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 A question concerning uniqueness JmsNxn Long Time Fellow Posts: 410 Threads: 81 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 Long Time Fellow Posts: 410 Threads: 81 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: 664 Threads: 23 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 « Next Oldest | Next Newest »

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