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Exponential Mean and Beyond
#1
Question 
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 Exp-Mean(2,4), because 2^4 = 4^2.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#2
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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#3
(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? Exp-mean(2,3) ~ 2.419. Exp-mean(2,4) = ln(16)/Lambert_W0(ln(16)) ~ 2.745. Does anyone know of any series for this, like a series for exp-meaan(x,2)?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#4
(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? Exp-mean(2,3) ~ 2.419. Exp-mean(2,4) = ln(16)/W0(ln(16)) ~ 2.745. Does anyone know of any series for this, like a series for exp-meaan(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 Tongue , 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 semi-operators 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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#5
Question 
What about using commutative hyper-operations to do means? Then the means would already be commutative.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#6
(06/13/2022, 01:18 AM)Catullus Wrote: What about using commutative hyper-operations 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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#7
Question 
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 Exp-Mean3(a^b,b^a) = k^k^(log(k,log(k,a))+log(k,log(k,b))). For some k, such as √(2)?
Then Exp-Mean3(a,b) = k^k^(log(k,log(k,))+log(k,log(k,))).
If a and are complex, then Exp-Mean3(a,b) = k^(log(k,)*log(k,)).
If and b are real, then Exp-Mean3(a,b) = k^(log(k,a)*log(k,b)/(b*a)).
What about defining Exp-Mean4(a,b) = sexp((slog(a)+slog(b))/2), for some base such as the Tetra-Euler Number?
This is my 100th post!  Smile
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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