Rational operators (a {t} b); a,b > e solved

[bo198214]

\(
f(t) = a\, \{t\}\, b =
\left\{
\begin{array}{c l}
\exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) + \exp_\eta^{\alpha -t}(b)) & t \in (-\infty,1]\\
\exp_\eta^{\alpha t-1}(\exp_\eta^{\alpha 1-t}(a)*b) & t \in [1,2]\\
\end{array}
\right.
\)

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 \( f\circ g \).