$A (x)$ is the generating function for partitions. $B(x)=\sum_{n=0}^{\infty}b_nx^n $

$$b_n =\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$$

Find the truncation of $A (x) B(x)$ to degree $10$.

This is what I have so far:

$$A (x) = \frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)\cdots}$$

I believe $B(x) = \frac{1}{(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots}$

$$A(x)B(x) = \frac{1}{(1+x)(1-x)(1+x^2)(1-x^2)(1+x^3)(1-x^3)\cdots}$$

Currently $A(x)B(x)$ is a power representation. My goal is to turn $A(x)B(x)$ into a power series (where it is infinite), then find the polynomial where it goes up to degree $10$.

My question is: 1) Is what I have currently correct?


2) How would I use geometric sums to compute the power series expansion of $A(x)*B(x)$? I actually asked this here: text

  • $\begingroup$ @bof: oh, yes, that is a typo. i will fix that. $\endgroup$ – Mathy Person Dec 21 '14 at 2:35
  • $\begingroup$ I believe $B(x)=(1-x)(1-x^2)(1-x^3)(1-x^4)\cdots$. $\endgroup$ – bof Dec 21 '14 at 2:37
  • $\begingroup$ @bof: how did you get that? $\endgroup$ – Mathy Person Dec 21 '14 at 2:38
  • $\begingroup$ @bof: if that is true, then $A(x)*B(x)$ = 1 $\endgroup$ – Mathy Person Dec 21 '14 at 2:42
  • $\begingroup$ I was wrong. I misremembered something from combinatorics class. Actually $(1-x)(1-x^2)(1-x^3)(1-x^4)\cdots$ is the generating function for (number of partitions of $n$ into an even number of distinct parts) minus (number of partitions of $n$ into an odd number of distinct parts). Sorry! $\endgroup$ – bof Dec 21 '14 at 3:08

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