Monday, 25 March 2013

Gibbs Phenomenon

One shortcoming of Fourier series today known as the Gibbs phenomenon was first observed by H. Wilbraham in 1848 and then analyzed in detail by Josiah W. Gibbs (1839-1903). We will start with an example.

Example. Consider the function

\begin{displaymath}f(x) = \left\{ \begin{array}{lll}
1 & 0 \leq x \leq \pi \\
-1 & -\pi \leq x < 0 .
\end{array} \right.\end{displaymath}

Since this function is odd, we have an = 0, for $n \geq 0$. A direct calculation gives

\begin{displaymath}b_n = \frac{2}{\pi} \frac{(1 - (-1)^n)}{n},\;\;\mbox{for $n \geq 1$}.\end{displaymath}

for $n \geq 1$. The Fourier partial sums of f(x) are

\begin{displaymath}f_{2n-1}(x) = \frac{4}{\pi}\left(\sin(x) + \frac{\sin(3x)}{3} +\ldots+

The main Theorem implies that this sequence converges to f(x) except at the point x0 = 0, which a point of discontinuity of f(x). Gibbs got interested to the behavior of the sequence of Fourier partial sums around this point.

Looking at the graphs of the partial sums, we see that a strange phenomenon is happening. Indeed, when x is close to the point 0, the graphs present a bump. Let us do some calculations to justify this phenomenon.
Consider the second derivative of f2n-1, which will help us find the maximum points.

\begin{displaymath}f'_{2n-1}(x) = \frac{4}{\pi}(\cos(x) + \cos(3x)+\ldots+\cos((2n+1)x)).\end{displaymath}

Using trigonometric identities, we get

\begin{displaymath}\pi \sin(x) f'_{2n-1}(x) = 2\sin(2nx).\end{displaymath}

So the critical points of f2n-1 are

\begin{displaymath}2nx = \pm \pi, \pm 2\pi,....,\pm (2n-1)\pi.\end{displaymath}

Since the functions are odd, we will only focus on the behavior to the right of 0. The closest critical point to the right of 0 is $\displaystyle \frac{\pi}{2n}$. Hence

\begin{displaymath}f_{2n-1}\left(\frac{\pi}{2n}\right) = \frac{4}{\pi}\left(\sin...
...}{3} +\ldots+ \frac{\sin(\frac{(2n-1)\pi}{2n})}{(2n-1)}\right).\end{displaymath}

In order to find the asymptotic behavior of the this sequence, when n is large, we will use the Riemann sums. Indeed, consider the function $\displaystyle F(x) = \frac{\sin(x)}{x}$ on the interval $[0,\pi]$, and the partition$\left\{{\displaystyle \frac{k\pi}{n}}, \;\;k \in [1,n]\right\}$ of $[0,\pi]$. So the Riemann sums

\begin{displaymath}\frac{\pi}{n}\left(\frac{\sin\left(\displaystyle \frac{\pi}{2...
{\displaystyle \frac{(2n-1)\pi}{2n}}\right)\end{displaymath}

converges to $\displaystyle \int_{0}^{\pi} F(x)dx$. Esay calculations show that these sums are equal to

\begin{displaymath}\frac{\pi}{2}f_{2n-1}\left(\displaystyle \frac{\pi}{2n}\right).\end{displaymath}


\begin{displaymath}\lim_{n \rightarrow \infty} f_{2n-1}\left(\frac{\pi}{2n}\right) = \frac{2}{\pi}\int_{0}^{\pi}\frac{\sin(x)}{x}dx.\end{displaymath}

Using Taylor polynomials of $\sin(x)$ at 0, we get

\begin{displaymath}\frac{2}{\pi}\int_{0}^{\pi}\frac{\sin(x)}{x}dx = 2 - \frac{\pi^2}{9} + \frac{\pi^4}{300} - \frac{\pi^6}{17640}+\ldots\end{displaymath}

i.e. up to two decimals, we have

\begin{displaymath}\frac{2}{\pi}\int_{0}^{\pi}\frac{\sin(x)}{x}dx = 1,18 .\end{displaymath}

These bumps seen around 0 are behaving like a wave with a height equal to 0,18. This is not the case only for this function. Indeed, Gibbs showed that if f(x) is piecewise smooth on $[-\pi,\pi]$, and x0 is a point of discontinuity, then the Fourier partial sums will exhibit the same behavior, with the bump's height almost equal to

\begin{displaymath}0,09 \Big(f(x_0+) - f(x_0-)\Big).\end{displaymath}

To smooth this phenomenon, we introduce a new concept called the $\sigma$-approximation. Indeed, let f(x) be a function piecewise smooth on $[-\pi,\pi]$ and fN(x) its Fourier partial sums. Set

\begin{displaymath}S_N(x) = \frac{a_0}{2} + \sum_{n=1}^{n = N} (a_n\sigma_n \cos(nx) + b_n\sigma_n \sin(nx)),\end{displaymath}


\begin{displaymath}\sigma_n = \frac{\sin\left(\displaystyle \frac{n\pi}{N}\right)}{\displaystyle \frac{n\pi}{N}}, \;\;\; n = 1,2,\ldots,N,\end{displaymath}

which are called the $\sigma$-factors. To see that the sequence of sums $\{S_N(x)\}$ better approximate the function f(x) than the Fourier partial sums $\{f_N(x)\}$, we use the following result:

Theorem. We have

\begin{displaymath}S_N(x) = \displaystyle \int_{\displaystyle -\pi/N}^{\displaystyle \pi/N} f_N(x+t)dt.\end{displaymath}

Proof. We have

...ft(\displaystyle \frac{n\pi}{N}\right)}{n} =

Similarly, we have

\begin{displaymath}\frac{N}{2\pi}\int_{-\frac{\pi}{N}}^{\frac{\pi}{N}}\sin(n(x+t))dt = \sin(nx)\sigma_n.\end{displaymath}


... \sum_{n=1}^{n=N} (a_n\cos(nx)\sigma_n + b_n

which yields the conclusion above.

Using the above conclusion, we can easily see that indeed the sums $\{S_N(x)\}$ approximate the function f(x) in a very smooth way. On the graphs, we can see that the Gibbs phenomenon has faded away.

Example. The picture

shows how the $\sigma$-approximation helps fade away the Gibbs phenomenon for the function

\begin{displaymath}f(x) = \left\{ \begin{array}{lll}
1 & 0 \leq x \leq \pi \\
-1 & -\pi \leq x < 0
\end{array} \right.\end{displaymath}

Note that in this case, we have

\begin{displaymath}f_N(x) = \frac{4}{\pi}\left(\sin(x) + \frac{\sin(3x)}{3} + ....+\frac{\sin(N-1)x}{N-1}\right),\end{displaymath}


\begin{displaymath}S_N(x) = \frac{4}{\pi}\left[\sin(x)\sin\left(\frac{\pi}{N}\ri...
...N-1)^2}\sin((N-1)x) \sin\left(\frac{(N-1)\pi}{N}\right)\right].\end{displaymath}
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