# Generating Functions of Partitions?

Show that $2(1-x)^{-3} [(1-x)^{-3} + (1+x)^{-3}]$ is the generating function for the number of ways to toss $r$ identical dice and obtain an even sum.

Workings:

I'm not too sure on this problem.

With the $-3$ I'm guessing that the dice are 6 sided.

I suppose that for the sum to be even there could be an even number of odd numbers on the die, an even number of even numbers shown on the die or an odd number of even numbers shown on the die.

But I'm not sure how to use this info into getting a generating function.

Any help will be appreciated.

• Expanding this out gives you $4 + 12x + 48x^2 + 112x^3 + ...$. Something is wrong here. – Andrew Szymczak Jan 20 '15 at 7:26
• What does it mean that you're not too sure on this problem? Is the problem correctly stated? I can't think of any way to interpret the problem such that the given answer is correct. – mjqxxxx Jan 26 '15 at 23:24
• I agree with @mjqxxxx and suggest you reformulate the question in order to get a helpful answer. See also the comment section of my answer. Best regards, – Markus Scheuer Jan 27 '15 at 0:27

Note: The formulation of the problem seems to be incorrect. Nevertheless here's a hint for a 6-sided die.

The generating function for tossing one die once is $x^1+x^2+\ldots+x^6$, the exponent of $x$ indicating the result of the throw. Let's denote with $A(x)$ the generating function for tossing $r$ dice.

We observe, applying the formula for finite geometric series:

\begin{align*} A(x)&=(x^1+x^2+\ldots+x^6)^r\\ &=x^r(1+x+\ldots+x^5)^r\\ &=x^r\left(\frac{1-x^6}{1-x}\right)^r \end{align*}

Since we are interested in even results only, we are looking for a generating function which contains only the even powers of $x$ in $A(x)$.

This job is done with

\begin{align*}\frac{1}{2}\left(A(x)+A(-x)\right)\tag{1}\end{align*}

Note: (1) is valid for all generating functions $A(x)=\sum_{n=0}^{\infty}a_nx^n$, since \begin{align*} \frac{1}{2}\left(A(x)+A(-x)\right)& =\frac{1}{2}\left(\sum_{n=0}^{\infty}a_nx^n+\sum_{n=0}^{\infty}a_n(-x)^n\right)\\ &=\sum_{n=0}^{\infty}a_n\frac{x^n+(-x)^n}{2}\\ &=\sum_{{n=0}\atop{n\equiv 0(2)}}^{\infty}a_nx^n\\ &=\sum_{n=0}^{\infty}a_{2n}x^{2n} \end{align*}

• Can you clarify why $\frac{1}{2}\left(A(x)+A(-x)\right)$ picks only event results? – Trismegistos Jan 26 '15 at 14:37
• @Trismegistos: I've added a note to my answer to clarify this aspect. Best regards, – Markus Scheuer Jan 26 '15 at 17:05
• You are giving a generating function for the number of ways to throw a particular (even) result, for a fixed number of dice. I.e., the coefficient of $x^k$ is the number of ways to throw the exact result $k$. The original question seems to be different: the coefficient of $x^k$ should be the number of ways to throw any even result with $k$ dice. – mjqxxxx Jan 26 '15 at 17:12
• @mjqxxxx: Although your interpretation seems to be reasonable, the resulting generating function is $\frac{3x}{1-6x}$. Since the generating function stated in the question is different, OP should correct and reformulate the problem in order to expect a helpful answer. Do you agree? – Markus Scheuer Jan 27 '15 at 0:23
• @MarkusScheuer It turns out that there was a typo in the book. So I need to shy why $2(1-x)^{-3} [(1-x)^{-3} + (1+x)^{-3}]$ is incorrect. – ineedanewnames Jan 27 '15 at 23:26