The exponential generating function of a sequence $a[n]$ is:

$\displaystyle \text{EG}(a;x) = \sum_{n=0}^\infty a[n] \frac{x^n}{n!}$

My question is about exponential generating functions of binary sequences $b[n]$, i.e. for which the codomain is $\{0,1\}$.

We know for all such $b[n]$, the exponential generating function $\text{EG}(b;x)$ is absolutely monotonic, and furthermore that it is an entire function with no poles, since the function $\exp(x)$ is the special case where $b[n] = 1$ for all $n$. We can compare any two such $\text{EG}(b_1;x)$ and $\text{EG}(b_2;x)$ by saying that $\text{EG}(b_1;x) \leq \text{EG}(b_2;x)$ iff there exists some real $r$ such that the inequality holds for all real $x > r$.

As a result, given two binary sequences $b_1[n]$ and $b_2[n]$, we can say that $b_1[n] \precsim b_2[n]$ iff $\text{EG}(b_1;x) \leq \text{EG}(b_2;x)$.

My questions:

  1. Is $\precsim$ a total order?
  2. If so, what is the order type of $\precsim$?
  3. There seems to be an initial segment of $\Bbb N$, given by the set of all binary sequences with finitely many $1$'s, ordered lexicographically. After this, is it dense?
  • 1
    $\begingroup$ (1) Define $b_1$ to be the sequence that has $1$s where the MacLaurin of $\sin(x)$ has a negative coefficient and zeros otherwise. Define $b_2$ to be the sequence that has a $1$ where that series has a positive coefficient and zeros otherwise. Then $EB(b_2;x)-EB(b_1;x)=\sin(x)$. $\endgroup$ – user647486 Mar 19 '19 at 23:21
  • $\begingroup$ Ah yes, very good - so it isn't a total order. Does seem to have an initial segment of $\omega$ though (and a final segment of $\omega^*$, which I forgot to mention). $\endgroup$ – Mike Battaglia Mar 19 '19 at 23:30
  • $\begingroup$ You might see this answer It deals with finite strings of digits and concludes the order type is $\omega \cdot (1+\Bbb Q)+1$ I believe you have this for the strings that are eventually $0$ and this reversed for the strings that are eventually $1$. In between we have all the strings that have infinitely many $1$s and infinitely many $0$s, which are the hard ones and most of them. $\endgroup$ – Ross Millikan Mar 20 '19 at 0:05

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