How can the green contour be graphed?

enter image description here

FourierF[a_, t_] := a.Table[Sin[ 2 Pi i t], {i, Length[a]}];

FourierAnim[a_, t_] :=
    A = Accumulate[a*Table[Cos[ 2 Pi i t], {i, Length[a]}]],
    B = Accumulate[a*Table[Sin[2 Pi i t], {i, Length[a]}]]},
   PrependTo[A, 0]; PrependTo[B, 0];
      {Circle[{A[[i]], B[[i]]}, a[[i]]],
       If[i != Length@a, 
        Line[{{A[[i]], B[[i]]}, {A[[i + 1]], B[[i + 1]]}}],
        {Red, Dashed, Line[{{A[[i]], B[[i]]}, {2, B[[i]]}}]}]

       (* next line needs to be fixed *)
       , {Green, Line[Table[{m, n}, {m, A[[i]], t}, {n, B[[i]], t}]]}
       (* end of section needing editing *)

      {i, Length@a}
     PlotRange -> {{-1.5, 3}, {-1, 1}}
    Plot[FourierF[a[[;; -2]], t - \[Tau]], {\[Tau], 2, 3}]

a = Table[(1 - (-1)^i)/i, {i, 64}]/Pi;
(* (1+(-1)^i)/i,{i,65} for sawtooth wave *)
(* ??? for triangle wave *)

  a[[;; j]], t], {t, 0, 1}, {j, 8, Length@a, 2}
(* {t,0,0.5},{j,9,Length@a,2} for sawtooth wave *)
(* ??? for triangle wave *)

The code is a modified version of the original; it demonstrates how the smooth motion of rotating circles can be used to build up any repeating curve.

The eventual solution was suggested by user Michael E2 in a comment from the second related thread posted below.

Can you make a list of the points and use Line? You seem to be able to calculate the points at each time (in order to draw the figure). Just make a Table of them over one period with suitable increment of time.

One of the early attempts of coding this was left uncommented, because it is the only one that produced visible related results. I now understand the error in Table and the array it generates, but this example is perfect for showing the confusion of a beginner. The solution will help me a lot in learning through examples.

Additional question:
What exactly needs to be changed in the formula 1-(-1)^i)/i in order to make (an approximation of) a triangle wave?


2 Answers 2


Generating the outline

By looking at the code we can infer that the position of the center of the outermost circle is given by

outline[a_, t_] := Module[{
   A = Accumulate[a Table[Cos[2 Pi i t], {i, Length[a]}]],
   B = Accumulate[a Table[Sin[2 Pi i t], {i, Length[a]}]]
   }, {Last[A], Last[B]}]

This code comes straight from the one you posted; each element in A and B are the $x$ and $y$ coordinates of the subsequent circles. The last elements represent the positions of the outermost circle.

We can now plot the outline using ParametricPlot:

a = Table[(1 - (-1)^i)/i, {i, 16}]/Pi;
ParametricPlot[outline[a, tmax], {tmax, 0, 1}]

Mathematica graphics

The above outline was generated as if there were sixteen circles in the model. You can change the number 16 to get another outline.

We can append this function to Show inside the original function to draw the outline together with the animation.

Show[..., ..., ParametricPlot[outline[a, tmax], {tmax, 0, t}]]

We have to make a small change in the call to Manipulate because we can't plot from 0 to 0, so we increase the lower bound slightly:

Manipulate[FourierAnim[a[[;; j]], t], {t, 0.001, 1}, {j, 1, Length@a, 1}]

enter image description here

The triangle wave

As for your second question about the triangle wave we have to look at the math. The Fourier series of the square wave is $$ f(x) = \frac{4}{\pi}\sum_{i=1,3,5,...}^{\infty}\frac{1}{i}\sin\left(\frac{i\pi x}{L}\right) $$ and the corresponding code that we used is

a = Table[(1 - (-1)^i)/i, {i, 16}]/Pi;
FourierF[a_, t_] := a.Table[Sin[ 2 Pi i t], {i, Length[a]}];

Note that the sum in the Fourier series only sums over odd indices. The 1-(-1)^i part captures this behavior, because it is zero for even i. The other part of the formula is 1/i, and this is the coefficient of the Sin function in the Fourier series. We don't have to match the constants, it's enough that what we have is proportional to the Fourier series.

The Fourier series of the triangle wave is $$ f(x) = \frac{8}{\pi^2}\sum_{i=1,3,5,...}^{\infty}\frac{(-1)^{(i-1)/2}}{i^2}\sin\left(\frac{i \pi x}{L}\right). $$ Like the square wave Fourier series it is a Sin Fourier series that only sums over odd indices. So we can surmise that our expression will have two parts just like above; one part to set the expression to zero for even indices, and one part to match the coefficient. We end up with this:

a = Table[(1 - (-1)^i) (-1)^(0.5 i - 0.5)/i^2, {i, 16}]/Pi;

where the two parts are (1 - (-1)^i) and (-1)^(0.5 i - 0.5)/i^2.

Putting this into our code, we get:

enter image description here

Because the expression sometimes takes on negative values I had to apply Abs to the radii of the circle:

Circle[{A[[i]], B[[i]]}, a[[i]] // Abs]

We can find the expressions corresponding to other Fourier series in the same way.

  • $\begingroup$ What an elegant solution! Thank you for the detailed walk-through, it is very useful and appreciated. Any idea how to make a triangle wave? $\endgroup$
    – Bo C.
    Jun 25, 2016 at 21:59
  • $\begingroup$ @BoC. I added a part about that. $\endgroup$
    – C. E.
    Jun 25, 2016 at 22:33
  • $\begingroup$ Amazing how the triangle formula accounts for phase inversion every other (odd) harmonic, in order to obtain the regular waveform from the current start point (3Pi/2 in trigonometric direction). By using (1 - (-1)^i)/i^2 all harmonics start in-phase, but the shape is rotated by Pi/2, making the time domain waveform unrecognizable. Theoretically, this phase change shouldn't affect the way generated tones sound. $\endgroup$
    – Bo C.
    Jun 26, 2016 at 7:45
  • $\begingroup$ How can phase be changed for sawtooth? The code for different waveforms and their Pi/2 phase change is to be inserted inside the Table function: a = Table[ ... ]/Pi; — square: (1-(-1)^i)/i,{i,64} | phase: (1-(-1)^i)(-1)^((i-1)/2)/i,{i,64} — triangle: (1-(-1)^i)(-1)^((i-1)/2)/i^2,{i,64} | phase: (1-(-1)^i)/i^2,{i,64} — saw: (1+(-1)^i)/i,{i,65} | phase: ??? $\endgroup$
    – Bo C.
    Jun 27, 2016 at 10:28
  • 1
    $\begingroup$ @BoC. This is not the appropriate site for the math. If you have further inquiries that are not about Mathematica, it is better that you ask on math.SE. $\endgroup$
    – C. E.
    Jun 27, 2016 at 10:45

This is an inefficient method based on an answer to this question. The other answers are relevant, particularly in relation to changing phase.

Exploiting C.E.'s answer (which I have voted for):

fourier[r_, n_] := 
 Module[{fun = 
        MapThread[#1 Exp[2 Pi I #2 x ] &, {r, Range[Length@r]}], {0, 
   par = Function[t, r.Table[Sin[2 Pi j t], {j, Length@r}]],
   tab, g, h}, tab = Table[fun[j], {j, 0, 1, 1/n}];
  g = Graphics[{White, MapThread[Circle[#1, Abs@#2] &, {Most@#, r}], 
       Red, Point[Most@#], Purple, PointSize[0.04], Point[Last@#], 
       Yellow, Line@tab[[All, -1]], Line@#, Dashed, Red, 
       Line[{Last@#, {5, #[[-1, 2]]}}]}, 
      PlotRange -> {{-5, 10}, {-4, 4}}] & /@ tab;
  h = Table[Plot[par[t - s], {s, 5, 7}], {t, 0, 1, 1/n}];
   Show[#1, #2, Background -> Black, ImageSize -> 500] &, {g, h}]
wave[n_, type_] := Which[
  type == "square", Table[(1 - (-1)^i)/i, {i, n}],
  type == "sawtooth", Table[(1 + (-1)^i)/i, {i, n}],
  type == "triangle", Table[(1 - (-1)^i) (-1)^((i - 1)/2)/i^2, {i, n}]]
fourieranim[type_, n_, steps_] := fourier[wave[n, type], steps];

The following are examples of 8 term series for square, sawtooth and triangular wavses:

enter image description here

  • $\begingroup$ Your version is very useful since I'm learning through examples here. The code from the question had a way of changing waveforms, from (approximately) square to sawtooth by modifying values according to the commented code. How can other waveforms be generated in your version? And how can the number of harmonics be modified (the initial j)? $\endgroup$
    – Bo C.
    Jun 22, 2016 at 11:23
  • $\begingroup$ @BoC. I suggest looking at documentation, e.g. reference.wolfram.com/language/tutorial/FourierTransforms.html. You can play around. Good luck :) $\endgroup$
    – ubpdqn
    Jun 22, 2016 at 11:28
  • $\begingroup$ Not just the Mathematica manual, but also a math course. The problem with this "triangle wave" rendition is, harmonics don't have amplitudes (circle radiuses) corresponding to the triangle wave function, that is, A_n=(1/n^2)A_1. I just managed to get a grip of the first example of the code, and thought that for an advanced user tracing a contour at the point where the dotted horizontal line starts would be a piece of cake: specifying the correct coordinates in the existing Plot function, or creating a new one. It's this that I want to learn. $\endgroup$
    – Bo C.
    Jun 23, 2016 at 5:48
  • $\begingroup$ @BoC. I am sorry if I have misunderstood. I sincerely apologize. I cannot delete it if you wish. I currently do not have time. I only posted this answer in the aim of you or others solving. I apologize again. I can happily delete if you wish $\endgroup$
    – ubpdqn
    Jun 23, 2016 at 5:52

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