• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 A question concerning uniqueness JmsNxn Ultimate Fellow Posts: 889 Threads: 110 Joined: Dec 2010 10/05/2011, 04:28 PM (This post was last modified: 10/05/2011, 04:52 PM by JmsNxn.) My question is simple, and I hope somebody has an answer because I am a little confused. How come Tetration has multiple possible extensions to the complex domain that are analytic, but exponentiation only has one? Is it possible to have an alternative extension for exponentiation that is still analytic? It would have a piecewise definition, the gamma variable is to distinguish it from regular exponentiation: $a^{\otimes_\gamma\,\,k} = a\,\cdot\,a\,\cdot... a$ k amount of times if $k \in N$ and then, for $0 $a^{\otimes_\gamma\,\,q} \neq a^q$ and $a^{\otimes_\gamma\,\,k + q} = a\,\cdot\,a\,\cdot\,a\,... \cdot\,a^{\otimes_\gamma\,\, q}$ Is it possible for $a^{\otimes_\gamma \,\,x}$ to be analytic? I'm sure it's possible to make alternative models, where $a^{\otimes_\gamma\,\,q}$ is defined arbitrarily, but is normal exponentiation the only analytic model? thanks for reading this and I hope someone can clarify As I think about it, I think this gets a little messy when we consider the laws of exponentiation: $a^{\otimes_\gamma\,\, b} \cdot a^{\otimes_\gamma\,\, c} \neq a^{\otimes_\gamma\,\, b+ c}$ because this would imply there is a number seperate from the square root that when squared returns a. This type of idea could extend to multiplication as well, insofar as we could define a new multiplication: $a\,\otimes_\gamma\,k = a_1 + a_2 + a_3 ... + a_k$ and $a\,\otimes_\gamma q \neq a \cdot q$ We could also go to addition, and define a different addition: $a \oplus_\gamma\, k = a + 1_1 + 1_2 + 1_3.... + 1_k$ and $a\,\oplus_\gamma\,q \neq a + q$ This may let us keep our exponentiation laws, though altered: $a^{\otimes_\gamma \,\,b}\,\,\otimes_\gamma\,\,a^{\otimes_\gamma\,\, c} = a^{\otimes_\gamma\,\,b\,\oplus_\gamma\,c}$ we'd also have a similar law for multiplication maybe: $(a\,\,\otimes_\gamma\,\, b)\,\oplus_\gamma\,(a\,\,\otimes_\gamma\,\, c) = a\,\,\otimes_\gamma\,\, (b\,\oplus_\gamma\,c)$ I wonder, are these suggested different operators incapable of being analytic? Or are the normal addition and normal multiplication and normal exponentiation the only ones that work? Thanks for any help, James BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 10/05/2011, 09:54 PM (This post was last modified: 10/05/2011, 10:05 PM by BenStandeven.) (10/05/2011, 04:28 PM)JmsNxn Wrote: My question is simple, and I hope somebody has an answer because I am a little confused. How come Tetration has multiple possible extensions to the complex domain that are analytic, but exponentiation only has one? Is it possible to have an alternative extension for exponentiation that is still analytic? It would have a piecewise definition, the gamma variable is to distinguish it from regular exponentiation: $a^{\otimes_\gamma\,\,k} = a\,\cdot\,a\,\cdot... a$ k amount of times if $k \in N$ and then, for $0 $a^{\otimes_\gamma\,\,q} \neq a^q$ and $a^{\otimes_\gamma\,\,k + q} = a\,\cdot\,a\,\cdot\,a\,... \cdot\,a^{\otimes_\gamma\,\, q}$ Is it possible for $a^{\otimes_\gamma \,\,x}$ to be analytic? I'm sure it's possible to make alternative models, where $a^{\otimes_\gamma\,\,q}$ is defined arbitrarily, but is normal exponentiation the only analytic model? So we would need $a^{\otimes_\gamma\,\,k + q} = a^k f(q)$ for $q$ from 0 to 1, and $k$ an integer, or in other words $a^{\otimes_\gamma\,\,x} = a^x f(frac(x))$; then using the same argument as for tetration, we get that $\log_a a^{\otimes_\gamma\,k+q} = k + \log f(q) = k + q + \log f(q)/a^q$. So $f(q)/a^q$ is analytic and takes value 1 at both 0 and 1. Then any function $f$ satisfying these conditions can be used to define a new "exponential" function; for example, we could have $a^{\otimes_\gamma\,\,x} = a^x cos(2 \pi x)$. Of course, f can also depend on $a$. Quote:thanks for reading this and I hope someone can clarify As I think about it, I think this gets a little messy when we consider the laws of exponentiation: $a^{\otimes_\gamma\,\, b} \cdot a^{\otimes_\gamma\,\, c} \neq a^{\otimes_\gamma\,\, b+ c}$ because this would imply there is a number seperate from the square root that when squared returns a. Yeah, the standard exponential functions are the only continuous functions on the reals which satisfy the full laws of exponentiation. Quote:This type of idea could extend to multiplication as well, insofar as we could define a new multiplication: $a\,\otimes_\gamma\,k = a_1 + a_2 + a_3 ... + a_k$ and $a\,\otimes_\gamma q \neq a \cdot q$ For this, the solution would be $a\,\otimes_\gamma x - a\,\times x = g(frac(x))$, where $g$ is zero at 0 and 1. Quote:We could also go to addition, and define a different addition: $a \oplus_\gamma\, k = a + 1_1 + 1_2 + 1_3.... + 1_k$ and $a\,\oplus_\gamma\,q \neq a + q$ For this, the solution would be $a\,\oplus_\gamma x = a + floor(x)\, \oplus_\gamma frac(x) = a + floor(x) + h(x)$, with $h(x)$ going from 0 to 1 on the unit interval. Here I am assuming that $h$ does not depend on $a$; if it did, it could also depend on $floor(x)$. Quote:This may let us keep our exponentiation laws, though altered: $a^{\otimes_\gamma \,\,b}\,\,\otimes_\gamma\,\,a^{\otimes_\gamma\,\, c} = a^{\otimes_\gamma\,\,b\,\oplus_\gamma\,c}$ we'd also have a similar law for multiplication maybe: $(a\,\,\otimes_\gamma\,\, b)\,\oplus_\gamma\,(a\,\,\otimes_\gamma\,\, c) = a\,\,\otimes_\gamma\,\, (b\,\oplus_\gamma\,c)$ Sure, just set $t(x) = x + h(x)$ for some 1-periodic function $h$, and define $a \oplus_\gamma b = h(h^{-1}(a) + h^{-1}(b))$, $a \otimes_\gamma b = h(h^{-1}(a) \times h^{-1}(b))$, and so on. I don't know if there is any other way to turn R into a field than this one, let alone a field with an exponential operator. JmsNxn Ultimate Fellow Posts: 889 Threads: 110 Joined: Dec 2010 10/06/2011, 12:15 AM Oh very very interesting! So I guess, the definitive property for exponentiation would be: $a\cdot a^x =a^{x+1}$ and no other function satisfies this requirement. I wonder, what exactly differentiates tetration such that there are multiple solutions given for $^b a$ when $a^{^b a} = \,\,^{b+1} a$ is the definitive property. I think it's because square roots, or root functions in general, are exponentiation by the multiplicative inverse. They are binded by multiplication, and there is only one number b (besides its negative sqrt) that when multiplied by itself gives a. However, super square roots (which there is only one value for every x) are not $^{\frac{1}{2}} a$ so this eliminates uniqueness by relation to super square roots. Instead if we want a different sort of parallel to squaring a squareroot involving (1/2) for tetration we can have $\exp_a^{\circ \frac{1}{2}}(\exp_a^{\circ \frac{1}{2}}(1)) = a$ but since $\exp_a^{\circ \frac{1}{2}}(x)$ is determined by the extension of tetration, we see there is no external reason limiting half values to one specific value. I guess this is a lot like how the one periodic function creates a separate super function for exponentiation. I think I'm beginning to understand why we have different solutions for tetration; I guess the real difficulty is defining what extra parameter makes it the solution. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 10/06/2011, 04:32 AM (This post was last modified: 10/06/2011, 03:58 PM by sheldonison.) (10/06/2011, 12:15 AM)JmsNxn Wrote: Oh very very interesting! So I guess, the definitive property for exponentiation would be: $a\cdot a^x =a^{x+1}$ and no other function satisfies this requirement. I wonder, what exactly differentiates tetration such that there are multiple solutions given for $^b a$ when $a^{^b a} = \,\,^{b+1} a$ is the definitive property...edit: this reply only applies for bases>eta, because for bases<=eta, the superfunction is real valued, so no Kneser mapping is necessary. For bases-2. - Sheldon Catullus Fellow Posts: 201 Threads: 45 Joined: Jun 2022 06/10/2022, 08:45 AM (This post was last modified: 07/12/2022, 05:01 AM by Catullus.) (10/05/2011, 04:28 PM)JmsNxn Wrote: My question is simple, and I hope somebody has an answer because I am a little confused.  How come Tetration has multiple possible extensions to the complex domain that are analytic, but exponentiation only has one?  Is it possible to have an alternative extension for exponentiation that is still analytic?Why, not how come. Why is better than how come. Regular exponentiation is not always analytic. For example, $0\uparrow x$ is not analytic at zero. Why a piecewise definition? Functions defined from piece-wise definitions are typically not analytic at the ends of pieces. Defining $a\uparrow\otimes_\gamma\uparrow b$ as $a\uparrow b*cos(b*\tau)$ might work. Defining $a\otimes_\gamma b$ as $a*b*cos(b*\tau)$ might work. Defining $a\oplus_\gamma b$ as $a+b+cos(b*\tau)$ might work. Though, these are not piece-wise definitions. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post [question] Local to global and superfunctions MphLee 8 305 07/17/2022, 06:46 AM Last Post: JmsNxn Uniqueness of fractionally iterated functions Daniel 7 296 07/05/2022, 01:21 AM Last Post: JmsNxn Universal uniqueness criterion? bo198214 57 108,797 06/28/2022, 12:00 AM Last Post: JmsNxn A random question for mathematicians regarding i and the Fibonacci sequence. robo37 1 3,966 06/27/2022, 12:06 AM Last Post: Catullus Question about tetration methods Daniel 17 662 06/22/2022, 11:27 PM Last Post: tommy1729 Math.Stackexchange.com question on extending tetration Daniel 3 2,199 03/31/2021, 12:28 AM Last Post: JmsNxn [Exercise] A deal of Uniqueness-critrion:Gamma-functionas iteration Gottfried 6 7,072 03/19/2021, 01:25 PM Last Post: tommy1729 Kneser method question tommy1729 9 12,104 02/11/2020, 01:26 AM Last Post: sheldonison A Notation Question (raising the highest value in pow-tower to a different power) Micah 8 12,734 02/18/2019, 10:34 PM Last Post: Micah Math overflow question on fractional exponential iterations sheldonison 4 10,479 04/01/2018, 03:09 AM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)