INVESTIGATION - The number e

The number $\displaystyle e$

To understand what the number $\displaystyle e$ is and some special properties of it, let's look at the original problem where $\displaystyle e$ was found.

In finance, when investing or borrowing, interest is often paid at a compounding rate. That is, the amount of interest will grow each period because it is calculated on both the initial amount and the interest from the previous period. The more often interest is compounded, the faster the account balance will grow.

Let's take the contrived scenario of $\displaystyle \$1$ borrowed at $\displaystyle 100$ per annum for $\displaystyle 1$ year.

How much would we owe after $\displaystyle 1$ year?

We would owe the $\displaystyle \$1$ borrowed plus the $\displaystyle \$1$ in interest, so $\displaystyle \$2$.

What about if the lender wanted to earn more interest, so calculated the interest owing and added it to the account twice a year?

At 6 months: Half the year has passed so we owe \displaystyle \frac{100%}{2}=50% in interest. This would be $\displaystyle \$0.50$. Added to our debt we now owe $\displaystyle \$1.50$.

At the end of the year: Now on a debt of $\displaystyle \$1.50$ we pay a further $\displaystyle 50$, which is $\displaystyle \$0.75$. Which leads us to a total debt of $\displaystyle \$2.25$.

This would be equivalent to multiplying our debt by $\displaystyle 150$ twice. That is, the final debt is $\displaystyle \$1\times \left(1+0.5\right)^2$.

What if the interest was charged quarterly? Here we would be increasing the initial debt by $\displaystyle 25$ four times over, which could be calculated as $\displaystyle \$1\times \left(1+0.25\right)^4$.

This leads us to the compound interest formula:

In our scenario the lender is quite greedy and uses an interest rate of $\displaystyle 100$ per year for $\displaystyle 1$ year and we are borrowing $\displaystyle \$1$. This would make the formula reduce to $\displaystyle F=\left(1+\frac{1}{n}\right)^{n}$.

We have seen as the frequency of the compounds increases the final amount due also increases. Suppose we had a lender who wanted to collect as much interest as possible. Is there a limit to how much the lender can collect? How much would they collect if they compound the interest every nanosecond of the year?

 

Activity 1

In order to find out if there is a limit to how much the lender can collect, let's start by looking for a pattern as the compounding period frequency increases.

1. Complete the following table and look for a pattern as the compounding period frequency increases:

Compounding period	$\displaystyle n$	Final debt, $\displaystyle F=\left(1+\frac{1}{n}\right)^n$
Annually	$\displaystyle 1$	$\displaystyle \$2$
Semi-annually	$\displaystyle 2$	$\displaystyle \$2.25$
Quarterly	$\displaystyle 4$
Monthly	$\displaystyle 12$
Fortnightly	$\displaystyle 26$
Weekly	$\displaystyle 52$
Daily
Hourly
Each second

2. Create a hypothesis for what will happen to the value of $\displaystyle \left(1+\frac{1}{n}\right)^{n}$ as the number of compounding periods increases.

3. Confirm or refute your hypothesis using the applet below. Either click the "Start Animation" button or slide the blue slide through the different values of $\displaystyle n$.

Created with Geogebra

 

Discussion questions

1. You have just glimpsed at a very important number in mathematics, Euler's number. $\displaystyle e=2.71828182859045...$. The number is defined as follows: $\displaystyle e=\lim_{n \rightarrow \infinity}\left(1+\frac{1}{n}\right)^n$
2. Do a little bit of research on this number, what can you find? Do the numbers settle into a repeating pattern or is $\displaystyle e$ an irrational number? How can we prove this?
3. How accurate was your calculation using seconds? How many decimal places is the value known to?
4. Are there other ways to calculate this number?
5. Where else does this number appear? While this investigation looks at compound interest, the number $\displaystyle e$ shows up in a surprising amount of mathematics across varied topics.
6. Euler was a very influential mathematician. See what you can learn about Euler and his time period in mathematics. Why was the constant named after him?

In our next investigation we will look at differentiating exponential functions and special properties for functions with a base $\displaystyle e$ involving calculus.

Other rates and the exponential function

In the example above we looked at compounding continuously at $\displaystyle 100$. What about other rates? For example what would compounding at $\displaystyle 5$ per year continuously be equivalent to as a once off annual rate?

Let's investigate the following limit for different values of $\displaystyle r$: $\displaystyle \lim_{n \rightarrow \infinity}\left(1+\frac{r}{n}\right)^{n}$

Activity 2

A convenient way to investigate this limit is by using a spread sheet or table in your calculator.

Step 1. Select a range of values of $\displaystyle r$ to investigate, such as those shown above, and enter these in the second row as shown. These can be changed later to investigate other values.

Step 2. To investigate the limit as $\displaystyle n$ increase, make a list of convenient values of $\displaystyle n$ getting larger in the first column.

Step 3. In cell B3 enter the formula to calculate $\displaystyle \left(1+\frac{r}{n}\right)^{n}$, where $\displaystyle r$ is the value in the first row and $\displaystyle h$ is the value in the first column. So for this cell we would have the formula =(1+B2/A3)^A3.

Then, so we can drag the formula, rather than rewrite it for each cell, we can lock in a column or row by placing the symbol $\displaystyle \$$ before a letter or number for a cell code to lock in the column or row respectively. Alter the formula in cell B3 to =(1+$B2/$A3)^$A3 and then drag the formula to the other cells in the table.

Once completed you should find for each of the values of $\displaystyle r$ the fraction approaches a limit and for $\displaystyle r=100$ it approaches our now familiar limit of $\displaystyle e\approx 2.7128$. But what value do the other expressions approach?

In fact $\displaystyle \lim_{n \rightarrow \infinity}\left(1+\frac{r}{n}\right)^{n}=e^r$. We can confirm numerically that this appears to be the case by creating a row under the current table in our spreadsheet, and evaluating $\displaystyle e^r$ for comparison.

Step 4. In the row under our current table enter the equation in the cell in column B, =exp(B2). This will evaluate $\displaystyle e^r$, drag the formula across the row and compare to the approximate limit found.

What this suggests is that $\displaystyle \lim_{n \rightarrow \infinity}\left(1+\frac{r}{n}\right)^{n}=e^r$. Can we show this algebraically?

$\displaystyle \lim_{n \rightarrow \infinity}\left(1+\frac{r}{n}\right)^{n}$	$\displaystyle =$	$\displaystyle \lim_{n \rightarrow \infinity}\left(\left(1+\frac{r}{n}\right)^{\frac{n}{r}}\right)^r$	Let $\displaystyle k=\frac{n}{r}$
	$\displaystyle =$	$\displaystyle \lim_{k \rightarrow \infinity}\left(\left(1+\frac{1}{k}\right)^{k}\right)^r$	See note below.
	$\displaystyle =$	$\displaystyle e^r$	Using the definition of $\displaystyle e$

 

Note: Since from our first investigation and the definition of $\displaystyle e$ we have $\displaystyle \lim_{k \rightarrow \infinity}\left(1+\frac{1}{k}\right)^{k}=e$.

This leads us to an alternate formula to calculate the future value in an account that is earning compound interest continuously.

Exploration

If $\displaystyle \$1000$ was invested at $\displaystyle 5$ per annum and interest was added to the account only at the end of the year, the value of the investment would be $\displaystyle F=\$1000\times 1.05=\$1050$. However, if the rate was compounded continuously throughout the year the investment would be worth:

$\displaystyle F$	$\displaystyle =$	$\displaystyle \$1000 \times e^{0.05}$
	$\displaystyle \approx$	$\displaystyle \$1051.27$

 

More generally, when modelling exponential growth (or decay), we are often interested in populations, investments and similar variables which are growing at continuous rate proportional to the current value. This leads us to the following general form for exponential models:

$\displaystyle A=A_0e^{kt}$

Where $\displaystyle A$ is the amount at a given time, $\displaystyle A_0$ is the initial amount and $\displaystyle k$ gives the continuous rate of growth for each unit of time $\displaystyle t$ for $\displaystyle k>0$ or the rate of decay for $\displaystyle k<0$.