Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my work I have met the function on the unit circle whose Fourier coefficients are $$ c_n=\frac{1}{|n|}\prod (d_k+1) $$ if $n=\pm\prod p_k^{d_k}$ is the decomposition of the integer $n$ into the product of prime numbers. The formal series of the function is $$ \sum_{n\in\mathbb Z}c_nz^n. $$ This function could have already appeared and I would much appreciate any references about it. (I am interested in properties of this function, but now it is not yet easy to say what I really need.)

share|improve this question
    
The function $d(n)=\prod_k (d_k+1)$ is the number of positive divisors of $n$. Also sometimes written $\tau(n)$. So your $c_n=\frac{d(|n|)}{|n|}$ –  Thomas Andrews Nov 5 '12 at 21:48
    
Thanks, I know this. I am interested in the function given by the series. –  Limanac Nov 5 '12 at 21:54
    
Ah, when you said "I have met the function on the unit circle," I assumed that meant you have a known function, and were trying to understand the coefficients. –  Thomas Andrews Nov 5 '12 at 21:58
    
Sorry if my words admit ambiguity. –  Limanac Nov 5 '12 at 22:01
    
your function is closely related to the logarithm of the modular form $\Delta$ (the discriminant) –  user8268 Nov 5 '12 at 22:15

1 Answer 1

up vote 0 down vote accepted

Doing a Google search for "sum d(n) x^n" gives http://en.wikipedia.org/wiki/Divisor_function as one of the first answers. Looking in there, I find "A Lambert series involving the divisor function is: $\sum_{n=1}^{\infty} q^n \sigma_a(n) = \sum_{n=1}^{\infty} \frac{n^a q^n}{1-q^n}$ where $\sigma_a(n) = \sum_{d|n} d^a$."

Setting $a=0$, $\sum_{n=1}^{\infty} q^n d(n) = \sum_{n=1}^{\infty} \frac{q^n}{1-q^n}$.

Your series, calling it $f$, is $f(z) = \sum_{n=1}^{\infty} z^n d(n)/n$. Differentiating, $f'(z) = \sum_{n=1}^{\infty} z^{n-1} d(n) =(1/z)\sum_{n=1}^{\infty} z^n d(n) = (1/z)\sum_{n=1}^{\infty} \frac{z^n}{1-z^n} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{1-z^n} $. To get $f(z)$ we integrate this term by term, using $f(z) = \int_0^z f'(y) dy$.

Letting $x = y^n$, $\int_0^z \frac{y^{n-1}}{1-y^n} dy =(1/n) \int_0^{z^n} \frac{dx}{1-x} =- \frac{\ln{1-y}}{n}]_0^{z^n} =- \ln(1-z^n)/n $ so $f(z) = -\sum_{n=1}^{\infty} \frac{\ln{1-z^n}}{n} = - \ln\prod_{n=1}^{\infty} (1-z^n)^{1/n} $.

I'm not sure where to go from here (assuming no errors, which has $p < .7$ in my estimation), so I'll leave it at this.

share|improve this answer
    
To get your formula, one can use the relation $f(z)=\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{z^{kn}}{kn}$. –  Limanac Nov 8 '12 at 17:56
    
For my question, one should take the real part. Thanks! –  Limanac Nov 8 '12 at 17:57

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.