Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading an engineering publication and attempting to follow the math and got stuck at this "easy to show but somewhat lengthy" step. The author starts with $$ x[n] = \sum_{k = 1}^{\infty} (-1)^{k} \frac{\sin \left( 2 \pi k M \frac{a}{b} n \right)}{\pi k} $$ where $a,b,M,k,n$ are integers. The claim is that when letting $$ \hat{b} = \frac{b}{\gcd(M,b)} $$ and if $\hat{b}$ is even, then $$ x[n] = \sum_{k=1}^{\hat{b}/2} \frac{(-1)^k}{\hat{b} \tan \left( \pi k / \hat{b} \right)} \sin \left( 2 \pi \frac{k \hat{M} a}{\hat{b}} n \right). $$ I tried splitting into partial sums but could come up with nothing absolutely convergent. I saw several questions that refer to complex analysis when dealing with these types of summation series, which I have no experience in. If that is the route required then I will grab a text. Wanted to know if there was some "easy" or "obvious" thing that I missed. I will note that the author said he used the "antisymmetry" of sine to accomplish this reduction.

share|cite|improve this question
Also, I should have noted that $a$ and $b$ are coprime. – dcdo Jan 19 '13 at 17:32

Not sure what your author is trying to do. The sum is simple to evaluate in terms of the given parameters as follows.

Write the sum as

$$f(y) =\sum_{k=1}^{\infty} (-1)^k \frac{\sin{\pi y k}}{\pi k} $$

Then if we take the derivative of $f$, the sum is easy to evaluate:

$$f'(y) = \sum_{k=1}^{\infty} (-1)^k \cos{\pi y k} $$

Write in terms of a complex exponential:

$$f'(y) = \Re{\sum_{k=1}^{\infty} (-1)^k e^{i \pi y k} } $$

This is simply a geometric series with sum

$$f'(y) = -\Re{\frac{e^{i \pi y}}{1+ e^{i \pi y}}}$$

which can be rewritten as

$$f'(y) = -\frac{1}{2}$$

Integrating with respect to $y$:

$$f(y) = -\frac{y}{2}+ C $$

with $C$ being a constant of integration, which may be found to be zero upon noting that $f(0)=0$. Therefore, using the constants provided in lieu of $y$, we get that the sum is

$$ x[n] = -\pi M \frac{a}{b} n $$

share|cite|improve this answer
The author is trying to derive the number of "tones" to be expected over one period of n. When $k$ goes from 1 to $\infty$ is looks like we should expect an infinite number of spectra, but he is arguing that the summation simplifies to $\hat{b}$ number of tones. Also, I am not sure I can take the derivative and compute in that manner since $x[n]$ is a discrete function and $f(y)$ is a continuous function. Perhaps I can better phrase the question? – dcdo Jan 19 '13 at 17:26
@dcdo: what I did was a trick that allowed me to manipulate the sum so I could evaluate it. $x[n]$ being discrete has no bearing; in fact, the final result still references $n$. As for the author, what is the publication? – Ron Gordon Jan 19 '13 at 17:54
"Noise Spectra of Digital Sine-Generators Using the Table-Lookup Method", by Soenke Mehrgardt, IEEE Transactions on Acoustics, Speech and Signal Processing, Volume 31, Issue 4, August 1983. I downloaded through IEEExplore through my university – dcdo Jan 19 '13 at 18:56
Thanks. I do not have access, but I'll try to see the context somehow. – Ron Gordon Jan 19 '13 at 18:58
I can take the equations and put them into MATLAB for a sanity check. I could not find it with a Google search, so apparently it is only downloadable from behind a paywall. I will report back the results. – dcdo Jan 19 '13 at 19:53

Your Answer


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.