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Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - Printable Version

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Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - Cherrina_Pixie - 06/14/2011

(06/08/2011, 11:47 PM)JmsNxn Wrote: However, I am willing to concede the idea of changing from base eta to base root 2.

That is to say if we define:



This will give the time honoured result, and aesthetic necessity in my point of view, of:
for all .

I like this also because it makes and potentially analytic over since 2 and 4 are fix points.

I also propose writing


I thought about this for some time and considered interpolation between arithmetic mean and geometric mean, coming to a rather curious result. The 'mean' function with fails to satisfy a property of means:

Define This yields the arithmetic mean for and the geometric mean for .
For ,




So it's not a 'true mean' in the sense that the scalar multiplication property fails. This result makes me doubt that the property may be satisfied for . Is there a way to rectify this issue, i.e. find a solution with and such that the property is satisfied?


RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - bo198214 - 06/14/2011

(06/14/2011, 04:22 AM)Cherrina_Pixie Wrote: I thought about this for some time and considered interpolation between arithmetic mean and geometric mean, coming to a rather curious result. The 'mean' function with fails to satisfy a property of means:

Define This yields the arithmetic mean for and the geometric mean for .

I just want to add the observation that:
and satisfy the modified property
.


RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - bo198214 - 06/14/2011

(06/14/2011, 09:17 AM)bo198214 Wrote: I just want to add the observation that:
and satisfy the modified property
.

And if we define


then we have for integers (and even non-integers) s=t and s=t-1:




RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - JmsNxn - 06/14/2011

Actually, I think if we use logarithmic semi-operators to notate this:

if

then:
for

This means, that multiplication isn't spreadable across [0,1], but logarithmic semi-operator multiplication is spreadable across [0,1]. Or put mathematically:


and


This should hold for complex numbers. Given the restriction on sigma. bo pretty much already noted this though, I just thought I'd give it a go Tongue.

I'm not sure if there's anything really interesting you can do with these averages.