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Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved)
#1
(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?
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Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - by Cherrina_Pixie - 06/14/2011, 04:22 AM

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