Timeline for Recursion for generating functions
Current License: CC BY-SA 4.0
9 events
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Jun 12, 2022 at 21:09 | comment | added | Marc | @Gupta If a power series $F$ has positive convergence radius then also any power series that can be reasonably expressed in terms of $F$ (for example as a rational function of $F$ and its derivatives and integrals and arbitrary compositions of those when they make sense) also will a positive radius of convergence. So if $G$ as zero radius of convergence, then is cannot be reasonably expressed in terms of $F$. And Gerald's example isn't just an isolated counter-example: this behaviour happens whenever $f(n)$ grows exponentially with $n$, so includes many interesting sequences. | |
Jun 12, 2022 at 18:04 | comment | added | Sam Hopkins | I think this question is fine but Gerald's answer is also very reasonable. I'm leaving it at that because the way you're responding is unpleasant. | |
Jun 12, 2022 at 17:59 | comment | added | Gupta | @SamHopkins Is saying an answer is wrong being aggressive? Am I suppose to pretend it is right? He is claiming that a power series that has a radius of convergence of 0 means somehow it can't have some algebraic relationships(He doesn't explicitly state why fails but the implication is that "it almost surely cannot work in any way"). But again, this is wrong because convergence doesn't mean anything in formal series. It's a non-sequitur. Sorry if pointing it out sounds aggressive but the truth doesn't need and shouldn't have a sugar coating so it is more palatable. | |
Jun 12, 2022 at 17:53 | comment | added | Sam Hopkins | Gupta, you're being pretty aggressive to someone who is just trying to help you answer your question. Gerald's answer looks very reasonable to me: often what's most useful about power series is moving between formal and analytic perspectives. In a literal sense, the answer to your question "can G(z) be expressed in terms of F(z)" is yes because the coefficients determine the power series and you defined the coefficients of G in terms of those of F. So ultimately you have to be willing to accept soft/subjective answers like Gerald's. | |
Jun 12, 2022 at 15:16 | comment | added | Gupta | @Buzz generating functions, though, are formal objects though and convergence is irrelevant except when using them to deduce other things about the series besides just algebraic manipulation(such as asymptotics). To prove I'm right I suggest reading page 21 from Analytic Combinatorics about P(z) = sum(n! z^k) which explicitly proves the "answer" is completely wrong(at least in terms of it's hypothesis that radius of convergence of 0 makes formal manipulation "unlikely". | |
Jun 12, 2022 at 10:49 | comment | added | Massimo Ortolano | More generally, I suspect that if one restricts $f$ to an integer range, there won’t probably be many useful convergent series. | |
Jun 12, 2022 at 1:57 | comment | added | Buzz | @Gupta "What does convergence have to do with formal power series?" Probably the most important reason for using generating functions is that a generating function turns an algebraic list of quantities into a function, which can be studied using the additional techniques of analysis. Recursion relations for coefficients become differential equations for the generating function, etc. | |
Jun 12, 2022 at 0:55 | comment | added | Gupta | What does convergence have to do with formal power series? Even then, what if we were to restrict f's growth rate? What if it is just a permutation of Z? Clearly when f(k) = k this works and G(z) = F(z) so it doesn't seem really unlikely. Of course we could get a better handle on convergence by using EGF. In fact, I'm only concerned with expressing recursion in terms of some transformation so unless there is some deep reason why it can't work except for a small finite number of cases(which would then be interesting to know) there is probably some expression that exists(although maybe complex) | |
Jun 12, 2022 at 0:31 | history | answered | Gerald Edgar | CC BY-SA 4.0 |