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

Suppose $f$ is a continuous periodic function and $S_Nf(x) = \sum^N_{n=−N} \hat f(n) e^{inx}$, where $$\hat f(n)= \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx} dx.$$

How can I show that $$\sum_{j=0}^{N-1}S_jf(x)= \int_{-\pi}^{\pi} \frac{\sin^2(\frac{1}{2}Ny)}{\sin^2(\frac{1}{2}y)}f(x-y)dy?$$

Supposedly it can be done using routine trigonometric manipulation, but I don't see it right away. Thank you in advance.

share|cite|improve this question
Are you missing a $2 \pi$ in your result? – Ron Gordon Apr 5 '13 at 9:27
up vote 2 down vote accepted

This relies on switching the order of summation and integration. For one particular value of $S_k$:

$$\begin{align}S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \sum_{n=-k}^k e^{i k (x-x')} \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1)(x-x')} - e^{-i k (x-x')}}{e^{i (x-x')} -1}\\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}}{\sin{\left[\frac{1}{2}(x-x')\right]}} \end{align}$$

Now we want to evaluate a sum over $k$ of $S_k$:

$$\begin{align}\sum_{k=0}^{N-1} S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{1}{\sin{\left[\frac{1}{2}(x-x')\right]}}\sum_{k=0}^{N-1}\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}\end{align}$$


$$\begin{align}\sum_{k=0}^{N-1}\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}&= \Im{\left[\sum_{k=0}^{N-1}e^{i\left[\left(k+\frac{1}{2}\right)(x-x')\right]}\right]} \\ &= \Im{\left[e^{i\left[\frac{1}{2}(x-x')\right]} \sum_{k=0}^{N-1}e^{i\left[k(x-x')\right]}\right]}\\ &=\Im{\left[e^{i\left[\frac{1}{2}(x-x')\right]}\frac{e^{i N (x-x')}-1}{e^{i(x-x')}-1}\right]}\\ &= \Im{\left[e^{i N (x-x')/2} \frac{\sin{[N (x-x')/2]}}{\sin{[(x-x')/2}]} \right]} \\ &= \frac{\sin^2{[N (x-x')/2]}}{\sin{[(x-x')/2}]}\end{align}$$


$$\sum_{k=0}^{N-1} S_k = \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{\sin^2{[N (x-x')/2]}}{\sin^2{[(x-x')/2}]}$$

The stated result follows, save for the factor of $2 \pi$.

share|cite|improve this answer
Thank you Ron. I think there is a bit of redundancy in your solution. It seems that you turn exponential into sin and then sin back into exponential. – Cantor Apr 5 '13 at 10:08
You're welcome. Not sure what you mean, though: I just wanted to keep things simple and sum exponentials so I could use simple formulas for geometric series. – Ron Gordon Apr 5 '13 at 11:12
Oh, I see what you mean; I could have just used the telescoping sum has I not converted to sines. Sure, but I also like to stop at intermediate steps like that to make sure things look OK. – Ron Gordon Apr 5 '13 at 13:37
There are no telescopic sums. But, one could write \begin{split} S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \sum_{n=-k}^k e^{i k (x-x')} \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1)(x-x')} - e^{-i k (x-x')}}{e^{i (x-x')} -1}\\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1/2)(x-x')} - e^{-i (k+1/2) (x-x')}}{e^{i (x-x')/2} -e^{-i (x-x')/2}}, \end{split} where in the last step we multiplied and divided by $e^{-i (x-x')/2}$ to make exponentials symmetric. [continued in the next comment] – Cantor Apr 5 '13 at 18:25
Then sum over $k$ noting that if $y=x-x'$: $$\sum_{k=0}^{N-1} e^{i (k+1/2)y} - e^{-i (k+1/2) y} = \frac{e^{iNy}-1}{e^{iy/2}-e^{-iy/2}} - \frac{e^{-iNy}-1}{-e^{iy/2}+e^{-iy/2}} = \frac{e^{iNy}-2+e^{-iNy}}{e^{iy/2}-e^{-iy/2}} $$ Then we put the expression back into the integral and convert exponentials into sin functions. – Cantor Apr 5 '13 at 18:26

The following identity is known as the Fejér kernel and is itself a sum of Dirichlet kernels (for derivations see this thread Why is the Fejér Kernel always non-negative?):

$$\frac{1}{N} \frac{\sin^2(\frac{1}{2}Ny)}{\sin^2(\frac{1}{2}y)}=\frac{1}{N}\sum^{N-1}_{j=0}\sum_{n=-j}^j e^{i n y}$$

Stick that in and change variable to $u=x-y$: \begin{aligned} \int_{-\pi}^{\pi} \frac{\sin^2(\frac{1}{2}Ny)}{\sin^2(\frac{1}{2}y)}f(x-y)dy &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\sum^{N-1}_{j=0}\sum_{n=-j}^n e^{i n y}f(x-y)dy\\ &=\sum^{N-1}_{n=0}\sum_{j=-n}^ne^{i n x}\frac{1}{2\pi}\int_{-\pi+x}^{\pi+x} e^{-i n u}f(u)du\\ &=\sum^{N-1}_{j=0}\sum_{n=-j}^je^{i n x}\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i n u}f(u)du\\ &=\sum^{N-1}_{j=0}\sum_{n=-j}^je^{i n x}\hat f(n)\\ &=\sum^{N-1}_{j=0}S_{j}f(x) \end{aligned}

where the $2\pi$ periodicity of the integrand was used to drop $x$ from the limits.

share|cite|improve this answer

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.