Case study: continuous compounding as a limit

Continuous compounding is a limit

A yearly rate $r$, written as a decimal, paid $n$ times a year grows $1 to $(1 + r/n)^n$ in a year (case study 1.4.6). Continuous compounding is what this approaches as $n$ grows without end: a limit at infinity, as in lesson 2.1.3.

The limit has the shape lesson 2.1.3 warned about. The base $1 + r/n$ tends to 1 while the power $n$ grows without end, and the result is neither 1 nor infinite. The two pulls balance at $e^r$, just as $(1 + 1/n)^n \to e$.

How fast is the limit reached?

The table shows the gap between continuous and $n$-times-a-year compounding, at $r = 5\%$, per $1 over a year.

The gap is always positive: compounding $n$ times a year always falls short of continuous. The last row is the key. The gap times $n$ settles on a number, about 0.0013141: another limit at infinity.

So for large $n$ the gap is about $\frac{0.0013141}{n}$. Compound ten times as often and the gap shrinks about ten times. That number is $\frac{r^2 e^r}{2}$, which is 0.0013141 at 5%; lesson 2.7.2 shows where it comes from.

What it means in money

On $1,000,000 for a year at 5%, monthly compounding falls $109.20 short of continuous, and daily compounding $3.60 short. For most purposes daily and continuous compounding are interchangeable, which is why models use the simpler $e^{rt}$.

The gap grows with the horizon. Over 10 years, monthly compounding on $1,000,000 falls $1,711.77 short of continuous; over 30 years, $13,944.76. A pricing model that says "continuous" while the contract says "monthly" is off by that much.

Try other rates and horizons. The gap on $1,000,000 shrinks as compounding gets more frequent, while gap $\times\, n$ settles; a higher rate or a longer horizon makes both bigger.

What goes wrong in practice

The formula is simple; the inputs and the computer are where it fails.

Rates are decimals. Typing 5 for 5% gives $e^5 = 148.41$ continuously and $(1 + 5/12)^{12} = 65.34$ monthly: growth factors no deposit could earn. A factor far from 1 for a one-year deposit means the rate went in as a percentage.

Zero and negative rates. At $r = 0$ every convention gives exactly 1, and the gap is 0. Rates can also be negative: some central banks have set them below zero. The limit still holds, and the gap $\frac{r^2 e^r}{2n}$ is still positive because $r^2$ is.

At $-0.5\%$, $1 shrinks to 0.99501248 continuously and to 0.99501144 monthly. Monthly compounding still ends lower, by $1.04 on $1,000,000. The formula does need $1 + r/n > 0$: a loss of 100% or more in one period wipes out the balance.

Very large $n$ breaks the computer. The computer stores about 16 significant digits, so $1 + r/n$ loses most of $r/n$ when $n$ is huge. Lesson 2.1.1 met the same trap, and here the results go wrong long before $n$ looks extreme:

At $10^8$ the computed gap is negative, which is impossible. At $10^{15}$, $1 + r/n$ is stored as exactly 1, so the computed growth is 1: no interest at all.

Two habits avoid this. For continuous compounding, use math.exp(r * t) directly, never a huge $n$. For a huge $n$, write $(1 + r/n)^n = e^{n \ln(1 + r/n)}$ (lesson 1.4.3) and compute $\ln(1 + z)$ with math.log1p(z), which stays accurate for tiny $z$.

import math


def test_annual_gap():
    """At 5% compounded once, the gap is e^0.05 − 1.05"""
    expect_close(compounding_gap(0.05, 1), math.exp(0.05) - 1.05, what="compounding_gap(0.05, 1)")


def test_monthly_gap():
    """At 5% compounded monthly, the gap is about 0.0001092"""
    expect_close(compounding_gap(0.05, 12), 0.00010919849429114059, what="compounding_gap(0.05, 12)", rel=1e-6)


def test_zero_rate():
    """At a zero rate every convention grows $1 to exactly $1, so the gap is 0"""
    expect_close(compounding_gap(0.0, 12), 0.0, what="compounding_gap(0, 12)")


def test_negative_rate():
    """At −2%, monthly compounding still falls short of continuous, by about 0.00001635"""
    expect_close(compounding_gap(-0.02, 12), math.exp(-0.02) - (1 - 0.02 / 12) ** 12, what="compounding_gap(-0.02, 12)", rel=1e-6)
    assert compounding_gap(-0.02, 12) > 0, "the gap should be positive at a negative rate too"


def test_periods_for_one_millionth():
    """At 5%, 1315 periods bring the gap below one millionth"""
    expect_equal(periods_needed(0.05, 1e-6), 1315, what="periods_needed(0.05, 1e-6)")


def test_smallest_n():
    """The answer is the smallest n: one period fewer is not enough"""
    n = periods_needed(0.10, 1e-5)
    assert compounding_gap(0.10, n) < 1e-5, f"gap at your n = {n} is not below 1e-5"
    assert compounding_gap(0.10, n - 1) >= 1e-5, f"n − 1 = {n - 1} already works, so {n} is not the smallest"


def test_close_to_estimate():
    """The answer is close to the estimate r² e^r / (2 tol)"""
    estimate = 0.10**2 * math.exp(0.10) / (2 * 1e-5)
    n = periods_needed(0.10, 1e-5)
    assert abs(n - estimate) / estimate < 0.01, f"expected about {estimate:.0f}, got {n}"