I want to find a function $f(x,y)$ which can satisfy the following equation,

$\prod _{n=1} ^{\infty} \frac{1+x^n}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})} = exp [ \sum _{n=1} ^\infty \frac{f(x^n,y^n)}{n(1-x^{2n})}]$

  • I would like to know how this is solved.

(..though I landed into this through a different route (calculating Witten Index!), such expressions also occur in finite dimensional representation theory where a generating function for the character of the (anti)symmetric powers of a representation (the LHS) is written as a Plethystic exponential (the RHS) of the original (generally fundamental) representation...)

One can perturbatively check that the following function satisfies the above equation,

  1. $f(x,y) = \sqrt{x} (\sqrt{y} + 1/\sqrt{y}) + x (1 + y + 1/y) + x^{3/2} (y^{3/2} + 1/y^{3/2}) + x^2 (y^2 + 1/y^2)$

$\quad\quad\quad\quad\quad + \frac{(x y)^{5/2} }{(1 - \sqrt{x y})}(1 - 1/y^2) + \frac{(x/y)^{5/2}}{(1 - \sqrt{x/y})} (1 - y^2) $

The paper doesn't state any proof or explanation for how this was obtained but the above is order-by-order in $x$ checkable to be right after truncating the original equation at any finite value of $n$. (...I don't know how to check this keeping the full sum/product over $n$..)

Now I tried to do something obvious but it didn't work!

$\prod_{n=1}^{\infty} \frac{ (1+x^n) }{1+x^n -x^{\frac{n}{2}} \left(y^{\frac{n}{2}} + y^{-\frac{n}{2}}\right) } = \exp \left[ \sum_{n=1}^{\infty} \frac{ I_{ST}(x^n,y^n) } {n (1-x^{2n}) } \right] $

$\Rightarrow \sum_{n=1}^{\infty} \left[ \ln (1+x^n) - \ln(1-(\sqrt{xy})^n) - \ln\left(1- \left(\sqrt{\frac{x}{y}}\right)^n\right) \right] = \sum_{n=1}^\infty \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} $

Now we expand the logarithms and we have,

$\sum_{n=1}^{\infty} \left[ \sum_{a=1}^{\infty} (-1)^{a+1} \frac{x^{na}}{a} + \sum_{b=1}^{\infty} \frac{ (\sqrt{xy})^{nb} } {b} + \sum_{c=1}^{\infty} \frac{ (\sqrt{x/y})^{nc} }{c} \right] = \sum _{n=1}^\infty \frac{f(x^n,y^n)} {n(1-x^{2n})}$

$\Rightarrow \sum _{a=1} ^{\infty} \frac{1}{a} \left[ \sum _{n=1} ^{\infty} \left( (-1)^{a+1}x^{na} + (xy)^{\frac{na}{2}} + \left(\frac{x}{y}\right)^{\frac{na}{2}} \right) \right] = \sum _{n=1} ^\infty \frac{f(x^n,y^n)} {n(1-x^{2n})}$

By exchanging $a$ and $n$ (relabeling on the LHS) and matching the patterns on both sides and picking out the $n=1$ term one sees that one way this equality can hold is if,


$f(x,y) = (1-x^2) \sum _{a=1} ^{\infty} \left[ x^a + (xy)^{\frac{a}{2}} + (\frac{x}{y})^{\frac{a}{2}} \right]$

$\Rightarrow f (x,y) = (1-x^2) \left(-1 + \frac{1}{1-x} -1 + \frac{1}{1-\sqrt{xy}} - 1 + \frac{1}{1-\sqrt{\frac{x}{y}} } \right)$

But this solution is neither the one above which could be perturbatively checked to be true nor does it satisfy the original equation! Why?

After doing a series expansion of the above (using Series on Mathematica) one sees that this above derived equation (2) differs from (1) in having just one extra term of $x^2$. (...I would like to know what is wrong in the derivation that gives (2) this one extra term compared to the non-derivable but perturbatively checked correct answer (1)...)

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    $\begingroup$ Can you please give the exact reference to "a certain paper" ? $\endgroup$
    – F. C.
    Oct 5, 2012 at 18:48
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    $\begingroup$ write the product over $n$ of $F_{n}(x,y)$ as the exponent of the sum over $n$ of log $F_{n}(x,y)$ and then expand $F$ as a power series in $x$. $\endgroup$ Oct 5, 2012 at 19:14
  • $\begingroup$ @Carlo Beenakker Can elaborate a bit on what is the theorem you are using to identify $f$ - what you are suggest will make the LHS look like the RHS - but beyond that what? I need to solve many such kinds of equations and I would like to know what is the general principle being used. $\endgroup$
    – Anirbit
    Oct 6, 2012 at 21:26
  • $\begingroup$ @F.C I did not link the paper since its a paper whose central focus is something deep into physics and nothing to do with this one equation - and hence I thought it would be distracting to link to that huge paper - this equation and its solution I have quoted is just one line in that huge paper. $\endgroup$
    – Anirbit
    Oct 6, 2012 at 21:38
  • $\begingroup$ @Carlo Beenakker By what you are saying the equation will look like, $exp [ \sum _ {n=1} ^{\infty} log F(x^n,y^n) ] = exp [ \sum _ {n=1} ^{\infty} \frac{f(x^n,y^n)}{n(1-x^{2n})} ]$ Now how do you propose to identify $f(x,y)$ from a power expansion of $log F(x^n,y^n)$? $\endgroup$
    – Anirbit
    Oct 6, 2012 at 21:43

1 Answer 1


I will try to by more helpful than I was in the comments. First a general observation.

Your infinite product contains socalled Euler functions,


which have the Lambert series expansion


Part of your infinite product can be expanded in this way,

$P(x,y)\equiv\prod_{n=1}^{\infty}\frac{1}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})}= \frac{1}{\phi(\sqrt{xy})\phi(\sqrt{x/y})}=\exp\left[\sum_{n=1}^{\infty}\frac{1}{n}g(x^n,y^n)\right]$.

The function $g(x,y)$ is given by


This is not quite what you have written. I have difficulty verifying your expression. For example, the limit $x\rightarrow 0$ of the left-hand side of your expression is $1+\sqrt{x}(\sqrt{y}+1/\sqrt{y})$, but the same limit of the right-hand side is $1+\sqrt{x}(y+1/y)$.

UPDATE 1: Your corrected expression is still problematic; the left-hand side contains the infinite product

$\prod_{n=1}^{\infty}(1+x^n)=\exp\left[-\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}G(x^{n})\right]$, with $G(x)=x/(1-x)$.

The factor $(-1)^n$ in the sum over $n$ seems inconsistent with the right-hand side of your expression --- but in fact it is consistent (see update 2).

UPDATE 2: One more identity is needed, in addition to the Euler function identities, to complete the identification:


We then have, quite generally,


with the functions $F(x)=x/(1-x^2)$, $g(q)=q/(1-q)$.

Your equation (1) corresponds to $q_1=\sqrt{x/y}$, $q_2=\sqrt{xy}$.

I know, your function $f(x,y)$ looks much more lengthy, but it is really just $F(x)+g(\sqrt{xy})+g(\sqrt{x/y})$.

  • $\begingroup$ @Carlo Beenakker Thanks for your reference about Lambert functions. It seems to fit the kind of pattern that I want - let me read more about it and get back to you. Also it turns out that the original function (1) that I had typed was wrong (the original paper had it wrong!) and the correct function is what is now typed as (1). The function (1) satisfies the equation as far as one can check perturbatively. But I don't understand why the equation that I derived (2) neither satisfies the equation nor is that the same as (1) - but I thought that (2) was derived by a pretty trivial argument! $\endgroup$
    – Anirbit
    Nov 2, 2012 at 21:57
  • $\begingroup$ @Carlo Beenaker Thanks for your efforts. I did not understand your UPDATE. To which equation are you referring to? Firstly I would like to know how equation (1) can be derived since this (1) can at least perturbatively be checked to be right. Secondly I would like to know why the equation (2) is not solving the equation - I can't see anything obviously going wrong with its derivation. $\endgroup$
    – Anirbit
    Nov 4, 2012 at 19:27
  • $\begingroup$ @Carlo Beenakker Thanks for your updates. So using these Euler function identities you claim that equation (1) can be exactly derived. right? And what kind of series are these with fractional powers ? This is not a Laurant/Taylor series. And do you have any idea as to what is wrong with the derivation of (2). Why is that not working? $\endgroup$
    – Anirbit
    Nov 5, 2012 at 17:59
  • $\begingroup$ @Carlo Beenakker And can you give a reference which proves the two expansions that you have used? (..the Wiki link doesn't seem to prove it..) $\endgroup$
    – Anirbit
    Nov 5, 2012 at 18:02
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    $\begingroup$ the Euler function identity is just a Taylor series of $\ln\phi(q)$ around $q=0$; for the proof of the second identity I have asked MO for help, with success (see the answer to question 111648). $\endgroup$ Nov 6, 2012 at 15:30

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