Write code that finds the ripple of a function (existing code finds the wrong answer)

Question

I have the following code:

Clear["Global`*"];
u = 5;
\[Tau] = (1/2)*10^(-3);
c = 10^(-6);
r = 1000;
y = u*Sum[((1 - E^((-t + \[Tau] + 4 n \[Tau])/(c*r)))  HeavisideTheta[
        t - 4 n \[Tau]] HeavisideTheta[
        t - \[Tau] - 4 n \[Tau]]) - ((1 - E^((-t + (3 + 4 n) \[Tau])/(
         c*r))) HeavisideTheta[t - 4 n \[Tau]] HeavisideTheta[
        t - (3 + 4 n) \[Tau]]), {n, 0, Infinity}];

If I plot (Plot[y,{t,0,15*10^(-3)},GridLines->{{(2*6+3)*\[Tau],(2*6+1)*\[Tau]},{y/.t->(2*6+3)*\[Tau],y/.t->(2*6+1)*\[Tau]}}]) the solution y, I get the following picture:

Now, we can see that the maximum and the minimum of the function $y(t)$ is found at the following times: $t=(2\text{k}+3)\tau$ and $t=(2\text{k}+1)\tau$ where $k\in\mathbb{N}$.

Question: I want to find the absolute difference between the maximum and minimum when $t\to\infty$.

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My work:

Mathematically speaking, I want to find:

$$\lim_{\text{k}\to\infty}\left|\text{y}\left(\left(2\text{k}+3\right)\tau\right)-\text{y}\left(\left(2\text{k}+1\right)\tau\right)\right|\tag1$$

I know that the answer to the question must be:

$$\frac{5 (e-1)}{1+e}\approx2.31059\tag2$$

Coding this in Mathematica gave me:

In[1]:=FullSimplify[
 Limit[Abs[(y /. t -> (2*k + 3)*\[Tau]) - (y /. 
      t -> (2*k + 1)*\[Tau])], k -> Infinity]]

Out[1]=Interval[{0, 20}]

Which is wrong. What is my mistake?

Answer 1

There 2 typo in My work. First, let plot function y, local minimum and maximum in one plot as follows

Clear["Global`*"];
u = 5;
\[Tau] = (1/2)*10^(-3);
c = 10^(-6);
r = 1000;
y = u*Sum[((1 - E^((-t + \[Tau] + 4 n \[Tau])/(c*r))) HeavisideTheta[
        t - 4 n \[Tau]] HeavisideTheta[
        t - \[Tau] - 4 n \[Tau]]) - ((1 - 
         E^((-t + (3 + 4 n) \[Tau])/(c*r))) HeavisideTheta[
        t - 4 n \[Tau]] HeavisideTheta[t - (3 + 4 n) \[Tau]]), {n, 0, 
     Infinity}];
 Show[Plot[y, {t, 0, .01}], 
 ListPlot[{Table[({t, y} /. t -> (2*k + 1)*\[Tau]), {k, {1, 3, 5, 7, 
       9}}], Table[({t, y} /. t -> (2*k + 1)*\[Tau]), {k, {2, 4, 6, 
       8}}]} // N, PlotStyle -> {Blue, Red}]] 

Therefore, we have minimum for even k , and maximum for odd k (Note, that it is not $2k+3$ and $2k +1$ like in the code In[1]:). Let define local extremum as follows

dy = (y /. t -> (2*k + 1)*\[Tau]);

Minimal and maximal values at k=100, 101 are (Note, that there is no limit for dy at k -> Infinity it is why there is Out[1]=Interval[{0, 20}]!),

minmaxy = {Limit[dy, k -> 100], Limit[dy, k -> 101]}

Out[]= {(5 (-1 + 1/E^100 - 1/E^99 + E))/(-1 + E^2), (
 5 (-1 + E + E^102 - E^103))/(E^101 (1 - E^2))}  

We can simplifier last expression by hand neglected by 1/E^100, 1/E^99 and so on, finally we have

 miny = (5 (-1 + E))/(-1 + E^2); maxy = (
 5 (E^1 - E^2))/ (1 - E^2);

 maxy - miny // Simplify

Out[]= (5 (-1 + E))/(1 + E)