Notebooks · posts/compound-interest

Compound interest

Adam Zettel · September 22, 2026 · written with an agent

$1000 at 5% a year becomes $1628.89 in ten years if the interest is added once a year, and $1647.01 if it is added monthly. Write balance to get both from the same four numbers, and years_to_double to say how long doubling takes.

P r n times a year, t years A = ?

Interest added $n$ times a year is $r / n$ each time, applied $n t$ times over $t$ years. Getting that substitution right is most of the exercise, and the two numbers above are far enough apart that a test fails if you use $r$ and $t$ where $r / n$ and $n t$ belong.

Monthly beats annual by $1647.01 - 1628.89 = 18.12$ dollars here. Weekly beats monthly by less, daily by less again, and no matter how often the interest is added the balance never passes $P e^{r t} = $1648.72$ . Once the tests pass, try a few values of $n$ to see it approach that limit.

years_to_double is the question the doubling-time post answers: $t$ is in the exponent, and taking the logarithm of both sides brings it down.

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[1]lines 26–55

#| The balance after $t$ years: $A = P\left(1 + \frac{r}{n}\right)^{nt}$.
def balance(p, r, n, t) {
    return p
}

#| Years for money at annual rate $r$, compounded once a year, to double.
def years_to_double(r) {
    return 0.0
}

show balance(1000.0, 0.05, 1.0, 10.0)
show years_to_double(0.05)

test "$1000 at 5% for 10 years, compounded once a year, is $1628.89" {
    assert_near(balance(1000.0, 0.05, 1.0, 10.0), 1628.89, 0.01)
}
test "compounded monthly it is $1647.01" {
    assert_near(balance(1000.0, 0.05, 12.0, 10.0), 1647.01, 0.01)
}
test "no time, no interest: the balance is the principal" {
    assert_near(balance(250.0, 0.08, 4.0, 0.0), 250.0, 1.0e-12)
}
test "at 5% a year the money doubles in 14.21 years, and the balance then is 2P" {
    t = years_to_double(0.05)
    assert_near(t, 14.2067, 1.0e-3)
    assert_near(balance(1.0, 0.05, 1.0, t), 2.0, 1.0e-9)
}

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# SPDX-FileCopyrightText: 2026 The Viv authors