2009 (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^(yz). There is more than one tetration. in fact, there are infinitely many "tetrations": *The standard, righttoleft tetration (we talk about it the most) has: a^^(b1) = log_a(a^^b) adding 1 to tetrant is doing a to power of a^^b. We say this by saying "the righttolefttetration is a "superfunction" of a^x." *the lefttoright tetration: a^^(b1) = 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/nth 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^(y1)). basically as bo198214 explained. Extending THAT tetration to the reals is trivial (very easy). just set b as any base in b^(b^(x1)), and you automatically get b^^x that can be used for any x. But righttoleft 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 righttoleft tetrations, and we don't know if they are actually the one and the same righttoleft tetration. we are extending tetration to the complex numbers and trying to look for something that would make tetration unique. Enough broad, offtopic summaries here. RE: Working from right to left.  robo37  07/05/2009 Sorry for putting this in the wrong section BTW. Im only 14 so I dont understand all this advanced mathematics, but let me ask you this, is x^^0 x? Multiplication is as follows: X*5=X+X+X+X+X X*4=X+X+X+X X*3=X+X+X X*2=X+X X*1=X X*0=XX X*1=XXX X*2=XXXX X*3=XXXXX X*4=XXXXXX X*5=XXXXXXX And then going up to hyper 3 it becomes: X^5=X*X*X*X*X X^4=X*X*X*X X^3=X*X*X X^2=X*X X^1=X X^0=X/X X^1=X/X/X X^2=X/X/X/X X^3=X/X/X/X/X X^4=X/X/X/X/X/X X^5=X/X/X/X/X/X/X So, therefore hyper 4 should be: X^^5=X^X^X^X^X X^^4=X^X^X^X X^^3=X^X^X X^^2=X^X X^^1=X X^^0=X√X X^^1=X√X√X X^^2=X√X√X√X X^^3=X√X√X√X√X X^^4=X√X√X√X√X√X X^^5=X√X√X√X√X√X√X RE: Working from right to left.  bo198214  07/05/2009 (07/05/2009, 03:32 PM)robo37 Wrote: Sorry for putting this in the wrong section BTW. I put proper parentheses around and apply the law (x^a)^b = x^(a*b): Quote:So, therefore hyper 4 should be: You can continue that ad infinitum and get generally: x^^n = x^(x^(n1)) thats what Tetratophile already wrote. If you apply this to n=0 you get x^^0 = x^(x^(1)) = x^(1/x) = thats exactly your conjecture: Quote:X^^0=X√X If we put the other negative values into n: X^^(1) = X^(X^(2))=X^((1/x)*(1/x))= and so on as you write: Quote:X^^1=X√X√X but be aware that you must set parentheses otherwise its not clear how to interpret it, e.g. X√X√X can be either X√(X√X) (what is what you meant) but it can also be interpreted as (X√X)√X, whats surely not what you meant. Note also that the convention among mathematicians is to interpret x^x^x as x^(x^x) instead of what you would suggest (x^x)^x. RE: Working from right to left.  nuninho1980  07/05/2009 (07/05/2009, 03:32 PM)robo37 Wrote: So, therefore hyper 4 should be: it is incorrect! but yes it's correct here down: X^^0 = log X = 1 X^^1 = log log X = 0 X^^2 = log log log X = oo X^^n = res, n > 2 > 'res' is real number; n=2 > res = oo. else n < 2 > that variable is complex number. lool RE: Working from right to left.  bo198214  07/05/2009 (07/05/2009, 06:41 PM)nuninho1980 Wrote:(07/05/2009, 03:32 PM)robo37 Wrote: So, therefore hyper 4 should be: In his posting ^^ is not the right bracketed tetration as we usually use it but it is the leftbracketed tetration. RE: Working from right to left.  nuninho1980  07/05/2009 (07/05/2009, 06:57 PM)bo198214 Wrote: In his posting ^^ is not the right bracketed tetration as we usually use it but it is the leftbracketed tetration. sorry. ok. hi, robo37: aah! sorry but your symbolic "^^" (= ^rl^ (rl  right to left) isn't hyper 4 nor "normal" tetration because x^rl^n = x^(x^(n1)). RE: Working from right to left.  robo37  07/05/2009 My point is that x^^0 should be the same as x√x because it follows along with the pattern, just as it follows along with the pattern I talked about in my first post and probably many other patterns as well. Left to right tetration is on the line of best fit as to say while right to left tetration seems to be as out of place as the whole concept of working from right to left is altogether. RE: Working from right to left.  nuninho1980  07/05/2009 more reason... ^rl^  isn't hyper4 because: e.g.: additionhyper1 > unit(s)tounit(s) multiplicationhyper2 > digit(s)todigit(s) exponentiationhyper3 > scientific notation(s)toscientific notation(s) > tetrationhyper4 > tetrational notation(s)totetrational notation(s)* * tetrational notation (new)  10^^n^x (remember  scientific notation x*10^n) 