What Is e, and Why Does Growth Keep Finding It?

Here is the rule your cousin used. If the 100% is added n times a year, each time it is $\tfrac{100}{n}$%, so the money is multiplied by $1 + \tfrac{1}{n}$, and that happens n times:

$$\text{after one year} = 1{,}00{,}000 \times \left(1 + \frac{1}{n}\right)^{n}$$

This is Grade 10’s half-yearly and quarterly interest carried much further. Work out the first few rows yourself, then look at the table.

Interest added	n	After one year	More than the row above
every year	1	Rs 2,00,000.00
every half-year	2	Rs 2,25,000.00	Rs 25,000.00
every quarter	4	Rs 2,44,140.63	Rs 19,140.63
every month	12	Rs 2,61,303.53	Rs 17,162.90
every week	52	Rs 2,69,259.70	Rs 7,956.17
every day	365	Rs 2,71,456.75	Rs 2,197.05
every hour	8,760	Rs 2,71,812.67	Rs 355.92
every minute	5,25,600	Rs 2,71,827.92	Rs 15.26
every second	3,15,36,000	Rs 2,71,828.18	Rs 0.25

• a.From monthly to every second, n grows more than 26 lakh times. How much does the money grow?
• b.Look at the last column. Does it look as if the amounts could ever pass Rs 3,00,000?
• c.Predict: if the interest were added a million times every second, what would the amount be, to the nearest rupee?

A question about a never-ending process (“what if n keeps growing?”) has an answer only if the amounts close in on one number. Here they do, and that number, about 2.71828, is called e. But a table can’t prove there is a ceiling. Perhaps the amounts creep past Rs 3,00,000 when n is a crore crore. To be sure, we need to see where the money comes from.

Take the half-yearly case and multiply it out, as you multiply brackets in algebra (in lakh, so the deposit is 1):

$$\left(1 + \tfrac{1}{2}\right)\left(1 + \tfrac{1}{2}\right) = 1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{4}$$

Each piece is real money with its own story. The 1 is your lakh. The two halves are the plain interest of each half-year. The $\tfrac{1}{4}$ is interest earned on interest: the first half-year’s interest, earning 50% in the second half.

Now do it monthly: twelve brackets, $\left(1 + \tfrac{1}{12}\right)$ each. Multiplying them out means choosing, from every bracket, either the 1 or the $\tfrac{1}{12}$, and adding up every way of choosing. Sort the pieces by how many $\tfrac{1}{12}$s they took:

• Layer 0, no $\tfrac{1}{12}$ at all: just 1, your lakh.
• Layer 1, one $\tfrac{1}{12}$: there are 12 of them, one for each month’s plain interest. Together, $12 \times \tfrac{1}{12} = 1$ lakh. This layer is always exactly one lakh, however often interest is added: it is the 100%.
• Layer 2, two $\tfrac{1}{12}$s: one for every pair of months, the interest of the earlier month earning interest in the later one. There are $\tfrac{12 \times 11}{2} = 66$ pairs (the handshake count: 12 people, each shaking 11 hands, each handshake counted twice), each worth $\tfrac{1}{144}$ lakh. Together $\tfrac{66}{144}$ lakh, Rs 45,833.
• Layer 3, three $\tfrac{1}{12}$s: interest on interest on interest, one for every three months chosen, 220 of them, each $\tfrac{1}{1728}$ lakh: Rs 12,731.

And so on, up to layer 12. With n periods, layer k is the number of ways to choose k of the n periods (Grade 12 writes it ${}^{n}C_{k}$ or $\binom{n}{k}$) times $\left(\tfrac{1}{n}\right)^{k}$. Play with the layers before reading on.

Here is the reason each layer has a cap. Choosing k periods one after another can be done in n(n − 1)(n − 2)⋯(n − k + 1) ways, and each set of k periods is counted once for every order, $k! = k(k-1)\cdots 2 \times 1$ times. So

$$\text{layer } k = \frac{n(n-1)\cdots(n-k+1)}{k!} \times \frac{1}{n^{k}} = \frac{1}{k!} \times \frac{n}{n} \times \frac{n-1}{n} \times \cdots \times \frac{n-k+1}{n}.$$

Every fraction after the $\tfrac{1}{k!}$ is 1 or a little less than 1. So layer k is always less than $\tfrac{1}{k!}$ (except layers 0 and 1, which equal it). As n grows, each fraction like $\tfrac{n-1}{n} = 1 - \tfrac{1}{n}$ creeps up towards 1, so each layer creeps up towards its cap $\tfrac{1}{k!}$ and never reaches it. That is what the widget shows: layer 2 is Rs 25,000 half-yearly, Rs 45,833 monthly, Rs 49,863 daily, always under Rs 50,000.

So the whole amount, in lakh, is always less than the caps added up:

$$1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots$$

And that sum can’t be big. From layer 2 on, each cap is at most half the one before ($k!$ is at least $2^{k-1}$), so the caps add up to less than $1 + 1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$, which is 3 (halves added forever never pass the whole). Rs 3,00,000 can never be reached. Your cousin’s crore was never possible.

The caps add up to a number of their own. Adding the first few:

Caps up to	Sum
1 + 1	2
{} + 1/2	2.5
{} + 1/6	2.666 7
{} + 1/24	2.708 3
{} + 1/120	2.716 7
{} + 1/720	2.718 06
up to $1/10!$	2.718 281 8

This number is e:

$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = 2.718\,281\,828\,459\,045\ldots$$

and the interest puzzle says the same thing another way: $\left(1 + \tfrac{1}{n}\right)^{n}$ gets as close to e as you like when n is big enough, and never passes it. (Why does it get close, not just stay under? Because for a very large n the layers up to layer 10 are all within a hair of their caps, and the caps after them, $1/11! + 1/12! + \cdots$, add up to less than 0.000 000 03.) Like $\pi$, e is irrational: its decimals never end and never repeat.

Adding interest “at every instant” is called continuous compounding, and e is the most that 100% a year can ever give. A lakh at 100% compounded continuously becomes e lakh, Rs 2,71,828.18, in a year.

No bank pays 100%. What happens at a real rate, say 5% a year? Share it out n times: each period multiplies by $1 + \tfrac{0.05}{n}$. The layers work as before, except each chosen period now brings $\tfrac{0.05}{n}$ instead of $\tfrac{1}{n}$, so layer k has a cap of $0.05^{k}/k!$. The ceiling is

$$1 + 0.05 + \frac{0.05^{2}}{2} + \frac{0.05^{3}}{6} + \cdots = e^{0.05}.$$

That last step is worth seeing. Count the periods in a different unit: put n = 0.05m, so each period multiplies by 1 + 1/m and there are 0.05m of them:

$$\left(1 + \frac{0.05}{n}\right)^{n} = \left[\left(1 + \frac{1}{m}\right)^{m}\right]^{0.05},$$

and as m grows the square bracket closes in on e. So the ceiling is $e^{0.05}$, and the layers say that $e^{r}$ is also the sum $1 + r + r^{2}/2! + r^{3}/3! + \cdots$. So a yearly rate r compounded continuously multiplies the money by $e^{r}$ each year, and over t years by $e^{rt}$:

$$A = P e^{rt}.$$

Now the layers on a lakh at 5%: the lakh, Rs 5,000 of plain interest, then at most Rs 125 of interest on interest, Rs 2.08 in layer 3, and 3 paisa in layer 4. So at real rates the extra from compounding often is small.

Rs 1,00,000 after a year, in Rs	at 5%	at 10%	at 36%
yearly	1,05,000.00	1,10,000.00	1,36,000.00
quarterly	1,05,094.53	1,10,381.29	1,41,158.16
monthly	1,05,116.19	1,10,471.31	1,42,576.09
daily	1,05,126.75	1,10,515.58	1,43,307.51
continuous	1,05,127.11	1,10,517.09	1,43,332.94

What the table means: at 5%, daily interest beats yearly by Rs 127 on a lakh, and continuous beats daily by 36 paisa. Once interest is daily, it is continuous for every practical purpose. But at a high rate the gap is real: 36% a year added monthly (3% a month) costs a borrower Rs 42,576 a year on a lakh, not Rs 36,000.

The same trick turns any growth into any other period. Money growing 15% a year is multiplied by $1.15^{t}$ in t years. Since $1.15 = \left(1.15^{1/12}\right)^{12}$,

$$1.15^{t} = \left(1.15^{1/12}\right)^{12t} \approx 1.0117^{12t},$$

so 15% a year is the same as about 1.17% a month, not 15 ÷ 12 = 1.25%. In the same way, every growth can be written with e: $1.15 = e^{0.1398}$, so $1.15^{t} = e^{0.1398\,t}$. The number 0.1398 is the logarithm of 1.15 to the base e, the natural logarithm, written $\ln 1.15$ (logarithms themselves are topics 1 and 2). For any base b, $b^{x} = e^{x \ln b}$.

How long to double?

Grade 10 gave a quick rule: at 12% a year, a sum doubles in about six years. At a rate r compounded continuously, money doubles when $e^{rt} = 2$, that is when $rt = \ln 2 = 0.693$. So the doubling time is 0.693 ÷ r: with the rate R in percent, about 70 ÷ R years.

Rate	doubles, interest yearly	continuous	72 ÷ R	70 ÷ R
2%	35.0 years	34.7	36	35
6%	11.9	11.6	12	11.7
8%	9.0	8.7	9	8.75
12%	6.1	5.8	6	5.8

The rule of 72 is the yearly version: yearly interest is a little slower than continuous, which pushes 69.3 up, and 72 is easy to divide by 2, 3, 4, 6, 8, 9 and 12. It is close for rates around 8%. The rule of 70 suits low rates and anything that grows smoothly, like a population: a country growing 2% a year doubles in about 35 years.

Interest is people’s rule. Why should nature care about e?

Look at what continuous growth means. Money at 100% a year, compounded continuously, is at every moment growing at a speed equal to its size: when there are 2 lakh, it grows at 2 lakh a year; at 2.5 lakh, at 2.5 lakh a year. On a graph, the speed of growth is the steepness of the curve, and the size is its height. Bacteria in a dish of food, with no shortage of anything, behave the same way: twice as many bacteria make twice as many new ones each minute. So do radioactive atoms, the other way round: twice as many atoms means twice as many decaying each second. Whenever a thing grows or shrinks at a rate in step with its own size, steepness is a fixed number times height. Which curve does that?

Try these with the widget.

• a.With b = 2, drag the point from left to right. The height and the steepness both change. What about steepness ÷ height?
• b.Change b. Which base gives steepness ÷ height = 1?
• c.When steepness equals height, where does the tangent meet the axis?

Here is why, with no calculus. Take a tiny step h to the right on $y = b^{x}$. The height goes from $b^{x}$ to $b^{x+h} = b^{x} \times b^{h}$, so it rises by $b^{x}(b^{h} - 1)$, and the steepness is about

$$\text{steepness} \approx \frac{b^{x}(b^{h} - 1)}{h} = \text{height} \times \frac{b^{h} - 1}{h}.$$

The fraction doesn’t depend on x: that is part a. Every exponential curve has its steepness in a fixed ratio to its height. For b = 2 and smaller and smaller steps, $(2^{h} - 1)/h$ is 0.718, 0.696, 0.693… (steps of 0.1, 0.01, 0.001): the ratio is $\ln 2$.

Now part b: which b makes the ratio 1? We need $(b^{h} - 1)/h$ to come as close to 1 as we like as the step h shrinks: $b^{h}$ very nearly 1 + h, so b very nearly $(1 + h)^{1/h}$, and the nearer the tinier h is. Write h = 1/n:

$$b = \left(1 + \frac{1}{n}\right)^{n}.$$

The cooperative’s formula, for n as large as you like, so b is e. Check it from the other side with the sum for $e^{h}$ from the section on real rates: $e^{h} - 1 = h + h^{2}/2 + h^{3}/6 + \cdots$, so $(e^{h} - 1)/h = 1 + h/2 + h^{2}/6 + \cdots$, which closes in on exactly 1 as h shrinks. A bigger base gives a steeper curve and a bigger ratio, so no other base gives 1. The curve whose steepness equals its height is $y = e^{x}$. That is why growth keeps finding e: anything that grows at a rate in step with its size, at every moment rather than in jumps, is described by $e^{kt}$, with k the ratio of its speed to its size (negative for shrinking).

Carbon dating. A plant takes in a little radioactive carbon-14 while it lives. When it dies, the carbon-14 starts to decay, and the decay goes in step with how much is left. Physicists write what is left after t years as $e^{-t/8267}$: 8,267 years is the average lifetime of a carbon-14 atom. Half is left when $\tfrac{t}{8267} = \ln 2$, at 8267 × 0.693 = 5{,}730 years, the half-life, the doubling time in reverse. After 10,000 years, $e^{-10000/8267} = 0.298$: about 30% is left. A lab measures how much is left in a bone or a piece of charcoal and turns the fraction back into years.

Here is a twist where nothing grows. A scratch card wins with a chance of 1 in 50. You buy 50 cards. What is the chance you win nothing?

Each card loses with chance $\tfrac{49}{50} = 1 - \tfrac{1}{50}$, so all 50 lose with chance $\left(1 - \tfrac{1}{50}\right)^{50} = 0.364$. With a 1 in 100 chance and 100 cards, $0.99^{100} = 0.366$; with 1 in 1,000 and 1,000 cards, 0.368. These creep towards $\tfrac{1}{e} = 0.3679$, and for the same reason as the cooperative: $\left(1 - \tfrac{1}{n}\right)^{n}$ is the interest puzzle with the interest taken away instead of added, and it closes in on $e^{-1}$. So “a 1 in n chance, tried n times” still misses altogether about 37% of the time.

$e^{kt}$ describes growth only while the reason for it lasts: “twice as many make twice as many new ones”. Bacteria in a dish run out of food and space, and their curve bends over and flattens. A population’s growth slows as families get smaller. A deposit grows at $e^{rt}$ only while the rate stays fixed, and bank rates change. So an exponential model is a good description of the early part of a story, and a dangerous one to carry on for a hundred years.

• Why Is a Magnitude 8 Earthquake Not Just a Bit Bigger Than a 7? Logarithms as a scale for things that grow by multiplying.
• How Did a Book of Numbers Make Multiplying Easy? Napier’s and Briggs’s logarithm tables, where e was hiding before anyone named it.
• When Will Nepal’s Population Stop Growing? What happens when exponential growth bends over.
• How Much Rice Is on the Chessboard? (Grades 6–8) Doubling, step by step: growth in jumps.
• Can You Add Halves Forever and Still Get One? (Grades 6–8) The endless sum that kept the interest under Rs 3,00,000.
• How Many Handshakes at a Party of 30? (Grades 6–8) The pairs that make layer 2.
• Can a Computer Check Every Possible Chess Game? (Computing, Going Further 6–8) Exponential growth against a computer’s speed. (Coming soon.)
• Grade 10, Compound interest and Growth: the yearly, half-yearly and quarterly rules this page carries on to every instant.