1
$\begingroup$

An ordinary enumerator is given as $(1+x+x^2)^p$. This is being understood as follows:

There are 2 each of p kinds of objects.The ordinary enumerator for selecting none (or) one (or) both the objects of that kind is $(1+x+x^2)^p$

Similarly,how do i visualize (or) understand the enumerator
$(1+x^5+x^9)^{100}$ in plain english

$\endgroup$
  • $\begingroup$ It seems to me the example "plain English" statement relies on the term ordinary enumerator without giving a definition. At best the statement is a kind of illustrative example, but it doesn't seem all that clear (perhaps leading to your difficulty in applying it to a quite similar example). Should we back up and give an understandable account of generating functions? $\endgroup$ – hardmath Jun 17 '15 at 15:06
  • $\begingroup$ @hardmath The actual question was to find coeffecient of (x^23) in (1+x^5+x^9)^100.An explanation leading to it's answer would be really great. $\endgroup$ – Pradeep Jun 17 '15 at 15:12
  • $\begingroup$ We could adapt the methods of computation I used in this recent Answer to find that coefficient. The wording of the Question above suggests you want an interpretation of (all) coefficients, which probably has a large number of valid responses, while the algebra of the polynomial, though somewhat tedious, has a well-defined result. $\endgroup$ – hardmath Jun 17 '15 at 15:20
  • $\begingroup$ @hardmath In the link mentioned above, how did we convert $(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2$ to $(\frac{1 - x^{21}}{1 - x})^2$. Is there a formula ? $\endgroup$ – Pradeep Jun 17 '15 at 15:42
  • $\begingroup$ This is just the usual formula for a geometric sum. It doesn't apply in any obvious way to your problem. $\endgroup$ – hardmath Jun 17 '15 at 15:45
2
$\begingroup$

Relating that polynomial to your own example, we can understand $(1+x^5+x^9)^{100}$ as follows:

There are $9$ each of $100$ kinds of objects. The ordinary enumerator for selecting none or five or all nine of the objects of that kind is $(1+x^5+x^9)^{100}$.

Here's another interpretation:

Consider three-sided dice whose faces have $0$, $5$, or $9$ pips, and roll $100$ such dice. Then the coefficient of $x^k$ in $(1+x^5+x^9)^{100}$ is the number of ways to get a total of $k$.

$\endgroup$
  • 1
    $\begingroup$ The second example was great.Thanks..!! $\endgroup$ – Pradeep Jun 17 '15 at 16:51
  • $\begingroup$ @Pradeep I find the second example a lot easier to understand too! $\endgroup$ – Théophile Jun 17 '15 at 18:07

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.