0
$\begingroup$

Let $a(q)$ denote the Borwein function $$a(q)=\sum_{m,n=-\infty}^\infty q^{n^2+nm+m^2}.$$ In this research paper the author has obtained the series expansion for $a(q)a(q^4)$. I want the series expansion of $a^2(q)a^2(q^4)$. I have squared the series and obtained an expression, but that series seems to be very long and not expressible in powers of $q$. But I want a series representation in powers of $q$.

This is in reference to my M.Sc project and I need some guidance at the moment. Kindly help me out or at least point me in the right direction. At present I am not sure how to proceed.

$\endgroup$
7
  • 4
    $\begingroup$ You already have a power series in $q$: $a^2(q)a^2(q^4)$. Expanding it you'll get a complicated expression for the coefficients. Then you can look at those indexed in lmfdb.org, your modular form will be a linear combination of the $f(dz)$ for $f$ of same weight and level $| \frac{l}d$, you will be able to identify from the first few coefficients, if you are lucky there will be no complicated cusp forms. $\endgroup$ – reuns Mar 30 at 8:32
  • $\begingroup$ @reuns I am not sure what you mean. Could you kindly explain ? $\endgroup$ – kk1997 Mar 30 at 10:20
  • $\begingroup$ $a(q)$ is a power series, so is $a(q^4)$, and the product of power series is a power series. $\endgroup$ – reuns Mar 30 at 10:23
  • $\begingroup$ @reuns Is there any way to obtain the series expression without the use of modular forms ? $\endgroup$ – kk1997 Mar 30 at 10:29
  • $\begingroup$ What exactly do you mean by "series seems to be very long and not expressible in powers of q"? Please show us your work and explain what you mean by this statement because the square of a power series in q is obviously a power series in q. Also, where exactly in the linked PDF is the series you refer to located? $\endgroup$ – Somos Mar 30 at 21:55
3
$\begingroup$

The same group of authors work out the answer to your question in theorem 2 of this paper, so you can simply cite their result.

As mentioned in the comments, the answer gets more complicated because it doesn't just involve the divisor function, but also the fourier coefficients of newforms of weight 4 on $\Gamma_0(6)$ and $\Gamma_0(12)$. You can check a proof of the same result in the language of (quasi-)modular forms in the article Evaluation of convolution sums and some remarks on cusp forms of weight 4 and level 12 by B. Ramakrishnan and B. Sahu.

$\endgroup$

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.