Means and intermediate operations (was: Rational operators (a {t} b); a,b > e solved)
#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.
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Messages In This Thread
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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