05/15/2014, 08:53 PM

To justify a bit more what I said notice that :

Let D be the differential operator.

This follows from

Now D exp^[1/2](x) is finally smaller than A * exp^[1/2](x) for any A > 1 BECAUSE

exp^[1/2](x) < exp(A x)

Take the logarithmic derivative on both sides :

( D exp^[1/2](x) ) / exp^[1/2](x) < A

QED.

Ofcourse we know D exp^[1/2](x) > exp^[1/2](x) / x from the simple consideration of a Taylor series.

This proves the previous post of me was correct.

regards

tommy1729

Let D be the differential operator.

This follows from

Now D exp^[1/2](x) is finally smaller than A * exp^[1/2](x) for any A > 1 BECAUSE

exp^[1/2](x) < exp(A x)

Take the logarithmic derivative on both sides :

( D exp^[1/2](x) ) / exp^[1/2](x) < A

QED.

Ofcourse we know D exp^[1/2](x) > exp^[1/2](x) / x from the simple consideration of a Taylor series.

This proves the previous post of me was correct.

regards

tommy1729