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$$\prod\limits_{i=1}^{\infty}\frac{1}{1-yx^{i}}=\sum\limits_{k=0}^{\infty}\frac{y^k x^k}{(1-x)(1-x^2)...(1-x^k)}$$

I know its combinatorial proof through "inspection",but is it true when $x,y$ are not integers?are there any non-combinactorial proof of this indentity?

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  • $\begingroup$ my teacher prove it under the condition of $x,y$ being integer $\endgroup$ – user510013 Dec 8 '17 at 8:04
  • $\begingroup$ Hmm, what kind of proof of that identity would rely on $x,y$ being integers? $\endgroup$ – anon Dec 8 '17 at 16:56
  • $\begingroup$ Hmm, what kind of proof of that identity would rely on $x,y$ being integers? Also, what is the proof by inspection? It may be more familiar if $y$ were replaced by $q$, as then it would be consistent with the fact this is a so-called $q$-identity. (It has a name, which I don't remember offhand.) $\endgroup$ – anon Dec 8 '17 at 22:36
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That is an equality of generating functions. The number of partitions of $n$, $p(n)$, is given by the number of ways for writing $n$ as the sum of some $1s$, some $2s$, some $3s$... In other terms,

$$p(n)= [z^n](1+z+z^2+z^3+\ldots)(1+z^2+z^4+\ldots)(1+z^3+z^6+\ldots)\cdots $$ or, by setting $p(0)=1$, $$ \sum_{n\geq 0} p(n)\,z^n=\frac{1}{1-z}\cdot\frac{1}{1-z^2}\cdot\frac{1}{1-z^3}\cdot\ldots = \prod_{h\geq 1}\frac{1}{1-z^h}. $$ If we introduce a tag-variable $y$, the coefficient of $x^n$ in $\prod_{i\geq 1}\frac{1}{1-y x^i}$ is a polynomial in the $y$-variable whose evaluation at $y=1$ returns $p(n)$. The coefficient of $y^a x^n$ in such product is the number of partitions of $n$ with exactly $a$ parts. Of course, if a partition has $a$ parts it is related to a number $\geq a$. Given a partition with $a$ parts, by transposing its Ferrer diagram we get a partition with each part having cardinality $\leq a$ and at least one part with cardinality $=a$. This is enough to explain the equality between the shown generating functions.

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