Thread Rating:
  • 1 Vote(s) - 5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Pseudoalgebra
#3
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
Reply


Messages In This Thread
Pseudoalgebra - by tommy1729 - 10/05/2016, 12:21 PM
RE: Pseudoalgebra - by tommy1729 - 10/08/2016, 12:22 PM
RE: Pseudoalgebra - by tommy1729 - 10/13/2016, 02:32 AM
RE: Pseudoalgebra - by tommy1729 - 10/19/2016, 08:47 AM
RE: Pseudoalgebra - by sheldonison - 10/23/2016, 09:17 PM



Users browsing this thread: 1 Guest(s)