Click on the following links to open an interactive graph.

**MathStudio Link: The M&M Experiment.**

- This is a simulator for the M&M population experiment. Instead of M&Ms that can be up or down we have a row of boxes that can contain a 0 or 1. M&M down is a 0, M&M up is a 1.
- The experiment starts with an initial number boxes all in 1 except the first box, which counts how many 1 we have in that row.
- The second row is constructed from the first by randomly choosing between 0 or 1. Again, the first box counts how many 1 we got.
- The third row is constructed from the second, if we had a box with a 0, it remains a 0. If we had a 1, we randomly choose a between 0 or 1. As before, the first box counts the number of 1 in that row.
- We keep going in this way till we have a number of rows given by the constant in
**Trials**. - We can also add a fix number of boxes on each round, we call these added boxes the immigrants.

Finally, we can generalize the M&M experiment by changing the mortality coefficient from 50% to any percent from 0 to 100.

- The simulation has one button and four slider:
- The button
**Run**runs the simulation with the chosen parameters. - The slider
**Initial**fixes the initial amount of M&Ms. Default**Initial = 30** - The slider
**Imm**fixes the number of immigrants M&M added on every trial. Default**Imm = 0** - The slider
**Trials**sets the number of rounds we have in the simulation. Default**Trials = 10** - The slider
**Mortality**sets the mortality percent. Default**Mortality = 50%**

We also can plot the first column with the counts of boxes with 1.

**MathStudio Link: Going Viral Simulation.**

This is a simulator of a viral infection spreading in a population with the following rules:

- The total population where the virus can spread is fixed at 100.
- Each person in the population has a fixed Identification Number in the interval \( [1,100]\).
- At day zero there is only one person infected.
- Each day, each infected person pass the virus to only one, random, person in the population.
- Infected persons remain always infected.
- Nobody dies.

**MathStudio Link: Direction Field.**

**MathStudio Link:**

**Picard Iteration vs Taylor Expansion: Linear Equations.**

- We graph in
**purple**the functions \(y(t)\) solutions of the initial value problem

\[

y^{\prime}(t) = 2 \,y(t) +3,

\qquad

y(0) = 1.

\] - The slider
**Function**turns on-off the graph of the solution \(y(t)\). - We graph in
**blue**approximate solutions \(y_n\) of the differential equation constructed with the Picard iteration up to order \(n=10\). The slider**Picard_App_Blue**turns on-off the Picard approximate solution. - We graph in
**green**the \(n\)-order Taylor expansion centered \(t=0\) of the solution of the differential equation up to order \(n=10\). The slider**Taylor_App_Green**turns on-off the Taylor approximation of the solution.

We conclude that the **Picard iteration is identical as the Taylor expansion** for solutions of the **linear** differential equation above.

**MathStudio Link:**

**Picard Iteration vs Taylor Expansion: Non-Linear Equations – Explicit Solution.**

- We graph in
**purple**the functions \(y(t)\) solutions of the initial value problem

\[

y^{\prime}(t) = y^2(t),

\qquad

y(0) = -1.

\] - The slider
**Function**turns on-off the graph of the solution \(y(t)\). - We graph in
**blue**approximate solutions \(y_n\) of the differential equation constructed with the Picard iteration up to order \(n=5\). The slider**Picard_App_Blue**turns on-off the Picard approximate solution. - We graph in
**green**the \(n\)-order Taylor expansion centered \(t=0\) of the solution of the differential equation up to order \(n=5\). The slider**Taylor_App_Green**turns on-off the Taylor approximation of the solution.

We conclude that the **Picard iteration is a different and better approximation than the Taylor expansion** for solutions of the **non-linear** differential equation above.

**MathStudio Link:**

**Picard Iteration vs Taylor Expansion: Non-Linear Equations – No Explicit Solution.**

- In this second example we study approximate solutions of the initial value problem

\[

y^{\prime}(t) = y^2(t) + t,

\qquad

y(0) = -1.

\] - In this case we do
**not**have an explicit expression for the solution \(y(t)\). We only have the approximate solutions. - We graph in
**blue**approximate solutions \(y_n\) of the differential equation constructed with the Picard iteration up to order \(n=5\). The slider**Picard_App_Blue**turns on-off the Picard approximate solution. - We graph in
**green**the \(n\)-order Taylor expansion centered \(t=0\) of the solution of the differential equation up to order \(n=5\). The slider**Taylor_App_Green**turns on-off the Taylor approximation of the solution.

We conclude, one more time, that the **Picard iteration is a different (hopefully better) approximation than the Taylor expansion** for solutions of the **non-linear** differential equation above.

**MathStudio Link: Beating and Resonance on an LC-series Circuit.**

- An LC-series circuit with a voltage source is described by Krichhoff equation

\[

L \, I'(t) + \frac{1}{C} \int I(t)\, dt = V(t).

\]- Consider the case that \(V(t) = L \sin(\omega t)\). In this case, computing one time derivative and dividing by \(L\) we get

\[

I” + \omega_0^2 I = \omega \cos(\omega t).

\] where \(\displaystyle\omega^2 = \frac{1}{\sqrt{LC}}\). The solution with initial conditions \(I(0)=0\) and \(I'(0)=0\) is

\[

I(t) = \frac{t}{2}\,\sin(\omega t).

\] - Consider the case that \(V(t) = L \sin(\nu t)\). In this case, computing one time derivative and dividing by \(L\) we get

\[

I” + \omega_0^2 I = \nu \cos(\nu t).

\] where \(\displaystyle\omega^2 = \frac{1}{\sqrt{LC}}\). The solution with initial conditions \(I(0)=0\) and \(I'(0)=0\) is

\[

\tilde I(t) = \frac{\nu}{(\omega^2-\nu^2)}\bigl( \cos(\nu t)-\cos(\omega t) \bigr)

\]

- Consider the case that \(V(t) = L \sin(\omega t)\). In this case, computing one time derivative and dividing by \(L\) we get
- Click on the interactive graph link here to see how the function changes when \(\nu \to \omega_0\), exhibiting the beating phenomenon.

**MathStudio Link: Dirac Delta Sequences.**

We show a few different sequence of functions that have the same Dirac’s delta as their limit. The sequences are:

- In
**red**: \(\displaystyle\Bigl. \delta_n(t) = n \,\bigl(u(t) – u(t- \frac{1}{n})\bigr)\). - In
**blue**: \(\displaystyle\Bigl.\delta_n(t) = \frac{n}{2} \,\bigl(u(t+\frac{1}{n}) – u(t-\frac{1}{n})\bigr)\). - In
**green**: \(\displaystyle\Bigl.\delta_n(t) = \sqrt{\frac{n}{\pi}} \, e^{-n^2 t^2}\). - In
**purple**: \(\displaystyle\Bigl.\delta_n(t) = \frac{1}{\pi} \frac{n}{(1 + n^2 t^2)}\). - In
**gray**: \(\displaystyle\Bigl.\delta_n(t) = \frac{\sin(n t)}{\pi \,t}\).

**Convolution Graphs.**

We compute the convolution of the functions \(f\) and \(g\),

\[

(f*g)(t) = \int_0^t f(\tau) \,g(t-\tau)\, d\tau,

\] and we plot \(f\) in

**blue**, \(g\) in

**green**, and \(f*g\) in

**red**.

We see that the convolution is a measure of (although not equal to) the area of the overlap of the two functions, \(f\) and \(g\), which is shown in **gray**.

**MathStudio Link:****Convolution: Example 1**(Slow).

In this graph we choose

\[

f(x) = u(x) -u(x-1),

\qquad

g(x) = u(x) -u(x-1).

\]**MathStudio Link:****Convolution: Example 2**.

In this graph we choose

\[

f(x) = u(x) \, e^{-x},

\qquad

g(x) = u(x) \sin(x).

\]**MathStudio Link:****Convolution, Example 3.**(Slow).

In this first graph we choose

\[

f(x) = u(x) -u(x-1),

\qquad

g(x) = 2\,u(x) \, e^{-x}.

\]

**MathStudio Link: BVP and Eigenfunctions.**

In the first picture we show the solution to the BVP

\[

y”+ \pi^2 y =0,

\qquad

y(0)=1,

\quad

y(1)=-1

\]
This BVP has infinitely many solutions, given by

\[

y(x) = \cos(\pi x) + k \sin(\pi x),

\qquad

k\in \mathbb{R}.

\]

- We plot the fundamental solution \(y_1(x)= \cos(\pi x)\) in
**red**. - We plot the fundamental solution \(y_2(x)= \sin(\pi x)\) in
**red**. - We plot the solution \(y_k(x) = k \sin(\pi x)\) in
**purple**. - We plot the solution \(y(x) = \cos(\pi x) + k\sin(\pi x)\) in
**blue**.

In the second picture we plot the function

\[

y_n(x) = \sin(n\pi x),

\qquad

n\in \mathbb{R}.

\]
In the case that \(n\) is an integer, these functions are eigenfunctions solutions

\[

y” + \lambda y =0,

\qquad

y(0)=0,

\quad

y(1)=0.

\]
In this problem the eigenvalues are \(\lambda_n = (n\pi)^2\) for \(n = 1, 2, 3, \cdots\), and the eigenfunctions are the functions \(y_n\) above for \(n=1,2,3,\cdots\).