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:

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

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?
Reply
#2
(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
\( M(r_1^c,\dots,r_n^c)=M(r_1,\dots,r_n)^c \).
Reply
#3
(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:

\( M^s(r_1 (t) c, \dots, r_n (t) c) = M^s(r_1,\dots,r_n) (t) c \)
Reply
#4
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 Tongue.

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


Possibly Related Threads…
Thread Author Replies Views Last Post
  Could there be an "i" of the tetrative operations? 000_Era 1 426 07/12/2024, 06:26 AM
Last Post: MphLee
  Computing sqrt 2 with rational functions. tommy1729 0 670 03/31/2023, 11:49 AM
Last Post: tommy1729
Question Is the Tetra-Euler Number Rational? Catullus 1 1,208 07/17/2022, 06:37 AM
Last Post: JmsNxn
  Has anyone solved iterations of z+Γ(z)? Leo.W 5 4,874 01/07/2022, 08:15 AM
Last Post: JmsNxn
  Intresting ternary operations ? tommy1729 0 4,127 06/11/2015, 08:18 AM
Last Post: tommy1729
  Change of base formula using logarithmic semi operators JmsNxn 4 15,922 07/08/2011, 08:28 PM
Last Post: JmsNxn
  book: the theory of fractional powers of operators bo198214 2 11,432 06/22/2010, 07:30 PM
Last Post: Ztolk
  f( f(x) ) = exp(x) solved ! ! ! tommy1729 25 45,618 02/17/2009, 12:30 AM
Last Post: tommy1729
  Rational sums of inverse powers of fixed points of e jaydfox 14 35,480 11/23/2007, 08:22 AM
Last Post: bo198214
  A more consistent definition of tetration of tetration for rational exponents UVIR 21 45,367 10/21/2007, 10:47 PM
Last Post: UVIR



Users browsing this thread: 1 Guest(s)