I have the following statement in a paper:

Let $\Psi$ be the formal power series defined over the alphabet $\Omega$ and the log semiring by: $(\Psi, (a, b)) = -log(c((a,b)))$ for $(a,b) \in \Omega$, and let $S$ be the formal pwer series $S$ over the log semiring defined by: $S=\Omega^*+\Psi+\Omega^*$ (an alphabet is a finite set of symbols and $\Omega$ contains pairs of such symbols.)

$S$ is a rational power series as a +-product and closure of the polynomial power series $\Omega$ and $\Psi$.

What exactly is meant here? I know about the automata theoretical aspects, but I haven't heard of the notion "formal power series" over an alphabet and a semiring. How can I think of this?

The paper is at http://www.cs.nyu.edu/~mohri/pub/, [99], page 15 and page 17.

  • 1
    $\begingroup$ Is that a verbatim copy from the paper? $\endgroup$ – Mariano Suárez-Álvarez Jan 24 '11 at 5:02
  • $\begingroup$ Could you provide a link or a precise reference to the paper, please? $\endgroup$ – t.b. Jan 24 '11 at 5:04
  • $\begingroup$ yes. updated. thanks. $\endgroup$ – Felix Dombek Jan 24 '11 at 5:17

A forum power series is a "Symbolic series" and not necessarily one where the object the series is over is a "number".

The idea comes from group there where you have the operations of multiplication and addition in a ring. you can form the expressions $ax + bx^2 + cx^3$, etc... e.g., the polynomials or P[x] but note x may not necessarily be in the ring. We can still write such expressions and take x as a sort of place holder or symbol

Polynomial ring

Formal power series

The idea is an extension from the idea of a "normal" power series but applied to a "formal object". Those two links should clue you in on the exact meaning.

  • $\begingroup$ Sorry, the Wikipedia article didn't help me much :( $\endgroup$ – Felix Dombek Jan 24 '11 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.