Unit 3, lesson 37

The Natural Base e

Where the number e comes from, how compounding interest more and more often leads to it and to continuous growth, and why a table of values suggests a limit but cannot prove it.

  • Theory part 73
  • Practice part 74
  • Exponentials, e, and Logarithms

Put 1 dollar in an account that pays 100 percent interest per year. Paid once at the end of the year, the interest leaves you with 2 dollars. Paid in two halves, it leaves you with 2.25 dollars, and paid every day, with about 2.71 dollars. The total keeps growing as the payments get more frequent, yet it never passes a certain number near 2.718. That number is ee, and it turns out to be the most natural base for exponential functions.

Key terms

Compound interest

Definition Interest is compounded when the interest earned in each period is added to the balance, so that later interest is computed on the larger balance. If a principal P>0P > 0 earns interest at an annual rate r>0r > 0, written as a decimal, and the interest is compounded once a year, then after tt whole years the balance is

A=P(1+r)t.A = P(1 + r)^t.

Each year multiplies the balance by the growth factor 1+r1 + r.

In plain words Interest earns interest. The balance grows by the same percentage every year, so it grows exponentially, not linearly.

Example Put 1000 dollars in an account at 5 percent, compounded yearly. After 3 years the balance is

A=1000(1.05)3=1157.625,A = 1000(1.05)^3 = 1157.625,

about 1157.63 dollars. Simple interest, paid only on the original 1000 dollars, would give 1000+3⋅50=11501000 + 3 \cdot 50 = 1150 dollars. The extra 7.63 dollars is interest on interest.

Finite compounding

Definition Suppose the annual rate rr is compounded nn times per year, where nn is a positive integer. Each period then applies the rate rn\frac{r}{n}, and after tt years, that is after ntnt whole periods, the balance is

A=P(1+rn)nt.A = P\left(1 + \frac{r}{n}\right)^{nt}.

The number rr is called the nominal annual rate. Monthly compounding has n=12n = 12, quarterly has n=4n = 4 and daily is usually taken as n=365n = 365.

In plain words Split the year into nn equal pieces and charge a small share of the rate in each piece. More pieces give interest more chances to earn interest.

Example Here is 1000 dollars at a nominal rate of 6 percent after one year, so t=1t = 1 and the balance is 1000(1+0.06/n)n1000(1 + 0.06/n)^n.

CompoundingnnBalance in dollars
yearly111060.001060.00
quarterly441061.361061.36
monthly12121061.681061.68
daily3653651061.831061.83

Going from yearly to quarterly adds 1.36 dollars. Going from monthly to daily adds only 15 cents. Each step up in frequency helps less than the one before.

Numerical convergence

Definition An infinite list of numbers a1,a2,a3,…a_1, a_2, a_3, \dots converges to a real number LL if the distance ∣an−L∣\lvert a_n - L \rvert becomes smaller than any tolerance you choose and stays smaller, as long as nn is large enough. More precisely, for every ε>0\varepsilon > 0 there is a number NN such that ∣an−L∣<ε\lvert a_n - L \rvert < \varepsilon for every n>Nn > N. The number LL is the limit, and we write

lim⁡n→∞an=L.\lim_{n \to \infty} a_n = L.

Watching the computed values ana_n for larger and larger nn, and seeing more and more of their leading digits stop changing, is called observing numerical convergence.

In plain words The terms close in on one number and never wander away from it again. You can see this happening in a table as the leading digits settle down.

Example Take the 1 dollar account at 100 percent interest, compounded nn times in one year. Its balance is an=(1+1/n)na_n = (1 + 1/n)^n.

nn(1+1/n)n(1 + 1/n)^n rounded
1122
10102.593742.59374
1001002.704812.70481
100010002.716922.71692
10410^42.718152.71815
10510^52.718272.71827
10610^62.718282.71828

The values increase, and each time nn grows by a factor of 10 the distance to the number they approach shrinks by about a factor of 10, so roughly one more decimal place settles. The convergence is real but slow.

The number e

Definition The number ee is the limit

e=lim⁡n→∞(1+1n)n,e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n,

where nn runs through the positive integers. The limit exists because the values (1+1/n)n(1 + 1/n)^n increase with nn and all of them stay below 33, and an increasing list of real numbers that stays below a fixed bound always converges. Since e>1e > 1, the function f(x)=exf(x) = e^x is an increasing exponential function with domain all real numbers and range (0,∞)(0, \infty). It is called the natural exponential function. Books introduce ee in different ways, and this course defines it by the limit above. The calculus course starts instead from another property of the same number, that (eh−1)/h(e^h - 1)/h approaches 11 as hh approaches 00, in the lesson on special limits, and then proves the limit above, so both routes lead to the same number.

In plain words The number ee is what 1 dollar grows to in one year at 100 percent interest when the interest is compounded more and more often, with no upper limit on the frequency.

Example With P=1P = 1, r=1r = 1 and t=1t = 1, the compound interest formula gives exactly (1+1/n)n(1 + 1/n)^n. Monthly compounding pays

(1+112)12≈2.61304\left(1 + \frac{1}{12}\right)^{12} \approx 2.61304

dollars, and daily compounding pays

(1+1365)365≈2.71457\left(1 + \frac{1}{365}\right)^{365} \approx 2.71457

dollars. Even compounding every day leaves you less than 0.004 dollars short of ee.

Continuous compounding

Definition For a fixed principal PP, rate r>0r > 0 and time t>0t > 0, the balances under finite compounding approach a limit as the number of periods per year grows without bound,

lim⁡n→∞P(1+rn)nt=Pert.\lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt} = Pe^{rt}.

Continuous compounding is the model in which the balance after tt years is

A=Pert.A = Pe^{rt}.

To see where ee enters, put m=n/rm = n/r. Then

(1+rn)nt=[(1+1m)m]rt,\left(1 + \frac{r}{n}\right)^{nt} = \left[\left(1 + \frac{1}{m}\right)^{m}\right]^{rt},

and as nn grows so does mm, which pushes the bracket toward ee. The number mm need not be a whole number. That the bracket still approaches ee in that case is proved in calculus in the lesson on limits at infinity, and from there the whole power approaches erte^{rt}.

In plain words Continuous compounding is the ceiling that ever more frequent compounding approaches. It replaces the base 1+r/n1 + r/n and the exponent ntnt with the single clean expression erte^{rt}.

Example Compounding 1000 dollars continuously at 6 percent for one year gives

A=1000e0.06≈1061.84 dollars.A = 1000e^{0.06} \approx 1061.84 \text{ dollars.}

Daily compounding gave 1061.83 dollars, so the step from daily to continuous compounding is worth about half a cent.

The value of e

Definition To fifteen decimal places

e=2.718281828459045…,e = 2.718281828459045\ldots,

and e≈2.71828e \approx 2.71828 to five decimal places. The number ee is irrational, so it is not a ratio of two integers and its decimal expansion never ends and never repeats. That is a theorem with a proof, not something any table of digits could show. Every decimal you write down for ee is an approximation. A much faster way to compute it than compounding is the infinite sum

e=1+1+12!+13!+14!+⋯ ,e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots,

where k!=1⋅2⋅3⋯kk! = 1 \cdot 2 \cdot 3 \cdots k.

In plain words The number ee is a little less than 2.722.72. When you see 2.718282.71828, read it as a rounded stand-in for ee, not as ee itself.

Example The first eight terms of the sum give

1+1+12+16+124+1120+1720+15040=685252≈2.718254,\begin{aligned} &1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} \\[0.8em] &\quad + \frac{1}{5040} = \frac{685}{252} \approx 2.718254, \end{aligned}

which already agrees with ee to four decimal places. To get that close with (1+1/n)n(1 + 1/n)^n you need n≥48784n \ge 48784.

Approximation versus proof

Definition A table of computed values is numerical evidence. It can suggest that a limit exists and suggest its first digits, but it cannot prove either, because it shows only finitely many terms and each computed value may carry rounding error. A proof is an argument that covers every nn at once. For ee, what is proved is that for every positive integer nn

(1+1n)n<(1+1n+1)n+1<3.\begin{aligned} \left(1 + \frac{1}{n}\right)^n &< \left(1 + \frac{1}{n + 1}\right)^{n + 1} \\[0.8em] &< 3. \end{aligned}

Together with the completeness of the real numbers, one form of which says that an increasing list with an upper bound has a limit, this guarantees that ee exists. Both inequalities can be proved with the binomial theorem, from the later lesson on the binomial theorem and Pascal’s triangle.

In plain words A table tells you where to look. Only a proof tells you that the answer is really there.

Example A typical computer program that keeps about 16 significant digits returns about 3.0353.035 for (1+1/n)n(1 + 1/n)^n with n=1015n = 10^{15}, which is above the proven bound 33. The reason is that it stores 1+10−151 + 10^{-15} as roughly 1+1.11×10−151 + 1.11 \times 10^{-15}, and raising that slightly wrong number to the power 101510^{15} gives about e1.11≈3.035e^{1.11} \approx 3.035. With n=1016n = 10^{16} the same computer returns exactly 11, because 1+10−161 + 10^{-16} rounds to 11 before the power is taken. Both outputs are wrong. The proof shows that every value is at least 22 and below 33, so neither 3.0353.035 nor 11 is possible, however confident the computer looks.

Check yourself

  1. Find the balance after 2 years when 2000 dollars earns a nominal rate of 3 percent compounded monthly, and compare it with continuous compounding.
Show the answer

Monthly compounding has n=12n = 12 and nt=24nt = 24 periods, so

A=2000(1+0.0312)24=2000(1.0025)24≈2123.51 dollars.\begin{aligned} A &= 2000\left(1 + \frac{0.03}{12}\right)^{24} \\[0.8em] &= 2000(1.0025)^{24} \\[0.4em] &\approx 2123.51 \text{ dollars.} \end{aligned}

Continuous compounding gives 2000e0.06≈2123.672000e^{0.06} \approx 2123.67 dollars, about 16 cents more.

  1. How much must you deposit today, compounded continuously at 5 percent, to have 10000 dollars after 10 years?
Show the answer

The deposit PP must satisfy Pe0.05⋅10=10000Pe^{0.05 \cdot 10} = 10000, so

P=10000e−0.5≈6065.31 dollars.P = 10000e^{-0.5} \approx 6065.31 \text{ dollars.}
  1. A classmate computes (1+1/n)n(1 + 1/n)^n for n=10,100,1000n = 10, 100, 1000, rounds the results to 2.592.59, 2.702.70 and 2.722.72, and concludes that the limit is 2.722.72. What is wrong with this reasoning?
Show the answer

Three rounded values cannot settle the limit. The value at n=1000n = 1000 is 2.716922.71692, which only rounds to 2.722.72, and the limit e=2.71828…e = 2.71828\ldots is not 2.722.72 either. A table only suggests where the values are heading, and a proof that the values increase and stay below 33 is what shows that a limit exists at all.

How this lesson connects