# What is the reason for these jiggles when truncating infinite series?

Plotting the series $$\displaystyle y = \sum_{k} \frac{\sin kx }{k}$$

In the limit it would look like

Taking a finite number of terms, I want to understand what is the reason for the jiggling at the extremes, while there the jiggling in the middle is so small its not noticable.

I truncated the sum to $1,2,3\; \mbox{and}\;4$ terms but cannot deduce much of a reason.

The "jiggling" was noticeable here because the sum is linear in the limit, however, for an expression like $$p(x) = x\prod_k\Big(1-\frac{x^2}{k^2\pi^2}\Big)$$

Does the truncated expression oscillate back and forth the limit?

-
@Qiaochu so for the second part, the answer is no? As $\sin$ is a continouous function being approximated by continuous functions, the Gibbs phenomenon should not occur. –  kuch nahi Aug 6 '11 at 19:02
I don't understand the second question. –  Qiaochu Yuan Aug 6 '11 at 19:05
In the full product, $p(x)=\sin(x)$. The partial products are not periodic, and are unbounded as $x\to\infty$. They get better for a wider range of x as more terms are incorporated into the product, but there are no discontiuities at which to observe anything like a Gibbs phenomenon. –  robjohn Aug 6 '11 at 23:02
It is gratifying to see cited a Wikipedia article whose initial version I created in November 2003---a time when the few sensible people who'd heard of Wikipedia knew that it would never be of any value. –  Michael Hardy Aug 7 '11 at 0:51

As Qiaochu Yuan commented, this is called the Gibbs phenomenon. It happens at discontinuities because of the behavior of the Dirichlet kernel $$D_n(x)=\sum_{k=-n}^{n}e^{ikx}=\frac{\sin((n+\frac{1}{2})x)}{\sin(\frac{x}{2})}$$ When you truncate the Fourier series of a function, $f(x)$, at the $n^{th}$ term, you get back that function convolved with the Dirichlet kernel $$D_n*f(x)=\int_{-\pi}^\pi f(y) D_n(x-y) dy$$ Here are plots of the Dirichlet kernel for $n=3$ and its integral.
Note how the integral goes from $0$ to $1$, but it wiggles because of the wavy nature of the Dirichlet kernel. This wiggle is the root of the Gibbs phenomenon. As $n\to\infty$, the kernel approaches a periodic Dirac delta distribution and its integral has a steeper slope and smaller (but tighter and more numerous) wiggles.
I should mention that since we are using $e^{ikx}$ on $\mathbb{R}/2\pi\mathbb{Z}$ instead of $e^{2\pi ikx}$ on $\mathbb{R}/\mathbb{Z}$, we need to throw in a factor of $\frac{1}{2\pi}$ when convolving with $D_n$. This factor has been incorporated in the plot of the integral. –  robjohn Aug 7 '11 at 11:15