Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

How do i get the sum of such a sequence:

$1 + x^{-1} + x^{-3} + x^{-6} + ...$

where the exponents are actually sum of arithmetic progression. i.e.

$x^{-0} + x^{-(0 + 1)} + x^{-(0 + 1 + 2)} + x^{-(0 + 1 + 2 + 3)} + ...$

which can also be expressed as

$\sum_{i=0}^{\infty} x^{-\frac{i(i + 1)}{2}}$

?

Thanks.

share|improve this question
1  
Mathematica says: $\frac{1}{2} x^{1/8} \big( \frac{2}{x^{1/8}} + \text{EllipticTheta}\big[2, 0, \frac{1}{\sqrt{x}}\big]\big)$. –  Micah Milinovich Nov 1 '11 at 18:36
    
If there exists a formula (which sounds unlikely) for that one, then it could be modified to work for any arithmetic progression since it would just be a linear translation of the exponent, which would mean applying the function to $x^b$ and multiplying by $x^a$. Now, is there a closed-form formula? I doubt it. –  Will Sawin Nov 1 '11 at 19:12
    
oh wait that's not true sorry. –  Will Sawin Nov 1 '11 at 19:16
1  
I would bet that the function is not elementary but it'll take some effort to prove it. Are you really curious about this stuff or it is just a random question? –  fedja Nov 1 '11 at 23:48
1  
what markov chain gave you a theta function as solution?? –  john mangual Nov 2 '11 at 15:43
show 1 more comment

2 Answers

I think you can get it from the Jacobi triple product identity \[ \prod_{m=1}^\infty (1-x^{2m})(1-x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) = \sum_{n=-\infty}^\infty x^{n^2}y^{2n} \] We can set $x = q^{1/2}, y = q^{1/2}$. \[ \prod_{m=1}^\infty (1-q^{m})(1-q^{m+\frac{1}{2}})(1+q^{m-\frac{3}{2}}) = \sum_{n=-\infty}^\infty q^{\frac{n(n+1)}{2}} \] This is close to what you want.

Triple product identity and the theory of partition is still actively studied in many contexts.

share|improve this answer
    
I'm going to go out on a limb here and make a guess that expressing the infinite sum as an infinite product is not what OP wanted (but is as good as OP is going to get). –  Gerry Myerson Nov 2 '11 at 4:40
add comment

The Jacobi Theta function $\text{JacobiTheta2}(z,q)$ (in Maple's notation) is $\sum _{k=0}^{\infty }2 \cos \left(\left( 2k+1 \right) z \right) { q}^{\left( 2 k+1 \right) ^{2}/4}$. So what you have is $\text{JacobiTheta2}(0,\sqrt{1/x}) x^{1/8}/2 $ (for $|x| >1$).

share|improve this answer
add comment

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.