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