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 Rational operators (a {t} b); a,b > e solved bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/06/2011, 06:53 AM (06/06/2011, 02:45 AM)JmsNxn Wrote: $ f(t) = a\, \{t\}\, b = \left\{ \begin{array}{c l} \exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) + \exp_\eta^{\alpha -t}(b)) & t \in (-\infty,1]\\ \exp_\eta^{\alpha t-1}(\exp_\eta^{\alpha 1-t}(a)*b) & t \in [1,2]\\ \end{array} \right.$ But James, this is not analytic at $t=1$, if we reformulate: $ f(t) = a\, \{t\}\, b = \left\{ \begin{array}{c l} \exp_\eta^{\circ t}(\exp_\eta^{\circ-t}(a) + \exp_\eta^{\circ -t}(b)) & t \in (-\infty,1]\\ \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a)+\exp_\eta^{\circ -1}(b)) & t \in [1,2] \end{array} \right.$ We can say: $a\, \{t\}\, b = \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a) + h_b(t))$ where $h_b(t)=\left{\begin{array}{c l} \exp_\eta^{\circ -t}(b) & t\le 1\\ \exp_\eta^{\circ -1}(b) & t\in [1,2] \end{array}\right.$ $f$ is addition and composition of analytic functions, except this one function $h_b$. The whole function $f(t)$ can not be analytic. I wonder why it looks so smooth. But then on the other hand there is a general problem with semioperators (note that your operator is 1 off the standard notation, i.e. {t}=[t+1]): (a [0] 0 = 1) a [1] 0 = a a [2] 0 = 0 a [3] 0 = 1 a [n] 0 = 1 for n>3 or (a [0] 1 = 2) a [1] 1 = a+1 a [2] 1 = a a [3] 1 = a for n>2 As soon as one defines a [t] 1 = a for t > 2, then the whole analytic function is already determined to be a, i.e. it must also be a for t=1 which is wrong. Hence an analytic t |-> a[t]1 will not be constant but somehow meandering between the a's, which is somehow really strange. But I see you gracefully avoided that problem by just defining it for a,b > e PS: 1. $g(t) = a\, \} t \{ \, b$, This notation is ambiguous, compare $\{ a \} t \{ b \} + c$. Please invent a better one! 2. $\exp_\eta^{\alpha t}$, not \alpha but \circ belongs in the exponent: $\exp_\eta^{\circ t}$. This notation is derived from the symbol for function composition $f\circ g$. « Next Oldest | Next Newest »

 Messages In This Thread Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 04:39 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:34 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 06:02 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 07:03 AM RE: Rational operators (a {t} b); a,b > e solved - by nuninho1980 - 06/06/2011, 05:16 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 06:53 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 08:47 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:23 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 06/06/2011, 11:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:44 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:28 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 07:47 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 08:43 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/07/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/07/2011, 06:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 04:54 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 07:31 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/08/2011, 08:32 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/08/2011, 09:14 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 01:50 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 11:47 PM RE: Rational operators (a {t} b); a,b > e solved - by Gottfried - 06/11/2011, 02:33 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/12/2011, 07:55 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/21/2016, 06:56 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 08/22/2016, 12:36 AM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/24/2016, 07:24 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/29/2016, 02:06 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 09/01/2016, 06:47 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:04 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:11 AM

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