The Number e and Continuous Growth

You’ll be able to

Explain where e comes from as a limit of increasingly frequent compounding
Use both the periodic and continuous compound-interest formulas correctly
Convert between a growth rate expressed as a base and one expressed as a continuous rate

Compounding more often, but not without limit

Invest $1 at a nominal 100% annual rate. Compounded once you have $2. Compounded quarterly you have (1 + 1/4)⁴ ≈ $2.441. Monthly gives (1 + 1/12)¹² ≈ $2.613, daily ≈ $2.7146, and every second ≈ $2.71828. The values increase but converge — they do not run away. That ceiling is e ≈ 2.718281828, and it is defined exactly as the limit of (1 + 1/n)ⁿ as n grows without bound.

Compound interest, periodic and continuous

A = P(1 + r/n)^(nt) · A = P·e^(rt) as n → ∞

r is the nominal annual rate as a decimal, n the compoundings per year, t the years. The second formula is the first one's limit, not a different model.

Worked example

Invest $2,000 at a nominal 6% for 10 years. Compare quarterly compounding with continuous compounding.

1.Quarterly: n = 4, so A = 2000(1 + 0.06/4)^(4·10) = 2000(1.015)⁴⁰.
2.(1.015)⁴⁰ ≈ 1.8140, so A ≈ 2000(1.8140) ≈ 3628.04.
3.Continuous: A = 2000·e^(0.06·10) = 2000·e^(0.6).
4.e^(0.6) ≈ 1.8221, so A ≈ 2000(1.8221) ≈ 3644.24.

Answer: Quarterly gives about $3,628.04 and continuous about $3,644.24 — a difference of only $16.20 over ten years, about 0.4%. Continuous compounding is the ceiling, and quarterly is already close to it; the gap between compounding schedules is far smaller than most people expect.

Two ways to write the same growth

A quantity growing 7% per year can be written as (1.07)^t or as e^(kt). Both are legitimate; they are related by 1.07 = e^k, so k = ln(1.07) ≈ 0.0677. The base form makes the per-period rate obvious, and the e form makes the continuous rate obvious. Note that k ≈ 6.77% is slightly less than 7% — a smaller continuous rate compounds to the same annual result, because it is compounding constantly.

Converting between base and continuous rate

b^t = e^(kt) where k = ln b · equivalently b = e^k

k > 0 means growth (b > 1), k < 0 means decay (0 < b < 1). k is often called the continuous rate or the relative growth rate.

Checkpoint

A culture grows according to N(t) = 500·e^(0.23t) with t in hours. What is the hourly percent growth rate?

What makes e the natural choice

Among all exponential functions, y = eˣ is the unique one whose rate of change at every point equals its own value. Its graph has slope exactly 1 where it crosses the y-axis, while 2ˣ has slope about 0.69 and 3ˣ about 1.10 there. That property is why e appears throughout science without anyone choosing it: any process whose rate is proportional to its current amount — radioactive decay, cooling, unchecked population growth, continuously compounded interest — is described most simply in base e.

Tip

Worth memorizing: e ≈ 2.718, e² ≈ 7.389, e^0.5 ≈ 1.649, ln 2 ≈ 0.693, ln 10 ≈ 2.303. These make it easy to spot a calculator entry error before it propagates through a whole problem.

Worked example

A sample of 80 mg of a radioisotope decays to 62 mg in 5 days. Write a continuous-decay model and find the daily percent loss.

1.Model: A(t) = 80·e^(kt), with k negative for decay.
2.Use the data point: 62 = 80·e^(5k), so e^(5k) = 62/80 = 0.775.
3.Take logs: 5k = ln(0.775) ≈ −0.25489, so k ≈ −0.050978.
4.Daily factor: e^(−0.050978) ≈ 0.95030.
5.So each day about 95.03% remains, meaning about 4.97% is lost.

Answer: A(t) = 80·e^(−0.0510t) mg with t in days, losing about 4.97% per day. Equivalently A(t) = 80·(0.9503)^t — the same function written with the daily factor as the base.

Checkpoint

Which is the largest after 20 years for a $1,000 deposit at a nominal 5%?

Answer the 2 checkpoints as you read.