\( \operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(x))) \)
to compute a taylor series , we need to know that it is justified.
concretely that means we need to prove that
\( \ D^m {TommySexp_e}(z,x)= D^m \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(x))) \)
, where D^m is the mth derivative with respect to x(*) , holds.
( with respect to z should give the same result , but seems harder at first sight )
analytic continuation preserves periodicity , which is 2pi i.
that is also the reason why we apparantly cant extent this method to bases between eta and e^(1/2). ( hints : fourier , entire , think about it )
a further remark and actually request is that i would love to see a plot of all the solutions , including branches of
exp(x) = t
exp(exp(x)) = t
...
exp^[n] = t
where t = 0 +/- 1.895494239i is one of the nonzero fixpoint of 2sinh.
it might give me new insight or ideas.
thanks in advance.
regards
tommy1729
to compute a taylor series , we need to know that it is justified.
concretely that means we need to prove that
\( \ D^m {TommySexp_e}(z,x)= D^m \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(x))) \)
, where D^m is the mth derivative with respect to x(*) , holds.
( with respect to z should give the same result , but seems harder at first sight )
analytic continuation preserves periodicity , which is 2pi i.
that is also the reason why we apparantly cant extent this method to bases between eta and e^(1/2). ( hints : fourier , entire , think about it )
a further remark and actually request is that i would love to see a plot of all the solutions , including branches of
exp(x) = t
exp(exp(x)) = t
...
exp^[n] = t
where t = 0 +/- 1.895494239i is one of the nonzero fixpoint of 2sinh.
it might give me new insight or ideas.
thanks in advance.
regards
tommy1729