The Story of e:
The number \(e\) is a fundamental constant that appears naturally when we study systems that grow continuously, like populations, radioactive decay, or compound interest. It isn’t just a random number; it is an observed phenomenon of nature.
Let’s build the intuition for \(e\) with a simple story about earning interest.
1. The Scenario: The “Once-a-Year” Bank
Imagine you invest €1 in a bank that offers a fantastic 100% annual interest rate.
If the interest is applied only once at the end of the year, the calculation is simple: You get your initial €1 back, plus 100% of €1 (which is €1) in interest.
\[\text{Total} = 1 \times (1 + 100\%) = 1 \times (1 + 1) = 2\]You double your money. But what if we could get that interest sooner and reinvest it?
The Power of Compounding
Now, let’s say the bank agrees to compound the interest more frequently. Instead of giving you 100% at the end of the year, they’ll split the rate and apply it periodically.
Compounding Twice a Year: The bank applies half the interest (50%) every 6 months.
- First 6 months: You get \(1 \times (1 + 0.50) = 1.50\).
- Last 6 months: You earn 50% on your new total: \(1.50 \times (1 + 0.50) = 2.25\).
To generalize this behavior, let:
- \(y\) be the initial amount.
- \(x\) be the total growth rate for the year.
- \(n\) be the number of compounding periods.
Period 1: We take the starting amount (\(y\)) and add the interest for this first slice of time (\(\frac{x}{n}\)).
\[\text{Total}_1 = y + y\left(\frac{x}{n}\right) = y\left(1 + \frac{x}{n}\right)\]Period 2: We take the new total from Period 1 and apply the same interest rate to it.
\[\text{Total}_2 = \underbrace{y\left(1 + \frac{x}{n}\right)}_{\text{Previous Total}} + \underbrace{\left[ y\left(1 + \frac{x}{n}\right) \right] \cdot \frac{x}{n}}_{\text{New Interest}}\]Factoring out the common term \(y\left(1 + \frac{x}{n}\right)\):
\[\text{Total}_2 = y\left(1 + \frac{x}{n}\right) \cdot \left(1 + \frac{x}{n}\right) = y\left(1 + \frac{x}{n}\right)^2\]Period 3: Following the pattern, we multiply the previous total by the growth factor again.
\[\text{Total}_3 = y\left(1 + \frac{x}{n}\right)^2 \cdot \left(1 + \frac{x}{n}\right) = y\left(1 + \frac{x}{n}\right)^3\]Conclusion: For \(n\) periods, the formula naturally becomes:
\[\text{Total} = y \times \left(1 + \frac{x}{n}\right)^n\]Applying this back to our specific numbers (\(y=1, x=1, n=2\)):
\[1 \times \left(1 + \frac{1}{2}\right)^2 = 2.25\]By compounding more frequently, you earned an extra €0.25! This is “interest earning interest.”
2. The General Formula
Let’s generalize this. If we split the 100% annual rate into \(n\) equal periods, the interest rate per period is \(1/n\), and we compound it \(n\) times.
The formula for the total amount after one year becomes:
\[\text{Total} = \left(1 + \frac{1}{n}\right)^n\]What happens as we increase \(n\) and compound more and more frequently?
| Frequency | n | Calculation | Result |
|---|---|---|---|
| Annually | 1 | \((1 + 1/1)^1\) | 2.00000 |
| Semi-annually | 2 | \((1 + 1/2)^2\) | 2.25000 |
| Quarterly | 4 | \((1 + 1/4)^4\) | 2.44140 |
| Monthly | 12 | \((1 + 1/12)^{12}\) | 2.61303 |
| Daily | 365 | \((1 + 1/365)^{365}\) | 2.71456 |
| Every Second | 31m | \((1 + 1/31m)^{31m}\) | 2.71828 |
The Limit: Continuous Growth
Notice a pattern? As \(n\) gets larger, the growth slows down. Even though we are compounding millions of times, the total doesn’t shoot off to infinity. Nature puts a halt.
This “ceiling” is the special number \(e\). It represents the maximum possible efficiency of 100% growth applied continuously.
\[e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828...\]3. Generalizing to Any Rate (\(e^x\))
We found \(e\) assuming a 100% growth rate. But what if the rate is different, say \(x\) (where \(x=0.05\) for 5%, or \(x=2\) for 200%)?
We start with the same discrete formula, but now we use rate \(x\):
\[\text{Growth} = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n\]To see how this connects to \(e\), we use a substitution strategy.
Step A: The Substitution
We want the inside to look like the definition of \(e\) (\(1 + 1/m\)). Let’s define a new variable \(m\):
\[\frac{x}{n} = \frac{1}{m} \implies n = m \cdot x\]Step B: The Switch
Substitute these back into the original equation:
- Inside the bracket: \(\frac{x}{n}\) becomes \(\frac{1}{m}\)
- The exponent: \(n\) becomes \(m \cdot x\)
Step C: The Power Rule
Using the exponent rule \((a^{bc}) = (a^b)^c\), we can split the exponent to isolate the “growth engine” from the “rate”:
\[\text{Expression} = \left[ \left(1 + \frac{1}{m}\right)^m \right]^x\]Step D: The Result
As \(n \to \infty\), \(m\) also goes to \(\infty\). The term inside the brackets is the exact universal definition of \(e\).
\[\lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m = e\]Therefore, the entire expression simplifies to:
\[\text{Continuous Growth} = e^x\]Summary
- \(e\): The base unit of continuous change (approx 2.718).
- \(x\): The rate or intensity of that change.
- \(e^x\): Scaling that continuous change by your specific rate.