bounded derivatives and semi-group iso ( repost ?? )

[tommy1729]

sorry if this is a repost, im not sure.

The idea is we want for n E {0,1,2,3} , q E [0,1] and x positive real :

0 < (d/dx)^n exp^[q](x) =< (d/dx)^n exp^[q + h](x)

for any positive h such that q + h =< 1.

My fear is however this might not be analytic ?

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edit : incorrect statement removed

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If we relax the first equation to 

The idea is we want for n E {0,1,2,3} , q E [0,1] and x positive real :

0 =< (d/dx)^n exp^[q](x) =< (d/dx)^n exp^[q+h](x)

for any positive h such that q + h =< 1.

Then I think again hoosmand equation is the solution if it even has one.

Which makes me skeptic about the first equation having analytic solutions.


so intuitive simple bounded derivatives and semi-group iso seem to have no solutions.


Nice uniqueness criterion tears.

regards

tommy1729

[JmsNxn]

0 < (d/dx)^n exp^[q](x) =< (d/dx)^n exp^[q](x)

Sorry, Tommy. I can't understand your question because you've written \(a \le a\); when I know there's a typo somewhere in there. Could you elaborate?

 

Regards, James

[tommy1729]

0 < (d/dx)^n exp^[q](x) =< (d/dx)^n exp^[q](x)

Sorry, Tommy. I can't understand your question because you've written \(a \le a\); when I know there's a typo somewhere in there. Could you elaborate?

 

Regards, James

Thanks for warning.

It is an unmature idea at the moment but I edited.

The idea is we want for n E {0,1,2,3} , q E [0,1] and x positive real :

0 < (d/dx)^n exp^[q](x) =< (d/dx)^n exp^[q + h](x)

for any positive h such that q + h =< 1.


regards

tommy1729

[tommy1729]

I changed my mind based on investigation.

optimist now.


regards

tommy1729