Hoc can you find the sequence \( b_n \) from the sequence \( a_n \)?
Looking at the definition of entire function I saw that \( (|a_n|/n!)^{(1/n)} \) should converge to zero or \( (ln|a_n/n!|)/n \) to \( - \infty \)(for n that goeas to infinity), but for some sequences seems it does have a uggly behaviour...I guess that more the sequence grows fast and bigger are the problem...but sequences that grows very fast are exactly the ones defined using the Hyperops. I apologize if I made some errors.
Quote:Assume \( a_n \) is a sequence of complex numbers such that \( f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!} \) is entire. Then, there always exists \( b_n \) such that, \( g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!} \) is entire and Weyl differintegrable on all of \( \mathbb{C} \) and[...]And what happens if the sequence \( a_n \) gives you a non-entire \( f \)?
Looking at the definition of entire function I saw that \( (|a_n|/n!)^{(1/n)} \) should converge to zero or \( (ln|a_n/n!|)/n \) to \( - \infty \)(for n that goeas to infinity), but for some sequences seems it does have a uggly behaviour...I guess that more the sequence grows fast and bigger are the problem...but sequences that grows very fast are exactly the ones defined using the Hyperops. I apologize if I made some errors.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)