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If I have a generating function for $f(n)$ defined by

$g(x)=\displaystyle\sum_{n=0}^{\infty}f(n)x^n=\dfrac{P(x)}{Q(x)}$,

where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero function, how could I show that $f(n)$ is not more than exponential?

draks ...
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alo
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2 Answers2

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  1. Since $Q(0)\ne0$, $|Q(z)|\geqslant a$ for every $z$ in a neighborhood of $0$, say for every $|z|\leqslant\varepsilon$, for some $a\gt0$ and $\varepsilon\gt0$.
  2. Since $P$ is continuous, $|P(z)|\leqslant b$ for every $|z|\leqslant\varepsilon$, for some finite $b$.
  3. For every $n$, $$ f(n)=\frac1{2\mathrm i\pi}\oint\frac{P(z)}{Q(z)}\frac{\mathrm dz}{z^{n+1}}, $$ where the integral is over the circle of equation $|z|=\varepsilon$, whose length is $2\pi\varepsilon$.
  4. In particular, $$ |f(n)|\leqslant\frac1{2\pi}2\pi\varepsilon\frac{b}a\frac1{\varepsilon^{n+1}}=cK^n, $$ with $c=b/a$ and $K=1/\varepsilon$.
Did
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  • Here we go, you used the idea of the nth derivative of a function. Using the Cauchy's formula for the nth derivative. – Mhenni Benghorbal Jul 17 '12 at 20:22
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    @Mhenni: Sorry but I fail to see how your comment applies to my answer (which uses no derivative). – Did Jul 18 '12 at 06:50
  • What you used is the Cauchy formula for the nth derivative of $\frac{P(x)}{Q(x)}$ at the point $x=0$ divide bu $n!$. It is the formula $ f(n)=\frac{1}{n!}{\left(\frac{P(0)}{Q(0)}\right)}^{(n)} = \frac{1} {2\pi i} \int_C \frac { P(z) }{ Q(z) }\frac{dz}{z^{n+1}} $. – Mhenni Benghorbal Aug 11 '12 at 22:22
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If you look backward to your problem you see that $f(n)$ is nothing, but the nth derivative of your rational polynomial $\frac{P(x)}{Q(x)}$ divided by $ n! $. I refer you to my Ph.D thesis (section 6.2.1) which covers the topic of finding the nth derivative of any rational polynomial. In other words you get a closed form formula for the nth derivative which may help you to conclude you assertion:

https://docs.google.com/file/d/0B4FXAHVyGS9KMGRiNDMyNDctMmQ1NS00MDI3LTk2OWEtNzc3N2ZlNDVmYjJm/edit?hl=en_GB&pli=1

Mhenni Benghorbal
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    Once you have the partial fraction decomposition (as in your thesis), you hardly need explicit formulas for the derivatives; the standard geometric-series classification (as in anon's comment) is a much easier way of seeing the same thing. – Steven Stadnicki Jul 17 '12 at 19:44
  • My point was to give him an idea to prove his assertion by exploiting the nth derivative of his rational polynomial. – Mhenni Benghorbal Jul 17 '12 at 20:28