A question concerning uniqueness
#1
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 \( 0<q<1 \)
\( a^{\otimes_\gamma\,\,q} \neq a^q \)

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
#2
(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 \( 0<q<1 \)
\( a^{\otimes_\gamma\,\,q} \neq a^q \)

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.
#3
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.

#4
(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<eta, there are multiple superfunctions, one of which is entire, and the solutions are imaginary periodic, so the uniqueness discussion is much more complicated.

The uniqueness problem has basically been solved, and involves the behavior of sexp(z) in the complex plane as imag(z) increases. If you start with the Schroder function for exp(z), then that function has only one solution. Then the inverse Schroder function for exp(z) can be trivially turned into a unique complex valued superfunction for exp(z). Here, I'm using S for the Schroder function.
\( S^{-1}(\lambda z) = \exp(S^{-1}(z)) \),
\( \lambda = L\approx0.318+1.337i \), the fixed point for exp(z).

From there, there is one unique complex superfunction for exp(z). The superfunction has a complex period, and as imag(z) increases, and real(z) decreases, the superfunction decays to the fixed point.
\( \text{superfunction}(z)=S^{-1}(\exp(z L)) \)

Now, from there, there is exactly one unique real valued tetration, which converges to the complex valued superfunction(z+k), where k is a constant, as imag(z) increases. Here, \( \theta(z) \) is a 1-periodic periodic function that decays to a constant as imag(z) increases. \( \theta(z) \) has a singularities for integer values of z.
\( \text{sexp}(z)=\text{superfunction}(z+\theta(z)) \)

Any other different real valued sexp(z) function will have singularities as imag(z) increases, rather than converging to the complex superfunction. \( \text{sexp_{alt}(z)=\text{sexp}(z+\theta_{real}(z)) \), where \( \theta_{real}(z) \) is another different real valued 1-periodic function used to generate the proposed alternate sexp(z) solution from the correct sexp(z) solution. The important thing turns out to be that the alternative sexp(z) function is generated using \( \theta_{real}(z) \) which is required to be real valued. Because it is required to be real valued, it is easy to show that it grows exponentially as imag(z) increases.

There's a little more to it. I wrote a program that calculates theta(z), but it is tricky to rigorously prove the algorithm converges, although I have some good heuristic arguments. Kneser's approach is to use a unique Riemann mapping, which can be shown to be mathematically equivalent to the theta(z) approach, see this post. Kneser's Riemann mapping rigorously guarantees the existence of one unique solution real valued sexp(z). The net result is that there is one unique real valued sexp(z) function, that is well behaved in the complex plane and doesn't have any singularities for real(z)>-2.
- Sheldon
#5
(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, [Image: svg.image?0\uparrow%20x] 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 [Image: svg.image?a\uparrow\otimes_\gamma\uparrow%20b] as [Image: svg.image?a\uparrow%20b*cos(b*\tau)] might work.
Defining [Image: svg.image?a\otimes_\gamma%20b] as [Image: svg.image?a*b*cos(b*\tau)] might work.
Defining [Image: svg.image?a\oplus_\gamma%20b] as [Image: svg.image?a+b+cos(b*\tau)] might work.
Though, these are not piece-wise definitions.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\


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