Watching (1 + 1/n)^n Approach e

Compute $\left(1 + \frac{1}{n}\right)^n$ for $n = 1$, $n = 2$ and $n = 12$. What do you notice?

Let $n = 1$: $\left(1 + \frac{1}{1}\right)^1 = 2^1 = 2$ = $2$

Let $n = 2$: $\left(1 + \frac{1}{2}\right)^2 = 1.5^2 = 2.25$ = $2.25$

Let $n = 12$ (use a calculator): $\left(1 + \frac{1}{12}\right)^{12} \approx 1.083333^{12} \approx 2.61304$ = $\approx 2.61304$

Compare with larger $n$: $n = 365$ gives $\approx 2.71457$, $n = 1\,000\,000$ gives $\approx 2.71828$ = The values grow more and more slowly

Answer: The results $2$, $2.25$, $2.61304$, ... keep increasing, but they level off near $e \approx 2.71828$ and never go past it.

Compounding More and More Often

You put 1000 dollars in an imaginary account that pays 100% interest per year. How much do you have after one year if the interest is added yearly, monthly or daily?

Use the compound interest formula: $A = P\left(1 + \frac{r}{n}\right)^{n}$ with $P = 1000$, $r = 1$ (100%), $t = 1$ year = $A = 1000\left(1 + \frac{1}{n}\right)^{n}$

Yearly: $n = 1$: $1000 \cdot 2^1 = 2000$ = 2000 dollars

Monthly: $n = 12$: $1000 \cdot \left(1 + \frac{1}{12}\right)^{12} \approx 1000 \cdot 2.61304$ = about 2613.04 dollars

Daily: $n = 365$: $1000 \cdot \left(1 + \frac{1}{365}\right)^{365} \approx 1000 \cdot 2.71457$ = about 2714.57 dollars

Find the limit: Adding interest every instant gives $1000 \cdot e \approx 1000 \cdot 2.71828$ = about 2718.28 dollars

Answer: Yearly: 2000 dollars, monthly: about 2613.04 dollars, daily: about 2714.57 dollars. No matter how often the interest is added, the amount never passes about 2718.28 dollars, which is $1000 \cdot e$.

Evaluating the Natural Exponential Function

Let $f(x) = e^x$. Using $e \approx 2.71828$ and $e^2 \approx 7.389$, find $f(0)$, $f(1)$, $f(-1)$ and $f(2)$. Round to three decimal places.

Find $f(0)$: Any nonzero number to the power $0$ is $1$: $e^0 = 1$ = $f(0) = 1$

Find $f(1)$: $e^1 = e \approx 2.71828$ = $f(1) \approx 2.718$

Find $f(-1)$: $e^{-1} = \frac{1}{e} \approx \frac{1}{2.71828} \approx 0.36788$ = $f(-1) \approx 0.368$

Find $f(2)$: $e^2 = e \cdot e \approx 2.71828 \cdot 2.71828 \approx 7.389$ = $f(2) \approx 7.389$

Answer: $f(0) = 1$, $f(1) \approx 2.718$, $f(-1) \approx 0.368$, $f(2) \approx 7.389$. Each step of $1$ to the right multiplies the value by $e$.

Where Does y = e^x Lie?

Compare $2^x$, $e^x$ and $3^x$ at $x = 2$ and at $x = -1$. Use $e^2 \approx 7.389$ and $e^{-1} \approx 0.368$.

Evaluate at $x = 2$: $2^2 = 4$, $e^2 \approx 7.389$, $3^2 = 9$ = $4 < 7.389 < 9$

Evaluate at $x = -1$: $2^{-1} = 0.5$, $e^{-1} \approx 0.368$, $3^{-1} \approx 0.333$ = $0.333 < 0.368 < 0.5$

Evaluate at $x = 0$: $2^0 = e^0 = 3^0 = 1$ = All three meet at $(0, 1)$

Describe the graph: Because $2 < e < 3$, the value of $e^x$ is always between the other two = The curve of $e^x$ lies between $2^x$ and $3^x$

Answer: For $x > 0$, $e^x$ is above $2^x$ and below $3^x$; for $x < 0$ the order flips, but $e^x$ is still in the middle. All three curves cross the y-axis at $1$ and have the x-axis as horizontal asymptote.

A First Look at Continuous Compounding

1000 dollars is invested at 12% per year for one year. Compare compounding yearly, monthly, daily and continuously. For continuous compounding use $A = Pe^{rt}$ and $e^{0.12} \approx 1.1275$.

Yearly: $1000 \cdot 1.12 = 1120$ = 1120.00 dollars

Monthly: $1000 \cdot \left(1 + \frac{0.12}{12}\right)^{12} = 1000 \cdot 1.01^{12} \approx 1126.83$ = 1126.83 dollars

Daily: $1000 \cdot \left(1 + \frac{0.12}{365}\right)^{365} \approx 1127.47$ = 1127.47 dollars

Continuously: $A = Pe^{rt} = 1000 \cdot e^{0.12 \cdot 1} \approx 1000 \cdot 1.1275$ = 1127.50 dollars

Answer: The amounts are 1120.00, 1126.83, 1127.47 and 1127.50 dollars. More frequent compounding helps, but the gains get tiny, and the ceiling is $Pe^{rt}$. The lesson Continuous Growth and Decay uses this formula in detail.

Mistake: Thinking $\left(1 + \frac{1}{n}\right)^n$ grows without limit, or that it shrinks to $1$

Why: Two effects compete: $1 + \frac{1}{n}$ gets closer to $1$, but the exponent $n$ gets bigger. Neither effect wins completely.

Correct: Check with a table: $2$, $2.25$, $2.613$, $2.7146$, $2.71828$. The values increase but level off at $e \approx 2.71828$.

Mistake: Treating $e$ as a variable, like $x$

Why: Letters in algebra usually stand for unknowns, so students try to solve for $e$ or substitute other numbers for it.

Correct: $e$ is a fixed constant, just like $\pi$. In $y = e^x$ the variable is $x$; $e \approx 2.71828$ never changes.

Mistake: Computing $e^2$ as $2e \approx 5.437$

Why: An exponent means repeated multiplication, not multiplication by the exponent.

Correct: $e^2 = e \cdot e \approx 2.71828 \cdot 2.71828 \approx 7.389$.

Mistake: Writing $e = 2.71828$ or $e = 2.72$ as if it were exact

Why: $e$ is irrational, so no finite decimal and no fraction is exactly equal to it.

Correct: Write $e \approx 2.71828$ and round only at the end of a calculation, to the accuracy the question asks for.

Banks and Compounding Frequency

Banks state an annual rate, but they may add interest yearly, monthly or daily. The more often interest is added, the closer the result gets to the continuous limit based on $e$.

At 12% per year, 1000 dollars grows to 1120.00 dollars with yearly compounding and to about 1127.50 dollars with continuous compounding ($1000 \cdot e^{0.12}$).

Growing Cell Cultures

Cells in a lab culture divide at all times, not once a day, so biologists describe their growth with base $e$: $N = N_0 e^{kt}$.

A culture of 200 cells modeled by $N = 200e^{t}$ ($t$ in hours) has $200e^{1} \approx 544$ cells after 1 hour.

Gift Exchanges and 1/e

In a gift exchange where everyone draws a name from a hat, the chance that nobody draws their own name is almost exactly $\frac{1}{e}$ once the group has more than a few people.

For a class of 25 students, the chance that nobody draws their own name is about $0.368$, so a redraw is needed in about 63% of cases.

Q: Why is e called the natural base?

A: Because it appears on its own in nature and finance whenever growth happens all the time instead of in steps. In calculus, $e^x$ is the only exponential function whose slope at every point equals its own value, which makes formulas with base $e$ the simplest.

Q: Who discovered e?

A: Jacob Bernoulli found it around 1683 while studying compound interest, exactly as in this lesson. Leonhard Euler later named it $e$, proved many of its properties and computed it to many decimals, which is why it is called Euler's number.

Q: Is e the same as Euler's constant?

A: No. Euler's number $e \approx 2.71828$ is the base of natural exponential functions. Euler's constant $\gamma \approx 0.5772$ is a different number from number theory. Only $e$ is used in this topic.

Q: How do I use e on a calculator?

A: Most scientific calculators have an $e^x$ key. The calculator in this lesson has $e$ too: type e^0.5 for $e^{0.5}$, which gives about $1.6487$. Without an $e$ key, type 2.71828 instead, for example 2.71828^0.5.