In the coinductive calculus of streams (sensu Rutten) $\exp(rX) = 1/(1-rX)$.

Is there a similarly nice representation for $\exp(rX^2)$?

Edit: I've just received a downvote on this. I'm making a wild guess this is because of the weakness of the connection to category-theory, so I've removed the tag. It could also be that this is opaque if you don't know what stream calculus is. Not sure what to do about that. If you do downvote a comment would be helpful.

Edit 2: Maybe this will help:

Unfortunately I don't have a nice formula for you, but at least I can show you what I have.

If I have done my calculations correctly then we have $$(rX^2)^{\underline{n}} = \frac{(2n)!}{2^n} r^n X^{2n}.$$

Inserting this into formula (53) of Rutten, we get $$ \exp(rX^2) = \sum_{n=0}^{\infty} \frac{1}{n!} (rX^2)^{\underline{n}} = \sum_{n=0}^{\infty} \frac{(2n)!}{2^n n!} r^n X^{2n} \\ = \sum_{n=0}^{\infty} (2n-1)!! \, r^n X^{2n} \\ = 1 + rX^2 + 3r^2X^4 + 15r^3X^6 + 105r^4X^8 + 945r^5X^{10} + \cdots $$

  • Thanks, this seems correct. It may be the case that there isn't a nice formula, but I have asked a related question here which may lead to one. – alexpbell yesterday

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