Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

In general, how do I get Mathematica to symbolically expand various infinite series to an arbitrary degree of accuracy? I deal with q-hypergeometric series quite frequently, and specifically want to expand series like the following: $$\sum_{n=0}^{\infty}\frac{(-1)^nq^{6n^2}}{(q^3;q^3)_n(-q;q)_{3n}},$$ where $(a;q)_n$ denotes the standard QPochhammer symbol $\prod_{k=0}^{n-1}(1-aq^k)$. I would like to input a maximal degree, and for Mathematica to output the first few terms of the infinite series up until the specified degree. For example, if I inputted $13$, I would want Mathematica to output (I believe, doing this by hand) $q^6-q^7+q^{10}-q^{11}+q^{12}-q^{13}$. I cannot get the existing series functions to give something as clean as what I want.

One preliminary input I tried was

FunctionExpand[Sum[(-1)^n/QPochammer[q^3,q^3,n],{n,0,10}]],

which returned

1 - 1/(1 - 
  q^3) + 1/((1 - q^3) (1 - q^6)) - 1/((1 - q^3) (1 - q^6) (1 - 
    q^9)) + 1/((1 - q^3) (1 - q^6) (1 - q^9) (1 - q^12)) - 1/((1 - 
    q^3) (1 - q^6) (1 - q^9) (1 - q^12) (1 - q^15)) + 1/((1 - 
    q^3) (1 - q^6) (1 - q^9) (1 - q^12) (1 - q^15) (1 - 
    q^18)) - 1/((1 - q^3) (1 - q^6) (1 - q^9) (1 - q^12) (1 - 
    q^15) (1 - q^18) (1 - q^21)) + 1/((1 - q^3) (1 - q^6) (1 - 
    q^9) (1 - q^12) (1 - q^15) (1 - q^18) (1 - q^21) (1 - 
    q^24)) - 1/((1 - q^3) (1 - q^6) (1 - q^9) (1 - q^12) (1 - 
    q^15) (1 - q^18) (1 - q^21) (1 - q^24) (1 - q^27)) + 1/((1 - 
    q^3) (1 - q^6) (1 - q^9) (1 - q^12) (1 - q^15) (1 - q^18) (1 - 
    q^21) (1 - q^24) (1 - q^27) (1 - q^30)),

which is not the form of output that I want because the sum is not fully expanded as a polynomial in $q$. Also, if I insert an additional QPochhammer[-q,q,Infinity] out front, it leaves the $(-q;q)_{\infty}$ unexpanded because there is no degree specification. (This is why I want to implement the degree specification rather than settling for expanded partial sums.) These are two central issues that I'm not sure how to address.

share|improve this question
    
Post examples of what you've tried and the results vs what you require. More likely to get help that way. –  rasher Apr 8 at 2:39

1 Answer 1

Is this is what you are after?

Series[Sum[(-1)^n/QPochhammer[q^3, q^3, n], {n, 0, 10}], {q, 0, 
   10}] // Normal

(*  1 + q^6 + q^9   *)
share|improve this answer
    
I was about to say the same thing Series[Sum[(-1)^n q^(6 n^2)/( QPochhammer[-q,q,3n]QPochhammer[q^3,q^3,n]),{n,0,20}],{q,0,13}] –  chuy Apr 8 at 14:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.