Euler’s Method for Making Money

Chip Rollinson is a teacher at Buckingham Browne & Nichols School in Cambridge, Massachusetts. Today’s post is a note he sent to the AP Calculus Community bulletin board that I found interesting. I share it with you with his permission. I have done some minor editing. Chip wrote on February 5, 2015:

I had an epiphany today about the relationship between Euler’s Method and compounding growth. I had never made the connection before. I thought it was cool so I decided I needed to share it.

Consider the differential equation \frac{dA}{dt}=rA with the condition A(0) = P.  Solving this equation gives A\left( t \right)=P{{e}^{rt}}, but let’s ignore this for now.

Let’s look at this differential equation using Euler’s Method.

Leonhard Euler1707 - 1783

Leonhard Euler
1707 – 1783

Let’s start with a step size of \tfrac{1}{12} and use (0, P) as the “starting point.”

After one step, you arrive at the point \displaystyle \left( \tfrac{1}{12},P+\tfrac{r}{12}P \right)=\left( \tfrac{1}{12},P\left( 1+\tfrac{r}{12} \right) \right)

After 2 steps, you arrive at the point \displaystyle \left( \tfrac{2}{12},P+\tfrac{2}{12}P+\tfrac{{{r}^{2}}}{144}P \right)=\left( \tfrac{2}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{2}} \right)

After 3 steps, you arrive at the point \displaystyle \left( \tfrac{3}{12},P+\tfrac{3}{12}rP+\tfrac{3}{144}{{r}^{2}}P+P\tfrac{3}{1728}{{r}^{3}} \right)=\left( \tfrac{3}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{3}} \right)

After 4 steps, you arrive at the point \displaystyle \left( \tfrac{4}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{4}} \right)

And so on …

After 12 steps, you arrive at the point \displaystyle \left( 1,P{{\left( 1+\tfrac{r}{12} \right)}^{12}} \right)

After 120 steps, you arrive at the point \displaystyle \left( 10,P{{\left( 1+\tfrac{r}{12} \right)}^{120}} \right)

After 12t steps, you arrive at the point \displaystyle \left( t,P{{\left( 1+\tfrac{r}{12} \right)}^{12t}} \right)

Do these y-values look familiar? It’s the amount you’d have if you were compounding monthly with a yearly interest rate of r for a year, 10 years, and t years.

If the step size were \displaystyle \tfrac{1}{365} instead, you would arrive at the points

\displaystyle \left( 1,P{{\left( 1+\tfrac{r}{365} \right)}^{365}} \right), and \left( 10,P{{\left( 1+\tfrac{r}{365} \right)}^{3650}} \right), and \displaystyle \left( t,P{{\left( 1+\tfrac{r}{365} \right)}^{365t}} \right).

These y-values are the amount you’d have if you were compounding daily for a year, 10 years, and t years.

If the step size were \tfrac{1}{n}, the point after t years is \displaystyle \left( t,P{{\left( 1+\tfrac{r}{n} \right)}^{nt}} \right)

If n went to infinity, you would arrive at the point \displaystyle \left( t,P{{e}^{rt}} \right), the y-value for continuously compounding. This is the solution of  the differential equation mentioned above.

I’d never made this connection before, but it makes perfect sense now.

The actual problem that got me thinking about all of this was:

Suppose that you now have $6000, you expect to save an additional $3000 during each year, and all of this is deposited in a bank paying 4% interest compounded continuously.

This generates the differential equation: \displaystyle \frac{dA}{dt}=0.04t+3000

This was a fun one to do out. I started with a step size of 1/12 and then made them smaller. I derived the solution from Euler’s Method!


So writes Chip Rollinson. He included the following links to his computations: here and here

Then there were some interesting comments from others.

1. Mark Howell pointed out that if \displaystyle \frac{dy}{dx}=f\left( x \right), that is if the derivative is a function of x only, then Euler’s method is the same as a left Riemann sum approximation.

This would make a good exercise for your students to show given a derivative, a starting point, and a small number of steps. Try \frac{dy}{dx}={{x}^{3}}, and use Euler’s Method with 4 steps starting at (0, 0) and then a left-Riemann sum with 4 terms to approximate \int_{0}^{1}{{{x}^{3}}dx},  (Answer: \tfrac{9}{64})

2. Dan Teague went further and pointed out that this is really the development of the Fundamental Theorem of Calculus for a particular function. He contributed this development of the concept.

So Thank You Chip for this and the several other comments you’ve contributed to other post on this blog. And thank you Mark and Dan for your contributions.


 

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