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.

Let $e(k) = \exp \left(\frac{2 \pi i k}{N}\right)$ be a root of unity.

I wanted to numerically verify the Cauchy residue formula in Mathematica.
$$ \lim_{N \to \infty}\frac{1}{N}\sum_{k=0}^{N-1} \frac{e(k) }{e(k) - a} = \frac{1}{2\pi i} \oint_{|z|=1} \frac{dz}{z - a} = \mathbf{1}( |a| < 1)$$

This computes the winding number of the curve $|z|=1$ (counter-clockwise) around $a \in \mathbb{C}$.

Can this Riemann sum be evaluated exactly? I would like to know the leading-order correction of the Riemann sum to the integral .

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

Denote the lhs as $f_N(a)$. It can be regarded as a rational function of complex variable $a$. It's straightforward to check that $f_N(a e(i))=f_N(a)$, $i=0,\ldots,N-1$. Which means that $f_N$ is a rational function of $a^N$. From the other hand $\lim_{a\to\infty}f_N(a)=0\;$. This leaves the only possibility $$ f_N(a)=\frac c{\prod_{i=0}^{N-1}(e(i)-a)}=\frac c{1-a^N}. $$ Plugging $a=0$ gives $c=1$ so $f_N(a)=\frac1{1-a^N}$. Thus the difference between the sum and the integral is $f_N(a)-1=\frac {a^N}{1-a^N}$.

share|improve this answer
    
This is more clever than the way I was going. Very nice. (+1) –  user26872 Jul 9 '12 at 5:25
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.