10/13/2016, 02:32 AM
Im afraid the strategy fails.
For exp_b^[a] <*> and a < 1 we get
<*> @ = exp^[a] ( T x )
Where T Goes to 1 as x grows and for a >= 1 , T Goes to ln(b).
Proof sketch
S commutes with exp.
S(T x) = ln S ( T b^x) / ln(B).
= S ln ( T b^x) / ln(B)
= S ( ln T + ln B x ) / ln B
Maybe.
Still thinking ...
Regards
Tommy1729
For exp_b^[a] <*> and a < 1 we get
<*> @ = exp^[a] ( T x )
Where T Goes to 1 as x grows and for a >= 1 , T Goes to ln(b).
Proof sketch
S commutes with exp.
S(T x) = ln S ( T b^x) / ln(B).
= S ln ( T b^x) / ln(B)
= S ( ln T + ln B x ) / ln B
Maybe.
Still thinking ...
Regards
Tommy1729