4
$\begingroup$

How can I write $1-x-x^3+x^4+x^5+x^6-x^7 ....$ as a power series representation (i.e., a neat fraction such as $\frac{1}{1-x}$.

This stems from $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$.

I've been pondering this for a while, yet can't seem to think of any ways to solve this. Any hints?

EDIT: The polynominal with a few extra terms i: $1-x-x^3+x^4+x^5+x^6-x^7+2x^8-2x^9+2x^{10}-2x^{11}+3x^{12}-3x^{13}+3x^{14} ...$

$\endgroup$
  • 2
    $\begingroup$ What is the pattern here? $\endgroup$ – vadim123 Dec 21 '14 at 0:52
  • 1
    $\begingroup$ Would you mind writing out a few more terms to make a pattern more apparent $\endgroup$ – ClassicStyle Dec 21 '14 at 0:53
  • $\begingroup$ @vadim123: I'm not sure if there is any specific pattern. It was originally $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$, but I found some terms. $\endgroup$ – Mathy Person Dec 21 '14 at 0:53
  • $\begingroup$ @TylerHG: Sure, I don't mind. :) You can also reference my comment below. $\endgroup$ – Mathy Person Dec 21 '14 at 0:54
  • $\begingroup$ @TylerHG Ok. Added some more. It seems to alternate (positive, negative). Also, for the first 8 terms, the coefficient is 1. For the next 4 terms, the coefficient is 2. I'm assuming that for the next 2 terms after that, the coefficient is 3. $\endgroup$ – Mathy Person Dec 21 '14 at 1:06
0
$\begingroup$

This appears in the oeis, wherein it is given that the generating function is $$\prod_{k>0}1-x^{2k-1}=\prod_{k>0}\frac{1}{1+x^k}$$ There are also many references there, I highly recommend that link.

$\endgroup$
  • $\begingroup$ by k>0 , do you mean that it works for all k including 1 to infinity? $\endgroup$ – Mathy Person Dec 21 '14 at 1:18
  • $\begingroup$ That's correct; the products are each from $k=1$ on up. $\endgroup$ – vadim123 Dec 21 '14 at 1:23
  • $\begingroup$ It's not 'works for' - you need all those terms in the product. $\endgroup$ – Steven Stadnicki Dec 21 '14 at 1:24
  • $\begingroup$ @StevenStadnicki yes, that is what i meant to say, yet couldn't word it correctly. $\endgroup$ – Mathy Person Dec 21 '14 at 1:24
  • $\begingroup$ @vadim123: After using that: how would I turn $\frac{1}{(1+x)(1-x)(1+x^2)(1-x^2)(1+x^3)(1-x^3)....}$ into a generating function power series again (for powers past 15)? $\endgroup$ – Mathy Person Dec 21 '14 at 1:26

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.