Preface

Since the Gibbs phenomenon predicts overshoots and undershoots at the points of discontinuity that are proportional to the size of the jump, this section shows how this coefficient of proportionality, called Wilbraham's constant, can be numerically determined. Its derivation comes up in two cases: either when the point of discontinuity is within the interval or when it is at the end points of the main interval.

The idea behind its derivation is very similar in both cases. First, we consider the truncated finite Fourier series of the function responsible for the jump of discontinuity, differentiate it, and then determine critical points where the derivative is zero. Next substitution of these critical points into the finite Fourier series yields the finite Riemann sum for the sine integral \( \mbox{Si}(x) = \int_0^x \frac{\sin t}{t}\,{\text d}t \) in both cases. Taking the limit gives the numerical value of the undershoot/overshoot.

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Introduction to Linear Algebra with Mathematica

Glossary

Derivation of Wilbraham's constant

A periodic function of period 2ℓ from the Dirichlet theorem (that guarantees the pointwise convergence of the Fourier series to the given function) has on the interval [-ℓ,ℓ] a discrete number of discontinuities of finite jumps and the given interval [-ℓ,ℓ] can be divided into a finite number of subinterval so that on each of them the function is either monotonically increasing or decreasing. Such functions are called functions of finite (or bounded) variations. Discontinuities of such functions can occur in two cases: either within the interval or at the end points when the Fourier series periodically extends the given function. Since the Gibbs phenomenon can be observed only at the points of discontinuity, it is sufficient to consider only two cases when the function has a jump inside the interval (-ℓ,ℓ) or when the values of the function at end points are distinct.

Discontinuity in the middle of the interval

Without any loss of generality, we can assume that the function under consideration has only one finite jump within the interval (-ℓ,ℓ) because any function of finite variation can be represented as a finite sum of such functions. Let x0 denote an x-value where such a jump occurs; it is at most a matter of translation of coordinate system and perhaps a reflection to assume that x0 = 0. Let us assume that the left and right limits of f(x) at the point of discontinuity are \( -a = f(0-0) = \lim_{\varepsilon \to 0} \,f(-\varepsilon ) , \quad b = f(0+0) = \lim_{\varepsilon \to 0} \,f(\varepsilon ) \) (where limits were calculated for positive ε).

The derivation of Wilbraham's constant for a function having only one finite jump of value a+b at the origin can be broken down into the following steps.

  1. Construct a step function that has exactly the same jump of discontinuity as the given function at the origin.
  2. Subtract the step function found in the previous step from the given function to obtain the continuous function that we denote by g(x).
  3. Expand the step function into the Fourier series. We denote by \( \displaystyle S_N (x) = \frac{a_0}{2} + \sum_{k=1}^N a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell} \) its partial finite Fourier sum with N+1 terms. Since g(x) is a periodic continuous function on the interval (-ℓ,ℓ), its Fourier series does not exhibit the Gibbs phenomenon.
  4. Determine critical points of the finite Fourier series SN(x) by equating its derivative to zero and solving the corresponding transcendent equation.
  5. Identify the critical point that is closest to the origin (point of discontinuity) and where the function SN(x) attains maximum or minimum.
  6. Evaluate the finite Fourier sum SN(x) at the above critical point.
  7. Show that this finite Fourier series at the critical point is actually a Riemann sum for sine integral    \( \mbox{Si}(x) = \int_0^x \frac{\sin t}{t}\,{\text d}t \) at point x = π.
  8. Find the limit when \( N\mapsto \infty \) to determine the numerical value of Wilbraham's constant.
We can observe such a function through the following step function:
\[ s(x) = \begin{cases} b , & \ \mbox{ for } \ x > 0 , \\ (b-a)/2 , & \ \mbox{ for } \ x=0 , \\ -a , & \ \mbox{ for } \ x < 0 . \end{cases} \]
This step function has jump of a+b at the origin and it can be represented as a linear combination of the Heaviside functions:
\[ s(x) = b\,H(t) - a\, H(-t) = \frac{a+b}{2} \left[ H(t) - H(-t) \right] + \frac{b-a}{2} \left[ H(t) + H(-t) \right] = \frac{a+b}{2} \left[ H(t) - H(-t) \right] + \frac{b-a}{2} \]
because the sum of two Heaviside functions is identically 1:
\[ H(t) + H(-t) = 1 . \]
Since the Heaviside function
\[ H(t) = \begin{cases} 1, & \ \mbox{ for } \ t > 0 , \\ 1/2, & \ \mbox{ for } \ t =0 , \\ 0, & \ \mbox{ for } \ t < 0 , \end{cases} \]
has jump of discontinuity of unit length at the origin, the difference of them, s(x) = H(x) - H(-x), will have a symmetric jump of value 2. Although an arbitrary discontinuous function may have asymmetric jump of value (a+b) at the origin, we can always extract from it the function \( \frac{a+b}{2} \left[ H(t) - H(-t) \right] + \frac{b-a}{2} \) so the rest will be a continuous function. Therefore, it is sufficient to consider the Fourier series for the difference of two Heaviside functions (with symmetric jump), which we denote by S:
\[ S(t) \equiv H(t) - H(-t) = \frac{4}{\pi}\,\sum_{n\ge 1} \frac{\sin \frac{(2n-1)\pi t}{\ell}}{2n-1} = \begin{cases} 1, & \ \mbox{ for } \ t > 0 , \\ 0, & \ \mbox{ for } \ t =0 , \\ -1, & \ \mbox{ for } \ t < 0 . \end{cases} \]
Then we plot the step function \( S(t) = H(t) - H(-t) \) along with its 20-term Fourier approximation for ℓ = 2:
    Fourier approximation to unit step function.
  
S[t_] = UnitStep[t] - UnitStep[-t]
Simplify[Integrate[Sin[k*Pi*t/L], {t, 0, L}]/L - Integrate[Sin[k*Pi*t/L], {t, -L, 0}]/L]
S20[t_] = 4/Pi*Sum[Sin[(2*k-1)*Pi*t/2]/(2*k-1), {k,1,20}]
Plot[{S[t], S20[t]}, {t, -2.5, 2.5}, PlotStyle -> {{Thick, Black}. {Thick, Orange}}, Ticks -> {{-2, -1, 1, 2}, {1.1789797, -1.1789797}}]

To understand where the finite Fourier sum attains maximum or minimum, we build its partial sum and differentiate it.

\[ S_N (t) = \frac{4}{\pi}\,\sum_{k= 1}^N \frac{\sin \frac{(2k-1)\pi t}{\ell}}{2k-1} \qquad\Longrightarrow \qquad S'_N (t) = \frac{\text d}{{\text d}t}\, S_N (t) = \frac{4}{\ell} \,\sum_{k= 1}^N \cos \frac{(2k-1)\pi t}{\ell} . \]
Using the Euler formula and temporal notation ω = πt/ℓ, we find the explicit formula for the partial sum:
\begin{align*} S'_N (t) &= \frac{4}{\ell} \,\sum_{k= 1}^N \cos (2k-1)\omega \\ &= \frac{4}{\ell} \,\Re \,\sum_{k= 1}^N e^{{\bf j}(2k-1)\omega} = \frac{1}{\sin (\omega )}\,\cos (N\omega )\,\sin (N\omega ) . \end{align*}
Mathematica confirms:
Sum[Cos[(2*k - 1)*omega], {k, 1, n}]
Cos[n omega] Csc[omega] Sin[n omega]
Therefore, the partial Fourier sum SN(x) has critical points at
\[ t_j = \frac{\ell}{2N} \left( 1 + 2j \right) \qquad \mbox{and} \qquad t_i = \frac{\ell}{N}\,i , \qquad j=0, \ i,j=\pm 1, \pm 2, \ldots . \]
Next, we evaluate (of course, with the aid of Mathematica) the second derivative at these critical points:
\[ S''_N (t_j ) = \frac{N}{\sin \left( \frac{\pi (1+2j)}{2N} \right)} , \qquad j=0, \pm 1, \pm 2, \ldots , \]
Simplify[SF2[Pi*(1 + 2*j)/2/n], Assumptions -> Element[j, Integers]]
n Csc[(\[Pi] + 2 j \[Pi])/(2 n)]
and
\[ S''_N (t_i ) = -\frac{N}{\sin \left( \frac{\pi \,i}{N} \right)} , \qquad i = \pm 1, \pm 2, \ldots \pm(N-1). \]
Simplify[SF2[Pi*i/n], Assumptions -> Element[i, Integers]]
-n Csc[(i \[Pi])/n]
So at these points the partial Fourier sums attain alternating maximum and minimum. The first closest point to the origin is t0 = ℓ/2/N. Hence at this point, the partial Fourier sum becomes
\[ S_N \left( \frac{\ell}{2N} \right) = \frac{4}{\pi}\,\sum_{k= 1}^N \frac{\sin \frac{(2k-1)\pi}{2N}}{2k-1} = \frac{2}{\pi} \,\sum_{k= 1}^N \frac{\sin \frac{(2k-1)\pi}{2N}}{(2k-1)\pi /(2N)} \, \frac{\pi}{N} . \]
This last sum is a Riemann sum for the integral
\[ \frac{2}{\pi} \int_0^{\pi} \frac{\sin t}{t}\,{\text d}t = 1.1789797444721672702\ldots , \]
because Mathematica has a dedicated command SinIntegral for such integral:
N[SinIntegral[Pi]*2/Pi, 20]
Since this number corresponds to the total variation of the partial sum, Wilbraham's constant will be half of the latter:
\[ w = \left( \frac{2}{\pi} \,\mbox{Si}(\pi ) - 1 \right) \frac{1}{2} = \left( \frac{2}{\pi} \int_0^{\pi} \frac{\sin t}{t}\,{\text d}t - 1 \right) \frac{1}{2} = 0.08948987223608362\ldots , \]
w = N[(SinIntegral[Pi]/Pi*2 - 1)/2]
where Si(x) = \( \int_0^x \frac{\sin t}{t}\,{\text d}t \) is the sine integral (special function).
N[SinIntegral[Pi]/Pi, 20]
To confirm that the maximum oscillation occurs at the point above, we can observe another critical point. At this point, the partial sum will approach a smaller value than 2w. In addition, the second ripple will be about 1.59, which is seen from the above picture.

Discontinuity at the end points

Suppose that the given function f(x) on the interval [-ℓ, ℓ] has not equal values at the end points; suppose that the values of the function at end poinst are f(-ℓ) = -a and f(ℓ) = b, where a ≠ b.
  1. Identify a linear function l(x) that has the same values at the end points as the given function.
  2. Add the linear function found in the previous step to the given function to obtain the continuous function that we denote by φ(x)
  3. Expand the linear function into the Fourier series. We denote by SN(x) its partial finite Fourier sum with N terms. Since φ(x) is continuous at the end points of the interval (-ℓ,ℓ), its Fourier series does not exhibit the Gibbs phenomenon.
  4. Determine critical points of the finite Fourier series SN(x) by equating its derivative to zero and solving the corresponding transcendent equation.
  5. Identify the critical point that is closest to the end point and where the function SN(x) attains maximum or minimum.
  6. Evaluate the finite sum SN(x) at the above critical point.
  7. Show that this finite Fourier series at the critical point is actually a Riemann sum for sine integral.
  8. Find the limit when \( N\mapsto \infty \) to determine the numerical value of Wilbraham's constant.
We can observe such a fuction if we subtract from f(x) the linear function
\[ l(x) = \frac{b+a}{2}\, x + \frac{b-a}{2} , \]
the result will be a continuous function at end points. Therefore, we can consider only a linear function
\[ x = - \frac{2\ell}{\pi} \,\sum_{k\ge 1} \frac{(-1)^k}{k} \, \sin \left( \frac{k\pi x}{\ell} \right) . \]
Its finite partial sum
\[ S_N (x) = \frac{2\ell}{\pi} \,\sum_{k= 1}^N \frac{(-1)^{k+1}}{k} \, \sin \left( \frac{k\pi x}{\ell} \right) \]
has critical points where its derivative is zero:
\[ S'_N (x) = 2 \,\sum_{k= 1}^N (-1)^{k+1} \cos \left( \frac{k\pi x}{\ell} \right) = - \csc \left( \frac{x\pi}{2\ell} \right) \cos \left( \frac{\left( N+1 \right) \pi \left(\ell + x \right)}{2\ell} \right) \sin \left( \frac{N\pi \left( \ell + x \right)}{2\ell} \right) . \]
Simplify[Sum[(-1)^(k + 1)*Cos[k*omega], {k, 1, n}], Assumptions -> Element[n, Integers]]
1/2 (1 + Cos[(1 + n) (omega + \[Pi])] + Sin[(1 + n) (omega + \[Pi])] Tan[omega/2])
Since the cosine is positive on the given interval, cosθ > 0 for -π/2 < θ < π/2, we need to equate the other two functions to zero and obtain two sequences of zeroes:
\[ \begin{split} x &= \frac{\ell (2j+1)}{N+1} - \ell \qquad \mbox{where} \quad \cos \left( \frac{\left( N+1 \right) \pi \left(\ell + x \right)}{2\ell} \right) =0 , \\ x &= \frac{2i\ell}{N} - \ell \qquad \mbox{where} \quad \sin \left( \frac{N\pi \left( \ell + x \right)}{2\ell} \right) =0. \end{split} \]
Evaluating the partial sum at the first critical point, we obtain
\[ S_N \left( \frac{\ell}{N+1} - \ell \right) = \frac{2\ell}{\pi} \,\sum_{k=1}^N \frac{(-1)^{k+1}}{k} \, \sin \frac{k\pi}{\ell} \left( \frac{\ell}{N+1} - \ell \right) = - \frac{2\ell}{\pi} \,\sum_{k=1}^N \frac{1}{k}\, \sin \left( \frac{k\pi}{N+1} \right) \]
because cos kπ = (-1)k. The negative of the last sum is actually the Riemann sum for the sine integral at point x = π again:
\begin{align*} -S_N \left( \frac{\ell}{N+1} - \ell \right) &= \frac{2\ell}{\pi} \,\sum_{k=1}^N \frac{1}{k}\, \sin \left( \frac{k\pi}{N+1} \right) = \frac{\pi}{N+1} \, \sum_{k=1}^N \frac{\sin \frac{k\pi}{N+1}}{k\pi /(N+1)} \\ &\approx \frac{2}{\pi} \,\mbox{Si}(\pi ) = \frac{2}{\pi} \int_0^{\pi} \frac{\sin x}{x}\,{\text d}x = \frac{1.85\ldots}{1.57\ldots} = 1.1789797\ldots . \end{align*}
N[SinIntegral[Pi]*2/Pi, 40]
1.178979744472167270232028845824909741464

Hence, we conclude that overshoot and undershoot at end points will be again equal to \( \frac{a+b}{2}\, w = \frac{a+b}{2}\, 0.0894899\ldots . \)

Let us consider the following piecewise continuous function on the interval [-2,2]:
\[ f(x) = \begin{cases} 2\,x^2 - 3\,x +2, & \ \mbox{ for } \ x > 0 , \\ 1/2, & \ \mbox{ for } \ x =0 , \\ -2\, x^3 + 4\,x -3, & \ \mbox{ for } \ x < 0 . \end{cases} \]
This function has one internal point x = 0 of finite jump of 5 units and discontinuities at end points because f(-2) = 5 and f(2) = 4. To eliminate discontinuity at the origin, we subtract from the given function the piecewise step function
\[ s(x) = \begin{cases} 2, & \ \mbox{ for } \ x > 0 , \\ -1/2, & \ \mbox{ for } \ x =0 , \\ -3, & \ \mbox{ for } \ x < 0 . \end{cases} \]
Then the resulting function \( g(x) = f(x) - s(x) \) will be continuous at the origin.
f[x_] = Piecewise[{{2*x^2 - 3*x + 2, x > 0}, {-2*x^3 + 4*x - 3, x < 0}}]
s[x_] = 2*UnitStep[x] - 3*UnitStep[-x]
g[x_] = f[x] - s[x]
Plot[{f[x], s[x]}, {x, -2.5, 2.5}, PlotStyle -> Thick]
Plot[g[x], {x, -2.5, 2.5}, PlotStyle -> Thick, PlotLabel -> "g(x) = f(x) - s(x)"]
Graphs of function f(x) and step function.
Difference g(x) of two functions.

However, the function g(x), when extended periodically from the interval [-2,2], will have discontinuities at these end points. Therefore, we need to add to it the linear function l(x) = 3x/2, so that the sum φ(x) = g(x) + l(x) will have the same values at the end points.

\[ \varphi (x) = \begin{cases} 2\,x^2 + \frac{3}{2}\,x , & \ \mbox{ for } \ x > 0 , \\ 0, & \ \mbox{ for } \ x =0 , \\ -2\, x^3 + \frac{11}{2}\,x , & \ \mbox{ for } \ x < 0 . \end{cases} \]
l[x_] = 3*x/2
phi[x_] = g[x] + l[x]
Plot[{g[x], l[x], phi[x]}, {x, -2.5, 2.5}, PlotStyle -> Thick]
Plot[{phi[x]}, {x, -2.01, 2.01}, PlotStyle -> {Thick, Green}]
Auxiliary functions, where φ(x) is in green.
Continuous function φ(x) in green.

Now we expand the two functions, piecewise continuous function f(x) and continuous function φ(x) into Fourier series over the interval [-2,2]. Upon evaluating integrals on the interval [-2,2]

\begin{align*} a_0 &= \frac{1}{2} \int_{-2}^2 f(x)\,{\text d} x = - \frac{4}{3} , \\ a_k &= \frac{1}{2} \int_{-2}^2 f(x)\,\cos \left( \frac{k\pi x}{2} \right) {\text d} x = \frac{96}{k^4 \pi^4} \left[ 1 - (-1)^k \right] + \frac{2}{k^2 \pi^2} \left[ 7 + 25 (-1)^k \right] , \\ b_k &= \frac{1}{2} \int_{-2}^2 f(x)\,\sin \left( \frac{k\pi x}{2} \right) {\text d} x = - \frac{16}{k^3 \pi^3} \left[ 1 + 5 (-1)^k \right] + \frac{1}{k\pi} \left[ 5 + (-1)^k \right] , \end{align*}
we obtain its Fourier series
\begin{align*} f(x) &= -\frac{2}{3} + 12\, \sum_{n\ge 1} \frac{1}{n^4 \pi^4} \, \cos \left( n\pi x \right) + 2 \sum_{k\ge 1} \frac{7+25(-1)^k }{k^2 \pi^2} \, \cos \left( \frac{k\pi x}{2} \right) \\ & \quad - 16 \,\sum_{k\ge 1} \frac{1 + 5(-1)^k}{k^3 \pi^3} \,\sin \left( \frac{k\pi x}{2} \right) + \sum_{n\ge 1} \frac{5 + (-1)^n}{n\pi} \, \sin \left( \frac{n\pi x}{2} \right) . \end{align*}
Mathematica confirms
f[x_] = Piecewise[{{2*x^2 - 3*x + 2, x > 0}, {-2*x^3 + 4*x - 1, x < 0}}]
Simplify[Integrate[f[t]*Cos[0*Pi*t/2], {t, -2, 2}]/2]
-4/3
Simplify[Integrate[f[t]*Cos[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
(-96 (-1 + (-1)^k) + 2 (7 + 25 (-1)^k) k^2 \[Pi]^2)/(k^4 \[Pi]^4)
Simplify[Integrate[f[t]*Sin[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
(-16 (1 + 5 (-1)^k) + (5 + (-1)^k) k^2 \[Pi]^2)/(k^3 \[Pi]^3)

Now we expand auxiliary functions into Fourier series that were added to the given function f(x) in order to make the continuous function φ(x):

\[ s(x) = -\frac{1}{2} + \sum_{k\ge 1} \frac{10}{(2k-1)\pi} \,\sin \left( \frac{(2k-1)\pi x}{2} \right) = \begin{cases} \phantom{-}3, & \ \mbox{ if } 0 < x < 2, \\ -2, & \ \mbox{ if } 2 < x < 4, \end{cases} \]
and
\[ l(x) = -6\,\sum_{k\ge 1} \frac{(-1)^k}{k\pi} \,\sin \left( \frac{k\pi x}{2} \right) = \frac{3}{2}\, x , \qquad -2 < x < 2, \]
because
Integrate[s[t]*Cos[0*Pi*t/2], {t, -2, 2}]/2
Integrate[l[t]*Cos[0*Pi*t/2], {t, -2, 2}]/2
Simplify[Integrate[s[t]*Cos[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
Simplify[Integrate[l[t]*Cos[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
Simplify[Integrate[s[t]*Sin[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
Simplify[Integrate[l[t]*Sin[k*Pi*t/2], {t, -2, 2}]/2, Assumptions -> Element[k, Integers]]
Their linear combination gives Fourier expansion of the piecewise continuous function that does not converge uniformly
\[ l(x) -s(x) = \frac{1}{2} -6\,\sum_{k\ge 1} \frac{(-1)^k}{k\pi} \,\sin \left( \frac{k\pi x}{2} \right) - \sum_{k\ge 1} \frac{10}{(2k-1)\pi} \, \sin \left( \frac{(2k-1)\pi x}{2} \right) = \begin{cases} \frac{3}{2}\,x -2 , & \ \mbox{ for } \ 0 < x < 2 , \\ 1/2, & \ \mbox{ for } \ x =0 , \\ \frac{3}{2}\,x +3 , & \ \mbox{ for } \ -2 < x < 0 . \end{cases} \]
Note that the general term in each series converges to zero linearly.
dc100[x_] = 1/2 - 6/Pi*Sum[(-1)^k *Sin[k*Pi*x/2]/k, {k, 1, 20}] - 10/Pi*Sum[Sin[(2*k - 1)*Pi*x/2]/(2*k - 1), {k, 1, 100}];
Plot[{l[x] - s[x], dc100[x]}, {x, -2.5, 2.5}, PlotStyle -> Thick, Ticks -> {{-2, -1, 1, 2}, {3.447449, -2.44745}}]
Plot[{l[x] - s[x], dc100[x]}, {x, -0.3, 0.3}, PlotStyle -> Thick, Ticks -> {{0}, {3.447449, -2.44745}}]
Plot[{3.447449, -2.44745, dc100[x]}, {x, 1.7, 2.3}, PlotStyle -> {Thick}, Ticks -> {{2}, {1.0894899, 2.1789797, 3.447449, 0, -0.0894899, -2.44745}}]
Fourier approximation with 100 terms of the piecewise continuous function.
Fourier series approximation near the origin.
Fourier series approximation near x = 2.

Finally, we plot the given function f(x) with its Fourier approximation containing 100 terms.

ff100[x_] = -2/3 + 192/Pi^4*Sum[Cos[(2*n - 1)*Pi*x/2]/(2*n - 1)^4, {n, 1, 50}] + 2/Pi^2 *Sum[(7 + 25*(-1)^k)/k^2*Cos[k*Pi*x/2], {k, 1, 100}] - 16/Pi^3*Sum[(1 + 5*(-1)^k)*Sin[k*Pi*x/2]/k^3, {k, 1, 100}] + 1/Pi*Sum[(5 + (-1)^k)*Sin[k*Pi*x/2]/k, {k, 1, 100}]; Plot[{f[x], ff100[x]}, {x, -2.3, 2.3}, PlotStyle -> Thick, Ticks -> {{0, -2, 2}, {3.447449, -2.44745}}]
Fourier approximation with 100 terms of the given function.
   ■

 

  1. Fay, T.H. and Kloppers, P.H., The Gibbs' phenomenon, International Journal of Mathematical Education in Science and Technology, 2001, Vol. 32, No. 1, pp. 73--89; doi:10.1080/00207390117151
  2. Hewitt, E. and Hewitt, R.E., The Gibbs--Wilbraham phenomenon: An episode in Fourier analysis, Archive for History of Exact Science, 1979, Vol. 21, No. 2, pp. 129--160.

 

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