7

If I have an ordinary generating function (OGF) as a rational polynomial:

$$ Q(y) = \frac{f_1(y)}{g_1(y)} $$

Which has a power series representation

$$ Q(y) = \sum_{n \geq 0} a_n y^n $$

How can I compute the exponential generating function (EGF) $$ Z(y) = \sum_{n \geq 0} b_n y^n / n! $$ as a rational polynomial? $$ Z(y) = \frac{f_2(y)}{g_2(y)} $$

As an explicit solved example, consider the case where $$ Q(y) = \frac{yv -1}{y(2+v) -1} $$ The intermediate terms are $$ a_n = 2 (v+2) ^ {n-1} $$ Giving as a final answer $$ Z(y) = \frac{2 e^{(v+2)y}}{v+2} $$

... ideally, I like to be able to do this by going from one rational polynomial to another without computing $a_n$. We can assume that I can factor the polynomial $g_1(y)$.

Manolito Pérez
  • 1,241
Hooked
  • 6,637

1 Answers1

6

What you want is the Borel transform. This is related to the inverse Laplace transform. Specifically, $Z(s)$ is the inverse Laplace transform of $Q(1/t)/t$.

Robert Israel
  • 448,999
  • This doesn't seem give the right answer - looking it up on Wolfram Alpha gives an extra factor of $v/(v+2)$. It is however close, am I doing something wrong? – Hooked May 20 '11 at 18:25
  • 1
    Your example wasn't quite right. In the Maclaurin series of $Q(y) = (y v - 1)/(y (v+2) - 1)$, the coefficient of $y^k$ is $a_k = 2 (v+2)^{k-1}$ for $k \ge 1$, but is 1, not $2/(v+2)$, for $k=0$. $Q(1/t)/t = (t-v)/(t(t-v-2)$, whose inverse Laplace transform is $Z(y) = (2 e^{(2+v)y}+v)/(v+2)$. The Maclaurin series for this is indeed $1 + 2 y + \frac{2(v+2)}{2!} y^2 +
    \frac{2(v+2)^2}{3!} y^3 + \ldots$.     – Robert Israel May 21 '11 at 00:06