2
$\begingroup$

Let $$f_b(x)=\sum\limits_{a=1 , (a,b)=1}^{b}\frac{1}{1-e^{2\pi i \frac{a}{b}}x}$$ For example:

$$f_6(x) = \frac{1}{1-e^{2\pi i \frac{1}{6}}x}+\frac{1}{1-e^{2\pi i \frac{5}{6}}x}$$

I'm wondering if there is a simple closed for for my function. For instance, if we get rid of the $(a,b)=1$ condition, we see that $$\sum\limits_{a=1 }^{b}\frac{1}{1-e^{2\pi i \frac{a}{b}}x}=\frac{b}{1-x^b}$$ Which is very nice. And when we add in the coprime condition, we are reducing the case to primitive roots of unity.

$\endgroup$
1
$\begingroup$

Define $$g_b(x) = \sum\limits_{a=1 }^{b}\frac{1}{1-e^{2\pi i \frac{a}{b}}x}. $$ Then it's clear that $$ g_b(x) = \sum_{d\mid b} f_b(x) $$ (where the sum is over all positive integers $d$ dividing $b$). By Mobius inversion, we conclude that $$ f_b(x) = \sum_{d\mid b} \mu(b/d) g_d(x) = \sum_{d\mid b} \mu(b/d) \frac d{1-x^d}. $$

$\endgroup$

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.