2
$\begingroup$

Suppose there is formal power series in infinite product form as follows: $$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}$$, where $a_d$ are positive integers. Consider the expression $$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}-1$$

Now, this expression has no 1 in it. So, when expressed as formal power series, the expression has the form $\sum_{n=1}^{\infty} b_nx^n$. Now, my question is can we write the above expression in an infinite product form as above by taking something out as common term. So, for example, observe that $b_1=\frac{a_1}{q-1}$. So, possibly can we write

$$\prod_{d\geq 1} \left(1+\frac{u^d}{q^d-1}\right)^{a_d}-1=\frac{a_1}{q-1}u\times (\text{an infinite product expression})$$ or, maybe in the form (Exponential generating function)$\times$ (an infinite product expression).

We note that the coefficients $b_n$ has the following form:

$$b_n=\sum_{\lambda \vdash n} \prod_{i=1}^{n} \frac{\binom{a_i}{m_i(\lambda)}}{(q^i-1)^{m_i(\lambda)}}$$

where, $\lambda\vdash n$ means $\lambda$ is a partiton of $n$ (hence the outer sum runs over all partitons of $n$), and $m_{i}(\lambda)$ denotes the number of times $i$ occur in the partition $\lambda$.

I know that my question probably seems a little confusing as it doesn't perfectly mention what exactly is it that I want to know, but anyway this is the best I could frame it. Thanks in advance for any kind of help.

$\endgroup$
1
$\begingroup$

The question is rather vague, and thus it may have multiple answers. Here is one.

Let $P$ denote the product of interest and let $Q:=\log(P)$, i.e. $$Q = \sum_{d\geq 1} a_d\log(1+\frac{u^d}{q^d-1}).$$ Then we have $$P-1=\exp(Q)-1=Q\frac{\exp(Q)-1}{Q}=Q\exp(\log(\frac{\exp(Q)-1}{Q})).$$ We have $$\log(\frac{\exp(Q)-1}{Q}) = \frac{1}{2}Q + \frac{1}{2}\sum_{k\geq 1} \frac{B_{2k}}{(2k)!k} Q^{2k},$$ where $B_{2k}$ are Bernoulli numbers, and thus $$P-1 = Q \sqrt{P} \prod_{k\geq 1} \sqrt{\exp( \frac{B_{2k}}{(2k)!k} Q^{2k} )}.$$ So we represented $P-1$ as some factor times the infinite product as it was requested by OP.

$\endgroup$

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.