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Say I have a generating function $\Phi_\mathcal{A}$ for the set of partitions $\mathcal{A}$ which have no parts congruent to 2 mod 4, and I have the generating function for $\Phi_\mathcal{B}$ for the set of partitions $\mathcal{B}$ in which the parts divisible by 4 occur at most once, can I multiply the 2 generating functions together to obtain the generating function for the intersection of $\mathcal{A}$ and $\mathcal{B}$ (i.e for the partitions that satisfy both requirements)? And more importantly, is this a technique that I can use for arbitrary such conditions?

The reason I ask is because my homework assignment contains several questions which require me to give the generating function for partitions that have some property AND some other property, and I wanted to know if this technique was a good/valid one to use for these problems.

Also, is there a similar technique for giving the generating function for integer partitions that have some property OR some other property?

Thanks!

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1 Answer 1

up vote 1 down vote accepted

Multiplying them won’t do the job: that gives you the convolution of the corresponding sequences. (Actually, you can see that multiplication isn’t right by considering what happens when $\mathcal A=\mathcal B$: then you’d have to have $(\Phi_{\mathcal A})^2=\Phi_{\mathcal A}\Phi_{\mathcal B}=\Phi_{\mathcal A\cap\mathcal B}=\Phi_{\mathcal A}$, which you can’t reasonably expect.

In this case notice that the conditions allow unrestricted odd parts and restrict the even parts to multiples of $4$ occurring once, giving you

$$\prod_{k\ge 1}\frac1{1-x^{2k-1}}\cdot\prod_{k\ge 1}\left(1+x^{4k}\right)=\prod_{k\ge 1}\frac{1+x^{4k}}{1-x^{2k-1}}\;.$$

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Thanks Brian! What I ended up getting was the following: $$\prod_{k\ge 1}\frac{1+x^{4k}}{(1-x^{4k-3})(1-x^{4k-1})}\;.$$which is equivalent to what you wrote above I think :) –  Nizbel99 Nov 14 '12 at 18:20
    
@user43552: You’re welcome! Looks equivalent to me. –  Brian M. Scott Nov 14 '12 at 21:05

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