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06/04/2022, 08:57 AM
(This post was last modified: 06/19/2022, 01:01 AM by Catullus.)
If you have two numbers a and b, you want to take the mean of them, you could do ssqrt(a^b), or ssqrt(b^a). Then you could do the square super root of those exponentiated together. But which order? You do both orders, and do the super square root of those exponentiated together. And you keep doing it until they converge, and call that the exponential mean or exp mean for short.
You could even do tetration means and pentation means and beyound.
Does anyone know of any closed forms for any of these means, When the means are not already communitive? Such as ExpMean(2,4), because 2^4 = 4^2.
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This sounds terrifyingly hard. Even showing something like this converges sounds terrifyingly hard, lol. Not sure how much luck you'll have, but keep me posted if you think of anything, lol.
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06/06/2022, 02:54 AM
(This post was last modified: 06/19/2022, 01:01 AM by Catullus.)
(06/05/2022, 11:36 PM)JmsNxn Wrote: This sounds terrifyingly hard. Even showing something like this converges sounds terrifyingly hard, lol. Not sure how much luck you'll have, but keep me posted if you think of anything, lol. Like how hard exactly? Exp time? Expmean(2,3) ~ 2.419. Expmean(2,4) = ln(16)/Lambert_W0(ln(16)) ~ 2.745. Does anyone know of any series for this, like a series for expmeaan(x,2)?
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06/06/2022, 03:37 AM
(This post was last modified: 06/06/2022, 03:52 AM by JmsNxn.)
(06/06/2022, 02:54 AM)Catullus Wrote: (06/05/2022, 11:36 PM)JmsNxn Wrote: This sounds terrifyingly hard. Even showing something like this converges sounds terrifyingly hard, lol. Not sure how much luck you'll have, but keep me posted if you think of anything, lol. Like how hard exactly? Exp time? Expmean(2,3) ~ 2.419. Expmean(2,4) = ln(16)/W0(ln(16)) ~ 2.745. Does anyone know of any series for this, like a series for expmeaan(x,2)?
Oh, I apologize.
What I mean, is I can bet this probably converges! But I have no idea how to do it!
I mean, this looks like something that would probably converge, and I don't doubt it converges. PROVING it converges sounds impossible , at least, that's what I meant.
I believe you'd have more luck looking at arithmetic/geometric means for general functions. Look at if \(f,g\) are reasonable functions does a similar process converge. This would mean, looking at:
$$
\begin{align}
F&=f^{1}(f(a,b)*f(b,a),b)\\
G&=f^{1}(a,f(a,b)*f(b,a))\\
&...\\
\end{align}
$$
What I mean to say, is that you could probably find literature on similar cases for general functions; especially how you can iterate and mix arithmetic and geometric means (this was a proposed solution of semioperators on this forum, the trouble being it didn't satisfy the Goodstein equation). I would look into the general theory of iterated arithmetic/geometric means; which does exist, and view how they show convergence. I know there is research on \(f(a,b)\)'s mean, and so on and so forth. It brushes path with iteration theory. I apologize but I'm not the one to ask. I'd google it, but I'm not even sure what to google...
EDIT:
I kinda went down a rabbit hole, but this seems a good place to start:
https://www.jstor.org/stable/pdf/1967353.pdf
You'd have to do much of the techniques here, but with general functions as opposed to the arithmetic/geometric mean stuff.
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06/13/2022, 01:18 AM
(This post was last modified: 06/13/2022, 10:21 PM by Catullus.)
What about using commutative hyperoperations to do means? Then the means would already be commutative.
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(06/13/2022, 01:18 AM)Catullus Wrote: What about using commutative hyperoperations to do means? Then the means would already be commutative.
true.
but why do you want a new average ?
what is the purpose ?
this reminds me of the matrix question
is A^B equal to exp(log(A) B) or exp(B log(A))
and both are defendable.
regards
tommy1729
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06/19/2022, 01:08 AM
(This post was last modified: 07/01/2022, 09:47 AM by Catullus.)
What if there are an odd amount of things to do the exponential mean of? Then how would you do it?
What if you defined the exponential mean as a mean such that ExpMean3(a^b,b^a) = k^k^(log(k,log(k,a))+log(k,log(k,b))). For some k, such as √(2)?
Then ExpMean3(a,b) = k^k^(log(k,log(k, ))+log(k,log(k, ))).
If a and are complex, then ExpMean3(a,b) = k^(log(k, )*log(k, )).
If and b are real, then ExpMean3(a,b) = k^(log(k,a)*log(k,b)/(b*a)).
What about defining ExpMean4(a,b) = sexp((slog(a)+slog(b))/2), for some base such as the TetraEuler Number?
This is my 100th post!
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