Example 1: We consider another related to the Heaviside function that is known as the signum function

Plot of the signum function on interval [-2, 2] using Mathematica standard Sign command. Plot[Sign[x], {x, -2.5, 2.5}, PlotStyle -> {Thickness[0.01], Purple}]

In order to expand the signum function into the Fourier series, we need to choose a value ℓ and expand this function in odd way periodically from the interval (−&wll;, ℓ).

We expect that the Fourier series will give a periodic expansion shown in the figure. g[x_] = Piecewise[{{1, 0 < x < 1}, {-1, 1 < x < 2}, {-1, -1 < x < 0}, {1, -2 < x < -1}, {1, 2 < x < 3}, {-1, 3 < x < 4}, {1, 4 < x < 5}}]; sign = Plot[g[x], {x, -2, 5}, PlotStyle -> {Thickness[0.01], Red}, AspectRatio -> 1/3]; point = Graphics[{Red, {Disk[{-1, 0}, 0.05], Disk[{0, 0}, 0.05], Disk[{1, 0}, 0.05], Disk[{2, 0}, 0.05], Disk[{3, 0}, 0.05], Disk[{4, 0}, 0.05]}}]; Show[point, sign]

However, if you want remove the point valuse on the horizontal axis, you can include plot option Exclusions → None.

We plot oddsign function without dots: g[x_] = Piecewise[{{1, 0 < x < 1}, {-1, 1 < x < 2}, {-1, -1 < x < 0}, {1, -2 < x < -1}, {1, 2 < x < 3}, {-1, 3 < x < 4}, {1, 4 < x < 5}}]; Plot[g[x], {x, -2, 5}, PlotStyle -> {Thickness[0.01],Purple}, AspectRatio -> 1/3, Exclusions -> None]

As a result, we obtain the periodic version of the signum function that we denote as oddsign; its sine Fourier series is

Since this function is odd, there is no difference between its Fourier series and Fourier sine series; that is why we plot it in blue. Next, we calculate its Fourier coefficients: So all Fourier cosine coefficients are zeroes, and Fourier sine coefficients do not depend on ℓ, so we have You can use instead the build-in Mathematica commands:
FourierSinCoefficient[Sign[x], x, n, FourierParameters -> {1,(π)/L}]
FourierCosCoefficient[Sign[x], x, n, FourierParameters -> {1,(π)/L}]

For ℓ = π, we get

We can also expand the signum function into complex Fourier series:

because Therefore, the corresponding sine Fourier sine becomess Of course, this series is exactly the same as (1.3). Our next step is to truncate the series (1.3) or (1.4) and show how the finite Fourier series converges to the signum function. First, we use animation:

We plot partial sums with n = 10, 100, and 1000 terms.

Partial Fourier sum with 10 tems.		Partial Fourier sum with 100 tems.		Partial Fourier sum with 1000 tems.

Its Fourier series exhibits the Gibbs phenomenon at points of discontinuity x = (2n+1)ℓ, n∈ℤ, in amount of 2w ≈ 0.1789797, where w = 0.5707963267948966 is the Wilbraham number (see section). Recall that the value of the overshoot or undershoot of the partial Fourier sum is equal to h w, where h is the jump of discontinuoity (which 2 in our case).

It is known that restoring a function from its Fourier coefficients is an ill-posed problem. The Gibbs phenomenon clearly indicates that it is impossible to recover the function in a neighbohood of the point of discontinuity. A wealth of information hidden in the Fourier series can be recovered upon utilization of a summability method. For practical applications, its regularization, can be obtained with Cesàro's summation:

Note that Cesàro's regularization requires only three additional arithmetic operations on each step compared with the standard summation: one multiplication by the difference (1−k/n), one division (k/n), and one subtraction. This usually does not increase the time of computation.

Finally, we demonstrate animation for Cesàro's approximation

To observe convergence of Cesàro's sums, we plot three of them with m = 10, 100, and 1000 terms.

Partial Cesàro's sum with 10 tems.		Partial Cesàro's sum with 100 tems.		Partial Cesàro's sum with 1000 tems.

⁎ ✱ ✲ ✳ ✺ ✻ ✼ ✽ ❋ ========================================

Now we demonstrate how Fourier approximation depends on the number of terms (graphs present 3 terms, 7 terms, 10 terms, and 20 terms, but you can choose any numbers instead). Even use Manipulate to interactively see how the series works: Mathematica even knows how to find the infinite sum: then the plot is simple: You can write this partial sum as a Mathematica function of two variables: Plot a few partial sums    ■

Example 2: Let us find Fourier coefficients for the delta-function of Dirac on an interval of length 2ℓ containing the origin: Therefore, we get a formal Fourier series for the delta-function Since the delta-function is a distribution but not a regular function, the series above cannot truly converge (its terms don’t approach zero). But we can determine its partial sums and graph the latter. We naturally denote such partial sums by Using Euler's formula we can rewrite the finite sum as This is a geometric progression that starts from \( e^{-N{\bf j}\theta} \) and ends at \( e^{N{\bf j}\theta} . \) We have powers of the same exponential factor \( e^{{\bf j}\theta} . \) The sum of a geometric series is known: This partial sum approaches the Dirac delta-function in weak sense, so for any smooth probe function f, we expect So we obtain a remarkable result that for any smooth function, when we expand it into Fourier series the sum of its cosine coefficients approaches \( f(0) = \frac{a_0}{2} + a_1 + a_2 + \cdots . \)

Using Mathematica, we plot some partial sums

 

Now we introduce a sequence of square pulses that approaches the Dirac delta-function: The graph of each δ_h(x) is a tall thin rectangle: height 1/h, base h, and area=1. When h approaches zero, δ_h(x) is squeezed into a very thin interval. The tall rectangle approaches (weakly) the delta function δ(x). The Fourier coefficients of δ_h(x) are Hence, the Fourier cosine coefficient is expressed through the normalized sinc function: where