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 Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) Cherrina_Pixie Junior Fellow Posts: 6 Threads: 3 Joined: Jun 2011 06/14/2011, 04:22 AM (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? bo198214 Administrator Posts: 1,416 Threads: 92 Joined: Aug 2007 06/14/2011, 09:17 AM (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 $M(r_1^c,\dots,r_n^c)=M(r_1,\dots,r_n)^c$. bo198214 Administrator Posts: 1,416 Threads: 92 Joined: Aug 2007 06/14/2011, 09:58 AM (This post was last modified: 06/14/2011, 10:04 AM by bo198214.) (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: $M^s(r_1 (t) c, \dots, r_n (t) c) = M^s(r_1,\dots,r_n) (t) c$ JmsNxn Ultimate Fellow Posts: 754 Threads: 104 Joined: Dec 2010 06/14/2011, 09:52 PM (This post was last modified: 06/14/2011, 09:53 PM by JmsNxn.) 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. « Next Oldest | Next Newest »

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