The number e
Tonight's idea
A bank offers 100% interest a year on 100 dollars. After a year you have 200. Pay 50% twice a year instead: \(100 \times 1.5^2 = 225\). Compounding more often pays more, because the second half-year's interest is earned on a bigger pile. Quarterly: \(100 \times 1.25^4 = 244.14\). Monthly: 261.30. Daily: 271.46.
| \(n\) times a year | 1 | 2 | 4 | 12 | 52 | 365 | 1000 | 1,000,000 |
|---|---|---|---|---|---|---|---|---|
| \((1 + 1/n)^n\) | 2 | 2.25 | 2.4414 | 2.6130 | 2.6926 | 2.7146 | 2.7169 | 2.71828 |
The numbers climb but close in on something just under 2.72:
\(e = \lim_{n \to \infty}\left(1 + \frac1n\right)^n = 2.718281828459\ldots\) (read: the number that \((1 + 1/n)^n\) gets ever closer to as \(n\) grows).
Interest at rate \(r\) for \(t\) years, compounded \(n\) times a year, gives \(P(1 + r/n)^{nt}\); let \(n \to \infty\) and it becomes \(Pe^{rt}\), continuous compounding. Populations, radioactive decay and cooling coffee all follow \(e^{rt}\), because their rate of change is proportional to the amount present.
A second route to \(e\) is the series \(e = 1 + 1 + \frac1{2!} + \frac1{3!} + \cdots\), whose terms shrink so fast that seven of them give 2.71806.
The real reason \(e\) matters: the slope of \(b^x\) at \(x = 0\) is about 0.693 for \(b = 2\) and 1.099 for \(b = 3\). For \(b = e\) it is exactly 1, and that makes \(e^x\) its own slope everywhere (exercise 4, and the next sheet).
Watch one: $1000 at 5% for 10 years
Yearly: \(1000 \times 1.05^{10} = 1628.89\). Monthly: \(1000(1 + 0.05/12)^{120} = 1647.01\). Continuously: \(1000e^{0.5} = 1648.72\). Monthly to continuous adds under two dollars.
Rule of 72. Money at \(r\%\) doubles in roughly \(72/r\) years. Continuous growth needs \(e^{rt} = 2\), so \(t = \ln 2/r = 0.693/r\), or \(69.3/r\) with \(r\) in percent. 72 is used because it has more divisors.
Your turn
- 100 dollars at 100% compounded every hour (\(n = 8760\)). Compute it and compare with the continuous answer \(100e\).
- Find the exact doubling time at 3%, 6% and 12% yearly compounding (solve \(1.03^t = 2\): \(t = \ln 2 / \ln 1.03\)) and compare each with the rule of 72.
- Add up \(1 + 1 + \frac1{2!} + \frac1{3!} + \cdots\) one term at a time. How many terms before the sum agrees with \(e = 2.71828\) to five decimal places?
- Show that \(e^x\) is its own slope: write the slope at \(x\) as \(\dfrac{e^{x+h} - e^x}{h}\), factor out \(e^x\), and use \((e^h - 1)/h \to 1\). Test it at \(x = 2\): compute \((e^{2.001} - e^2)/0.001\) and compare with \(e^2\).
The grade 6 sheet A line, a curve and a rocket raced linear against exponential growth. Tonight's \(e\) is the engine inside that rocket.
Stretch: the hat-check problem
\(n\) people hand in their hats, which are returned at random. For \(n = 3\), list the 6 orderings and find the 2 where nobody gets their own hat. For \(n = 4\), find the 9 such orderings out of 24. The exact probability is \(1 - 1 + \frac1{2!} - \frac1{3!} + \cdots \pm \frac1{n!}\). Compare with the series for \(e^{-1}\): what does it approach?
Answers
- \(100(1 + 1/8760)^{8760} = 271.81\), against \(100e = 271.83\). Two cents short.
- 3%: \(\ln 2 / \ln 1.03 = 23.45\) years (rule: 24). 6%: 11.90 (rule: 12). 12%: 6.12 (rule: 6). The rule is slightly generous at low rates and slightly stingy at high ones.
- Partial sums: 1, 2, 2.5, 2.66667, 2.70833, 2.71667, 2.71806, 2.71825, 2.71828. The sum through \(1/8!\) (nine terms) is the first to round to 2.71828.
- \(\dfrac{e^{x+h} - e^x}{h} = e^x \cdot \dfrac{e^h - 1}{h} \to e^x\). At 2: \((e^{2.001} - e^2)/0.001 = 7.3928\), and \(e^2 = 7.3891\). The gap shrinks as \(h\) does.
Stretch. \(n = 3\): 231 and 312 (231 means person 1 gets hat 2, person 2 gets hat 3, person 3 gets hat 1), so \(2/6 = 0.333\). \(n = 4\): 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321, so \(9/24 = 0.375\). The series is \(e^x\) at \(x = -1\), so the probability approaches \(1/e = 0.3679\); already \(44/120 = 0.3667\) at \(n = 5\). For any party bigger than a handful, the chance nobody gets their own hat is about 36.8%.