Algebra 2

The number e and the natural exponential function

If a bank added interest not every month or every day but at every instant, how much would a balance grow, and what number sits behind the answer?

You need

Compound interest and compounding periodsAlgebra 2

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Evaluating logarithms, common and naturalAlgebra 2 Derivatives of trig, exponential and log functionsGrade 11

Where you are 6 in Exponential functions

Functions13

Exponential functions6

Exponential growth and decay6 Compound interest and compounding periods Graphs of exponential functions

The number e and the natural exponential function

Transforming an exponential graph Writing an exponential function from a table or graph

Why this exists

Compound interest left a question hanging. A yearly rate r compounded n times a year multiplies a balance by \left(1 + \dfrac{r}{n}\right)^n each year, and that factor grows as n grows, but by less and less each time. Quarterly beats once a year, monthly beats quarterly, daily beats monthly. So what happens if the year is split into more and more periods without end, so that interest is added at every instant? Does the balance grow without limit, or does it settle on some ceiling? And if there is a ceiling, how is it computed?

The answer turns on one number, found by asking the simplest version of the question: 1 dollar at 100\% a year, compounded more and more often. That number is called e. It then turns out to do something no other base does. The curve y = e^{x} climbs at every point at a slope equal to its own height, which is why e, and not 10 or 2, is the base that calculus, physics, biology and finance reach for when something grows or shrinks continuously.

Definition

The number e is the number that

\left(1 + \frac{1}{n}\right)^{n}

settles toward as n grows without bound (written n \to \infty). To nine decimal places,

e \approx 2.718281828

It is irrational: its decimals go on for ever without repeating, so every decimal written for it is an approximation. The letter e was chosen by Leonhard Euler, who worked out many of its properties.

The natural exponential function is the exponential function with base e,

f(x) = e^{x}

sometimes written \exp(x). Like every exponential function with a base above 1, it is defined for every real x, its outputs are always positive, and it increases.

Continuous compounding means compounding at a yearly rate r with the number of periods n allowed to grow without bound. The balance it gives is the ceiling that P\left(1 + \dfrac{r}{n}\right)^{nt} approaches as n \to \infty, and the rate r is then called a continuous rate.

Intuition

Put 1 dollar in an account paying 100\% a year. Compounded once a year it becomes 2 dollars. Compounded twice a year, each half adds 50\%: 1.5^2 = 2.25 dollars. Compounded n times a year it becomes \left(1 + \dfrac{1}{n}\right)^n dollars. Here is what happens as n grows, to six decimal places:

n	\displaystyle \left(1 + \frac{1}{n}\right)^{n}	Short of e by
1	2	0.718282
2	2.25	0.468282
10	2.593742	0.124539
100	2.704814	0.013468
1{,}000	2.716924	0.001358
10{,}000	2.718146	0.000136
1{,}000{,}000	2.718280	0.000001

The values keep rising, so more compounding always helps, but they bunch up under a ceiling. Once n is large, multiplying n by ten cuts the shortfall to about a tenth. No amount of compounding pushes 1 dollar past about 2.72 dollars in a year: the ceiling is e.

Figure 1The values climb but never pass the dashed line at e. The steps along the axis are not equal steps in n, since after 1 and 2 each one multiplies n by ten or more. The vertical axis starts at 1.95.

Now picture the graph. Since 2 < e < 3, the curve y = e^{x} runs between y = 2^{x} and y = 3^{x}, above 2^{x} and below 3^{x} to the right of the y-axis, and the other way round to the left. All three pass through (0, 1), since any base to the power 0 is 1. What singles out e is how steeply its curve passes through that point. At (0, 1) the curve y = 2^{x} climbs a little less steeply than the line y = x + 1, whose slope is 1, and y = 3^{x} a little more steeply. Somewhere between 2 and 3 there is a base whose curve leaves (0, 1) at slope exactly 1, touching the line y = x + 1 there. That base is e.

Figure 2y = e^{x} runs between y = 2^{x} and y = 3^{x}. Of all the exponential curves, it is the one whose steepness at (0, 1) is exactly 1, the slope of the line y = x + 1.

Results

Continuous compounding. A principal P at a continuous rate r a year has after t years the balance

A(t) = Pe^{rt}

For a positive rate and a positive time, this is more than compounding n times a year at the same rate r gives, for every whole n, and it is the ceiling those balances approach as n grows.

The effective rate of a continuous rate. One year multiplies the balance by e^{r}, so the balance can be written A(t) = P\left(e^{r}\right)^t, and the percent the balance really grows in a year is

r_{\text{eff}} = e^{r} - 1

Continuous growth and decay. Any quantity that starts at a and changes continuously at the rate k per unit of time is

y = ae^{kt}

With k > 0 it grows, with k < 0 it decays, and each unit of time multiplies it by the factor e^{k}. The continuous rate k is not the percent change per unit of time; that is e^{k} - 1.

Slope equals height. At every point of y = e^{x}, the slope of the curve (the slope of the line that just touches it there, its tangent line) equals the height e^{x}. At (0, 1) the slope is 1. No other base has this property: e is the one base whose curve has slope 1 where it crosses the y-axis.

Figure 3Slope equals height. Every tangent line to y = e^{x} climbs the height of its point over a run of exactly 1, so it meets the x-axis one unit to the left.

Why it is true

Why \displaystyle \left(1 + \frac{1}{n}\right)^n settles. The lesson on compounding periods showed, using the binomial theorem, that for a positive rate the yearly factor \left(1 + \dfrac{r}{n}\right)^n grows as n grows. Take r = 1: the values always rise. They also never reach 3. In the same expansion, the term with \dfrac{1}{n^k} is \dfrac{1}{k!} times brackets \left(1 - \dfrac{j}{n}\right) that are each at most 1, so it is at most \dfrac{1}{k!}, and k! \geq 2^{k-1} because each factor after the first is at least 2. So

\begin{aligned} \left(1 + \frac{1}{n}\right)^{n} &\leq 1 + 1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-1}} \\ &< 1 + 2 = 3 \end{aligned}

A list of numbers that keeps rising but stays below a fixed ceiling must settle on some number at or below that ceiling. This is a basic property of the real numbers (they have no gaps), and the number these values settle on is e.

Why continuous compounding gives Pe^{rt}. Write m = \dfrac{n}{r}, so that \dfrac{r}{n} = \dfrac{1}{m} and nt = m \cdot rt. By the power of a power rule,

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

As n grows without bound, so does m (because r is fixed and positive), so the inner bracket settles toward e and the whole factor settles toward e^{rt}. The value m need not be a whole number, but the same settling happens for every large m, whole or not.

Why the slope at (0, 1) is 1. For a large n, e is very close to \left(1 + \dfrac{1}{n}\right)^n. Taking the nth root of both, e^{1/n} is very close to 1 + \dfrac{1}{n}. Call \dfrac{1}{n} = h: for a small h, e^{h} is very close to 1 + h, so the line from (0, 1) to (h, e^{h}) has slope

\frac{e^{h} - 1}{h} \approx \frac{h}{h} = 1

and the smaller h is, the closer the slope is to 1. A calculator agrees: with h = 0.001, the slope from (0, 1) is about 0.6934 for base 2, about 1.0992 for base 3, and about 1.0005 for base e.

Why slope equals height everywhere. Take any point (x, e^{x}) and a nearby point (x + h, e^{x + h}). By the product of powers rule and the distributive property,

e^{x + h} - e^{x} = e^{x}e^{h} - e^{x} = e^{x}\left(e^{h} - 1\right)

so the slope of the line through the two points is

\frac{e^{x + h} - e^{x}}{h} = e^{x} \cdot \frac{e^{h} - 1}{h}

As h shrinks the second factor approaches 1, so the slope approaches e^{x}, the height. The same algebra with any base b gives b^{x} times the slope of y = b^{x} at (0, 1); only for b = e is that extra factor exactly 1.

Worked examples

Example 1

Powers of e on a calculator

Find, to four decimal places, e^{1.5}, e^{-2} and e^{0.25}, and say what each means for the graph of y = e^{x}.

Solution

A calculator has an e^{x} key (often on the same key as the natural logarithm, met in the next lessons). Entering the powers,

e^{1.5} \approx 4.4817 \qquad e^{-2} \approx 0.1353 \qquad e^{0.25} \approx 1.2840

Each is the height of the curve at that x: the graph passes through about (1.5, 4.4817), (-2, 0.1353) and (0.25, 1.2840). The second is a reciprocal, by the meaning of a negative exponent: e^{-2} = \dfrac{1}{e^{2}} \approx \dfrac{1}{7.3891}. The third is a fourth root, \displaystyle e^{0.25} = \sqrt[4]{e}, the factor that, used four times, gives e. A quick check that the answers are sensible: e^{1.5} lies between 2^{1.5} \approx 2.83 and 3^{1.5} \approx 5.20, as it must.

Example 2

A deposit compounded continuously

Obiageli deposits 1{,}750 dollars at 6.2\% a year compounded continuously and leaves it for 9 years. Find the balance and the interest earned, to the nearest cent. Compare with 6.2\% compounded monthly and daily (a year of 365 days), and give the effective annual rate.

Solution

Continuous compounding uses A(t) = Pe^{rt} with P = 1750, r = 0.062 and t = 9. The exponent is rt = 0.062 \times 9 = 0.558, so

\begin{aligned} A(9) &= 1750e^{0.558} \\ &\approx 1750 \times 1.747175 \\ &\approx 3057.56 \end{aligned}

The balance is about 3{,}057.56 dollars, of which 3{,}057.56 - 1{,}750 = 1{,}307.56 dollars is interest. Keep e^{0.558} in the calculator rather than rounding it first.

For comparison, the compound interest formula gives 1750\left(1 + \dfrac{0.062}{12}\right)^{108} \approx 3053.17 dollars monthly and 1750\left(1 + \dfrac{0.062}{365}\right)^{3285} \approx 3057.41 dollars daily. Continuous compounding beats monthly by 4.39 dollars but daily by only 0.15 dollars: it is the ceiling, and daily compounding is already close to it. The effective annual rate is e^{0.062} - 1 \approx 0.06396, about 6.396\%, against about 6.379\% for monthly compounding.

Example 3

Comparing a continuous rate with a yearly one

One account pays 5.9\% a year compounded continuously. Another pays 6.05\% a year compounded once a year. Which pays more?

Solution

The nominal rates are on different schedules, so compare effective rates. The continuous account multiplies a balance by e^{0.059} each year, so its effective rate is

e^{0.059} - 1 \approx 0.060775

about 6.078\%. The yearly account’s effective rate is its quoted rate, 6.05\%. The continuous account pays slightly more, even though its quoted rate is lower. On 10{,}000 dollars for a year, the balances are 10000e^{0.059} \approx 10607.75 dollars and 10000 \times 1.0605 = 10605 dollars. Because both yearly factors stay the same every year, the continuous account stays ahead for any whole number of years.

Example 4

Working back to the deposit

How much must be deposited now at 4.8\% a year compounded continuously to have 15{,}000 dollars in 12 years?

Solution

Now A(12) = 15000 is known and P is not. With rt = 0.048 \times 12 = 0.576,

15000 = Pe^{0.576}

Divide both sides by e^{0.576}, which is allowed because a power of e is never zero. Dividing by e^{0.576} is the same as multiplying by e^{-0.576}, by the meaning of a negative exponent:

P = 15000e^{-0.576} \approx 8432.14

About 8{,}432.14 dollars. Check: 8432.14 \times e^{0.576} \approx 8432.14 \times 1.778909 \approx 15000.01, the extra cent coming from rounding P to the cent.

Example 5

A continuous growth model

A culture of bacteria is modeled by N(t) = 800e^{0.35t}, where t is the time in hours. (a) How many bacteria are there at the start, and after 6 hours? (b) By what percent does the culture grow each hour?

Solution

(a) At the start t = 0, and e^{0} = 1, so N(0) = 800. After 6 hours the exponent is 0.35 \times 6 = 2.1, and

N(6) = 800e^{2.1} \approx 800 \times 8.16617 \approx 6533

about 6{,}533 bacteria, rounding to a whole number since bacteria are counted.

(b) One hour multiplies the count by e^{0.35}, since e^{0.35(t + 1)} = e^{0.35t} \cdot e^{0.35} by the product of powers rule. Now e^{0.35} \approx 1.419, so the culture grows by about 41.9\% an hour, not 35\%. The continuous rate 0.35 describes growth that compounds at every instant, and compounding through the hour turns 35\% into about 41.9\%, just as monthly compounding turns a yearly 5.4\% into a little more. In the form N(t) \approx 800(1.419)^t, it is the familiar growth-factor model.

Example 6

A continuous decay model

Air pressure falls with height above sea level. A simple model is p(h) = 101.3e^{-0.12h}, where p is in kilopascals (thousands of pascals, a unit of pressure) and h is the height in kilometers. Find the pressure at a height of 8 kilometers, and the percent the pressure falls with each kilometer of height.

Solution

The exponent is negative, k = -0.12, so this is decay. At h = 8 the exponent is -0.12 \times 8 = -0.96, so

p(8) = 101.3e^{-0.96} \approx 101.3 \times 0.382893 \approx 38.8

about 38.8 kilopascals, under two fifths of the sea-level value of 101.3. Each kilometer multiplies the pressure by e^{-0.12} \approx 0.8869, and a factor of 0.8869 means a loss of 1 - 0.8869 = 0.1131, about 11.3\% per kilometer. As with growth, the continuous rate 0.12 and the percent change per unit, 11.3\%, are different numbers.

Example 7

Tangent lines from slope equals height

Use the fact that the slope of y = e^{x} at any point equals its height to find the tangent lines at x = 0 and at x = 2, and find where each meets the x-axis.

Solution

At x = 0 the point is (0, e^{0}) = (0, 1) and the slope is the height, 1. The line through (0, 1) with slope 1 is

y = x + 1

It meets the x-axis where x + 1 = 0, at x = -1.

At x = 2 the point is (2, e^{2}), about (2, 7.389), and the slope is again the height, e^{2}. In point-slope form,

\begin{aligned} y - e^{2} &= e^{2}(x - 2) \\ y &= e^{2}(x - 2) + e^{2} = e^{2}(x - 1) \end{aligned}

using the distributive property to take out the common factor e^{2}. It meets the x-axis where e^{2}(x - 1) = 0, and since e^{2} \neq 0 that is x = 1.

Both lines meet the axis exactly one unit to the left of the point they touch, and this is no coincidence: a line whose slope equals its height at the point of touching drops that height in a run of exactly 1. The figure in Results shows both. It also shows the curve lying above the first line, touching it only at (0, 1), which says e^{x} \geq x + 1 for every x; a proof of that waits for calculus.

Boundaries

e is never written exactly as a decimal. It is irrational, so 2.718 and even 2.718281828 are approximations. Use the calculator’s e^{x} key and round only at the end; rounding e or a power of it early spoils answers, the more so when the result is multiplied by a large principal.

No whole n reaches e. \left(1 + \dfrac{1}{n}\right)^n is below e for every whole number n; e is where the values settle, not a value they take. In the same way, no real account compounds at every instant: Pe^{rt} is the ceiling that frequent compounding approaches, and a bank that advertises continuous compounding computes with that formula.

The rate and the time must match. In Pe^{rt} and ae^{kt}, r or k is a decimal rate per unit of time, and t must be counted in that same unit: a rate per year with t in years, a rate per hour with t in hours. The continuous rate is not the percent change per unit of time, which is e^{r} - 1; the two are close only when r is small.

Slope equals height belongs to base e alone. For another base the slope of y = b^{x} is its height times a fixed number that is not 1 (about 0.6931 for base 2). And the statement is about y = e^{x} itself: for y = ae^{kx} the slope is k times the height.

Positive outputs only. e^{x} is defined for every real x, but it is never zero or negative, so an equation such as e^{x} = 0 or e^{x} = -4 has no solution.

Where this goes next

Undoing e^{x} asks “what power of e gives this number?”, and the answer is the natural logarithm, met in evaluating logarithms, common and natural. With it, a continuous model such as Pe^{rt} can be solved for the time or the rate, and any exponential ab^{t} can be rewritten with base e. In calculus the slope at a point becomes the derivative, and slope equals height becomes the statement that e^{x} is its own derivative, in derivatives of trig, exponential and log functions; the factor that other bases carry is worked out in derivatives of exponential functions. The settling of \left(1 + \dfrac{1}{n}\right)^n toward e is an early example of a limit, the idea at the center of Limits.

Exercises

Powers of e

Find e^{3} to four decimal places.
Find e^{-0.7} to four decimal places.
Find e^{2/3} to four decimal places.
Find 4e^{0.15} to four decimal places.

Continuous compounding

2{,}400 dollars is deposited at 3.5\% a year compounded continuously. Find the balance after 7 years.
8{,}000 dollars is invested at 5.25\% a year compounded continuously for 20 years. How much interest does it earn?
Find the effective annual rate of 7.5\% a year compounded continuously.
3{,}600 dollars is deposited at 4.2\% a year for 10 years. Find the balance compounded continuously and compounded monthly, and the difference between them.
Which pays more: 4.9\% a year compounded continuously, or 5\% a year compounded once a year? Justify with effective rates.
How much must be deposited now at 4.4\% a year compounded continuously to have 20{,}000 dollars in 18 years?

Continuous growth and decay

The number of fish in a lake is modeled by N(t) = 250e^{0.08t}, with t in years. Find N(5), and the percent the population grows each year.
The mass of a substance, in grams, is m(t) = 60e^{-0.03t}, with t in years. Find m(20), and the percent of the mass lost each year.
1{,}000 dollars is deposited at 2.9\% a year compounded continuously. Write A(t) and find A(15).
An account’s balance after t years is A(t) = 5000e^{0.047t}. Give the principal, the continuous rate and the effective annual rate, and find A(12).
Does y = 3e^{-0.4x} grow or decay as x increases? Find y at x = 0 and at x = 5.

The number e and slope

Find \displaystyle \left(1 + \frac{1}{n}\right)^n for n = 50 and n = 5{,}000 to six decimal places, and how far each is below e.
Find \displaystyle \left(1 + \frac{2}{n}\right)^n for n = 10 and n = 1{,}000 to six decimal places, and compare with e^{2}. What money question does this answer?
Find the tangent line to y = e^{x} at x = -1, and where it meets the x-axis.
With h = 0.0001, find \displaystyle \frac{b^{h} - 1}{h} to five decimal places for b = e, b = 2.5 and b = 2.8. What do the results say about where e lies?

Reasoning

Explain why compounding 6\% a year continuously gives more than compounding it daily, yet does not give unlimited growth.
Show that Pe^{rt} = P\left(e^{r}\right)^t, and explain what e^{r} - 1 means for an account.
Explain why e^{x} lies between 2^{x} and 3^{x} when x > 0, and why the order is reversed when x < 0.
Prove that the tangent line to y = e^{x} at x = a meets the x-axis at x = a - 1, for every a.

Answers

20.0855. The calculator gives e^{3} \approx 20.08554.
0.4966. \displaystyle e^{-0.7} = \frac{1}{e^{0.7}} \approx 0.49659.
1.9477. \displaystyle e^{2/3} = \sqrt[3]{e^{2}} \approx 1.94773.
4.6473. e^{0.15} \approx 1.161834, and 4 \times 1.161834 \approx 4.64734.
3{,}066.29 dollars. 2400e^{0.035 \times 7} = 2400e^{0.245} \approx 3066.29.
14{,}861.21 dollars. The balance is 8000e^{0.0525 \times 20} = 8000e^{1.05} \approx 22861.21, and 22861.21 - 8000 = 14861.21.
About 7.79\%. e^{0.075} - 1 \approx 0.07788.
Continuously 3600e^{0.42} \approx 5479.06 dollars; monthly \displaystyle 3600\left(1 + \frac{0.042}{12}\right)^{120} = 3600(1.0035)^{120} \approx 5475.05 dollars. The difference is 5479.06 - 5475.05 = 4.01 dollars.
4.9\% compounded continuously. Its effective rate is e^{0.049} - 1 \approx 0.05022, about 5.02\%, which beats 5\%.
9{,}058.76 dollars. P = 20000e^{-0.044 \times 18} = 20000e^{-0.792} \approx 9058.76.
N(5) = 250e^{0.4} \approx 372.96, about 373 fish. Each year multiplies the population by e^{0.08} \approx 1.0833, growth of about 8.33\% a year.
m(20) = 60e^{-0.6} \approx 32.93 grams. Each year multiplies the mass by e^{-0.03} \approx 0.9704, so about 2.96\% is lost each year (1 - e^{-0.03} \approx 0.02955).
A(t) = 1000e^{0.029t}; A(15) = 1000e^{0.435} \approx 1544.96 dollars.
Principal 5{,}000 dollars, continuous rate 0.047, that is 4.7\% a year; effective rate e^{0.047} - 1 \approx 0.04812, about 4.81\%. A(12) = 5000e^{0.564} \approx 8788.45 dollars.
It decays, since the rate -0.4 is negative: each unit of x multiplies y by e^{-0.4} \approx 0.67, which is less than 1. y(0) = 3e^{0} = 3 and y(5) = 3e^{-2} \approx 0.4060.
\displaystyle \left(1 + \frac{1}{50}\right)^{50} = 1.02^{50} \approx 2.691588, short of e by about 0.026694. \displaystyle \left(1 + \frac{1}{5000}\right)^{5000} \approx 2.718010, short by about 0.000272. A hundred times the n gives roughly a hundredth of the shortfall.
1.2^{10} \approx 6.191736 and 1.002^{1000} \approx 7.374312, against e^{2} \approx 7.389056. The values rise toward e^{2}. They are what 1 dollar becomes in one year at 200\% a year compounded n times, so the question is what continuous compounding at 200\% gives in a year: e^{2} dollars, by A = Pe^{rt} with r = 2, t = 1.
y = e^{-1}(x + 2), meeting the x-axis at x = -2. The point is (-1, e^{-1}) and the slope is the height e^{-1} \approx 0.3679, so y - e^{-1} = e^{-1}(x + 1), which is y = e^{-1}(x + 2); this is zero only at x = -2.
About 1.00005 for e, 0.91633 for 2.5, and 1.02967 for 2.8. Base 2.5 leaves (0, 1) at a slope below 1 and base 2.8 above it, and the slope from (0, 1) grows with the base, so the base with slope exactly 1, which is e, lies between 2.5 and 2.8.
Continuous compounding is the ceiling that \displaystyle \left(1 + \frac{0.06}{n}\right)^n climbs toward as n grows, and the values climb with every increase in n, so the ceiling e^{0.06} is above the daily value \displaystyle \left(1 + \frac{0.06}{365}\right)^{365}. But the values settle rather than growing without end: by the substitution \displaystyle m = \frac{n}{0.06} the factor is \displaystyle \left(\left(1 + \frac{1}{m}\right)^m\right)^{0.06}, and the inner bracket never passes e. So a year at 6\% can never multiply a balance by more than e^{0.06} \approx 1.0618.
By the power of a power rule, \left(e^{r}\right)^t = e^{r \cdot t} = e^{rt}, so P\left(e^{r}\right)^t = Pe^{rt}. Read this way, e^{r} is the factor that multiplies the balance each whole year, and a factor 1 + q means growth of q, so e^{r} - 1 is the percent the balance really grows each year: the effective annual rate.
For x > 0, raising a larger positive base to the same positive power gives a larger result, and 2 < e < 3, so 2^{x} < e^{x} < 3^{x}. For x < 0, write x = -s with s > 0: then \displaystyle 2^{x} = \frac{1}{2^{s}}, \displaystyle e^{x} = \frac{1}{e^{s}} and \displaystyle 3^{x} = \frac{1}{3^{s}}, and since 2^{s} < e^{s} < 3^{s} and taking reciprocals of positive numbers reverses an inequality, 3^{x} < e^{x} < 2^{x}.
At x = a the point is (a, e^{a}) and, since slope equals height, the slope is e^{a}. The tangent line is y - e^{a} = e^{a}(x - a), so y = e^{a}(x - a + 1). It meets the x-axis where e^{a}(x - a + 1) = 0; since e^{a} is never zero, x - a + 1 = 0, so x = a - 1.