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:

up vote 1 down vote accepted

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 Aug 13 at 9:58
  • I managed to get in touch with Prof. Rutten and he doubts whether there exists a closed expression for exp(X^2) that is built up from constants, sum, convolution product and convolution inverse. He says that adding the use of shuffle product and shuffle inverse might help, but even then he still doesn't see how to compute a closed form. So I have accepted your answer as (probably) the best we can do. – alexpbell Aug 15 at 23:38

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