Investigation: Defining Euler's number (e)
Interesting
In finance, often interest is added to the account every month. The next month, the amount of interest will be higher because it is calculated on both the initial amount and the interest from the previous month. The more often interest is added, the faster the account balance will grow.
The formula for compound interest is:
A=P\left(1+\frac{r}{n}\right)^{nt} ,
where A is the future value, P is the present value, r is the annual interest rate, n is the number of times interest is compounded, and t is the time in years.
Down to the nanosecond
Suppose we had a lender who wanted to collect as much interest as possible from their customer. They want to compound the interest every nanosecond of the year. There are 31556952000000430 nanoseconds in a year and that overwhelms most calculators. How could we otherwise calculate this? What number would you use for n?
Since the lender is being greedy, let's use an interest rate of 100\% per year for 1 year. This would make the multiplier of the present value \left(1+\frac{1}{n}\right)^{n}. What happens as the frequency of compounding increases?
Activity
In order to find out how to charge interest for every nanosecond of every day, let's start by looking for a pattern as the compounding period frequency increases.
1. Complete the table below to look for a pattern as the compounding period frequency increases
| Compounding Period | Annual | Quarter | Month | Biweekly | Week | Day | Hour |
|---|---|---|---|---|---|---|---|
| n | 1 | 4 | 12 | 26 | 52 | 365 | 8760 |
| \left(1+\frac{1}{n}\right)^{n} |
2. Create a hypothesis for what will happen to the value of \left(1+\frac{1}{n}\right)^{n} as the number compounding period increases from every year to every nanoseconds.
3. Confirm or refute your hypothesis using the GeoGebra animation below. Either click the "Start Animation" button or slide the blue slide through the different values of n.
Discussion questions
- If we took out a loan for \$1000, under the conditions above, how much would we owe after a year if interest was compounded annually? weekly? every second? every nanosecond?
- You have just discovered a very important number in mathematics, Euler's number. e=2.718281828.... Do a little bit of research on this number, what can you find? Where else does this number appear?
- Euler was a very influential mathematician. See what you can learn about Euler and his time period in mathematics.
- This may be the first time which you have seen the idea of convergence, see what you can find about this concept.
Practice questions
Question 5
The natural base $e$e (Euler’s number) is defined as $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$e=limn→∞(1+1n)n
The table shows the value of $\left(1+\frac{1}{n}\right)^n$(1+1n)n using various values of $n$n.
| $n$n | $\left(1+\frac{1}{n}\right)^n$(1+1n)n |
|---|---|
| $1$1 | $2$2 |
| $100$100 | $1.01^{100}=2.704813$1.01100=2.704813 ... |
| $1000$1000 | $1.001^{1000}=2.716923$1.0011000=2.716923 ... |
| $10000$10000 | $1.0001^{10000}=2.718145$1.000110000=2.718145 ... |
| $100000$100000 | $1.00001^{100000}=2.718268$1.00001100000=2.718268 ... |
Evaluate $\left(1+\frac{1}{n}\right)^n$(1+1n)n for $n=1000000$n=1000000, correct to six decimal places.
Which of the following is the closest approximation of $e$e?
$2.718280821$2.718280821
A$2.718281828$2.718281828
B$2.718281820$2.718281820
C$2.718281818$2.718281818
D
Question 6
It is possible to compute $e^x$ex using the following formula.
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\text{. . .}+\frac{x^n}{n!}+\text{. . .}$ex=1+x+x22!+x33!+x44!+. . .+xnn!+. . .
The more terms we use from the formula, the closer we get to the true value of $e^x$ex.
Use the first five terms of the formula to estimate the value of $e^{0.7}$e0.7.
Give your answer to six decimal places.
Use the $\editable{e^x}$ex key on your calculator to find the value of $e^{0.7}$e0.7.
Give your answer to six decimal places.
What is the difference between the two results?
Give your answer to six decimal places.