04/29/2014, 09:25 PM
Im a bit puzzled by the " -s ".
e^[3] = e^(e^e)
But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??
So I do not know what the "-s" means.
However your last formula seems to end that problem so it seems
\( \frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw) \)
Is the kind of formula we look for.
However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.
But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?
Actually I bet that this function is not analytic and I even bet sheldon agrees with me.
Intresting though. But I suspect you have mentioned this function before, not ?
regards
tommy1729
e^[3] = e^(e^e)
But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??
So I do not know what the "-s" means.
However your last formula seems to end that problem so it seems
\( \frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw) \)
Is the kind of formula we look for.
However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.
But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?
Actually I bet that this function is not analytic and I even bet sheldon agrees with me.
Intresting though. But I suspect you have mentioned this function before, not ?
regards
tommy1729
