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 Working from right to left. robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 06/30/2009, 09:09 PM (This post was last modified: 07/01/2009, 04:57 PM by robo37.) 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. 13-12=1 and 12-9=3, so (x+y)-(x+z)=y-z. 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. 48-36=16, so 8×6-8×4=8×2. 35-30=5, so 5×7-5×6=5×1. From this it's clear that (x×y)-(x×z)=x×(y-z). 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. 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. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/30/2009, 09:47 PM (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). Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 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. robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 07/05/2009, 03:32 PM 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=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 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/05/2009, 06:27 PM (This post was last modified: 07/05/2009, 06:30 PM by bo198214.) (07/05/2009, 03:32 PM)robo37 Wrote: 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? I put proper parentheses around and apply the law (x^a)^b = x^(a*b): Quote:So, therefore hyper 4 should be: X^^1=X X^^2=X^X X^^3=(X^X)^X = X^(X*X)= X^(X^2) X^^4=(X^(X^2))^X = X^(X^2*X) = X^(X^3) X^^5=(X^(X^3))^X = X^(X^3*X)=X^(X^4) You can continue that ad infinitum and get generally: x^^n = x^(x^(n-1)) thats what Tetratophile already wrote. If you apply this to n=0 you get x^^0 = x^(x^(-1)) = x^(1/x) = $\sqrt[x]{x}$ 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))= $\sqrt[x]{\sqrt[x]{x}}$ and so on as you write: Quote: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 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. nuninho1980 Fellow Posts: 96 Threads: 6 Joined: Apr 2009 07/05/2009, 06:41 PM (This post was last modified: 07/05/2009, 10:46 PM by nuninho1980.) (07/05/2009, 03:32 PM)robo37 Wrote: So, therefore hyper 4 should be: 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 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/05/2009, 06:57 PM (07/05/2009, 06:41 PM)nuninho1980 Wrote: (07/05/2009, 03:32 PM)robo37 Wrote: So, therefore hyper 4 should be: 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 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 In his posting ^^ is not the right bracketed tetration as we usually use it but it is the left-bracketed tetration. nuninho1980 Fellow Posts: 96 Threads: 6 Joined: Apr 2009 07/05/2009, 07:03 PM (This post was last modified: 07/06/2009, 12:25 AM by nuninho1980.) (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 left-bracketed 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^(n-1)). robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 07/05/2009, 07:41 PM 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. nuninho1980 Fellow Posts: 96 Threads: 6 Joined: Apr 2009 07/05/2009, 08:10 PM (This post was last modified: 07/05/2009, 08:10 PM by nuninho1980.) more reason... ^rl^ - isn't hyper4 because: e.g.: addition-hyper1 -> unit(s)-to-unit(s) multiplication-hyper2 -> digit(s)-to-digit(s) exponentiation‎-hyper3 -> scientific notation(s)-to-scientific notation(s) ---> tetration-hyper4 -> tetrational notation(s)-to-tetrational notation(s)* * tetrational notation (new) - 10^^n^x (remember - scientific notation x*10^n) « Next Oldest | Next Newest »

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