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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? « Next Oldest | Next Newest »

 Messages In This Thread Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - by Cherrina_Pixie - 06/14/2011, 04:22 AM RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - by bo198214 - 06/14/2011, 09:17 AM RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - by bo198214 - 06/14/2011, 09:58 AM RE: Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved) - by JmsNxn - 06/14/2011, 09:52 PM

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