# Tetration Forum

You're currently viewing a stripped down version of our content. View the full version with proper formatting.
(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:

$\vartheta(a,b,\sigma) = \exp_{2^{\frac{1}{2}}}^{\circ \sigma}(\exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(a) + h_b(\sigma))\\\\
[tex]h_b(\sigma)=\left{\begin{array}{c l}
\exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(b) & \sigma \le 1\\
\exp_{2^{\frac{1}{2}}}^{\circ -1}(b) & \sigma \in [1,2]
\end{array}\right.$

This will give the time honoured result, and aesthetic necessity in my point of view, of:
$\vartheta(2, 2, \sigma) = 2\,\,\bigtriangleup_\sigma\,\, 2 = 4$ for all $\sigma$.

I like this also because it makes $\vartheta(a, 2, \sigma)$ and $\vartheta(a, 4, \sigma)$ potentially analytic over $(-\infty, 2]$ since 2 and 4 are fix points.

I also propose writing

$a\,\,\bigtriangle_\sigma^f\,\,b = \exp_f^{\circ \sigma}(\exp_f^{\circ -\sigma}(a) + h_b(\sigma))\\\\
[tex]h_b(\sigma)=\left{\begin{array}{c l}
\exp_f^{\circ -\sigma}(b) & \sigma \le 1\\
\exp_f^{\circ -1}(b) & \sigma \in [1,2]
\end{array}\right.$

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 $\sigma = -1$ fails to satisfy a property of means: $mean(c*r_1,c*r_2,c*r_3, ..., c*r_n) = c*mean(r_1,r_2,r_3, ..., r_n)$

Define $M_f^\sigma(r_1,r_2,r_3, ..., r_n) = \exp_f^{\circ \sigma}\left(\frac{\exp_f^{\circ -\sigma}(r_1) + \exp_f^{\circ -\sigma}(r_2) + \exp_f^{\circ -\sigma}(r_3) + ... + \exp_f^{\circ -\sigma}(r_n)}{n}\right),\ \sigma \le 1$ This yields the arithmetic mean for $\sigma = 0$ and the geometric mean for $\sigma = 1$.
For $\sigma = -1$,

$M_{\sqrt{2}}^{-1}(1,2) = \exp_{\sqrt{2}}^{\circ -1}\left(\frac{\exp_{\sqrt{2}}^{\circ 1}(1) + \exp_{\sqrt{2}}^{\circ 1}(2)}{2}\right) = \log_{\sqrt{2}}\left(\frac{\sqrt{2} + 2}{2}\right) \approx 1.5431066$
$M_{\sqrt{2}}^{-1}(3,6) = \exp_{\sqrt{2}}^{\circ -1}\left(\frac{\exp_{\sqrt{2}}^{\circ 1}(3) + \exp_{\sqrt{2}}^{\circ 1}(6)}{2}\right) = \log_{\sqrt{2}}\left(\frac{\sqrt{2}^3 + 8}{2}\right) \approx 4.8735036 \ \approx \ 3.15824 * M_{\sqrt{2}}^{-1}(1,2) \ \not= \ 3.00000*M_{\sqrt{2}}^{-1}(1,2)$

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 $0 < \sigma < 1$. Is there a way to rectify this issue, i.e. find a solution $(f,\sigma)$ with $f > 1$ and $0 < \sigma < 1$ such that the property is satisfied?
(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 $\sigma = -1$ fails to satisfy a property of means: $mean(c*r_1,c*r_2,c*r_3, ..., c*r_n) = c*mean(r_1,r_2,r_3, ..., r_n)$

Define $M_f^\sigma(r_1,r_2,r_3, ..., r_n) = \exp_f^{\circ \sigma}\left(\frac{\exp_f^{\circ -\sigma}(r_1) + \exp_f^{\circ -\sigma}(r_2) + \exp_f^{\circ -\sigma}(r_3) + ... + \exp_f^{\circ -\sigma}(r_n)}{n}\right),\ \sigma \le 1$ This yields the arithmetic mean for $\sigma = 0$ and the geometric mean for $\sigma = 1$.

I just want to add the observation that:
$M^1$ and $M^2$ satisfy the modified property
$M(r_1^c,\dots,r_n^c)=M(r_1,\dots,r_n)^c$.
(06/14/2011, 09:17 AM)bo198214 Wrote: [ -> ]I just want to add the observation that:
$M^1$ and $M^2$ satisfy the modified property
$M(r_1^c,\dots,r_n^c)=M(r_1,\dots,r_n)^c$.

And if we define
$x (t) y = \exp^{\circ t}(\log^{\circ t}(x)+\log^{\circ t}(y))$

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

$M^s(r_1 (t) c, \dots, r_n (t) c) = M^s(r_1,\dots,r_n) (t) c$
Actually, I think if we use logarithmic semi-operators to notate this:

if $\bigtriangleup_{\sigma}\,\,\sum_{n=N}^{R} f(n)= f(N)\,\,\bigtriangleup_{\sigma}\,\,f(N+1)\,\,\bigtriangleup_{\sigma}\,\,...\,\,\bigtriangleup_{\sigma}\,\, f( R )$

then:
$M^{\sigma}(r_1,...,r_n) = (\bigtriangleup_{\sigma}\,\,\sum_{c=1}^{n}\,r_c)\,\,\bigtriangledown_{1+\sigma} \,\,n$ for $\R (\sigma) \le 1$

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

$M^{\sigma}(r_1,...,r_n)\,\,\bigtriangleup_{\sigma}\,\, a=M^{\sigma}(r_1\,\,\bigtriangleup_{\sigma}\,\, a,...,r_n\,\,\bigtriangleup_{\sigma}\,\, a)$
and
$
M^{\sigma}(r_1,...,r_n)\,\,\bigtriangleup_{\sigma + 1} \,\, a = M^{\sigma}(r_1 \,\,\bigtriangleup_{\sigma + 1} \,\, a,...,r_n\,\,\bigtriangleup_{\sigma + 1} \,\, a)$

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 .

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