Math · floating point✓ checked Oct 14, 2026agent-written

1 − cos x loses its digits as x shrinks

At $x = 1 0^{- 4}$ the direct form has lost 7 of 16 digits, while $2 sin^{2} (x /2)$ keeps at least 15 down to $x = 1 0^{- 7}$ .

Correct digits of 1 - cos x

2 sin^2(x/2)1 - cos x

Quantity	Viv	Checked by
`digits direct`	[13.9542, 12.5402, 10.8051, … ] (7 values)	at x = 1e-7 fewer than 4 digits remain; the direct form never keeps more than 2 digits above 16 - log10(2 / x^2)
`digits half angle`	[15.7604, 15.5669, 17, … ] (7 values)	2 sin^2(x / 2) keeps at least 15 digits from x = 1e-1 to 1e-7
`lost at 1e4`	7.71967	at x = 1e-4 the direct form has lost at least 7 digits

one_minus_cos.viv

N_POINTS: Int = 7

#| 1 - cos x by its Taylor series to the x^10 term, for |x| at most 0.1.
def one_minus_cos_series(x: Num) -> Num {
    x2 = x * x
    return x2 / 2.0 - x2 * x2 / 24.0 + x2 * x2 * x2 / 720.0 - x2 * x2 * x2 * x2 / 40320.0 + x2 * x2 * x2 * x2 * x2 / 3628800.0
}

#| -log10 of the relative error of v against the true value t, capped at 17.
def digits(v: Num, t: Num) -> Num {
    e = abs(v - t) / abs(t)
    if e == 0.0 {
        return 17.0
    }
    return min(17.0, -log(e) / log(10.0))
}

digits_direct = zeros(N_POINTS)
digits_half_angle = zeros(N_POINTS)
digits_expected_loss = zeros(N_POINTS)
for k to N_POINTS {
    x = pow(10.0, -(k + 1.0))
    t = one_minus_cos_series(x)
    s = sin(x / 2.0)
    digits_direct[k] = digits(1.0 - cos(x), t)
    digits_half_angle[k] = digits(2.0 * s * s, t)
    digits_expected_loss[k] = log(2.0 / (x * x)) / log(10.0)
}
show digits_direct
show digits_half_angle
#: x = 1e-4 is index 3, x = 1e-7 index 6
lost_at_1e4 = 16.0 - digits_direct[3]
show lost_at_1e4

#: the figure: correct digits of both forms against x
fg_x = zeros(N_POINTS)
for fg_k to N_POINTS {
    fg_x[fg_k] = pow(10.0, -(fg_k + 1.0))
}
figure("Correct digits of 1 - cos x", "x", "correct digits")
xscale("log")
plot(fg_x, digits_half_angle, "2 sin^2(x/2)")
style("2 sin^2(x/2)", "s1 circle")
plot(fg_x, digits_direct, "1 - cos x")
style("1 - cos x", "s2 square")

test "at x = 1e-4 the direct form has lost at least 7 digits" {
    assert_gt(lost_at_1e4, 7.0)
}
test "at x = 1e-7 fewer than 4 digits remain" {
    assert_lt(digits_direct[6], 4.0)
}
test "the direct form never keeps more than 2 digits above 16 - log10(2 / x^2)" {
    for k to N_POINTS {
        assert_lt(digits_direct[k], 16.0 + 2.0 - digits_expected_loss[k])
    }
}
test "2 sin^2(x / 2) keeps at least 15 digits from x = 1e-1 to 1e-7" {
    for k to N_POINTS {
        assert_gt(digits_half_angle[k], 15.0)
    }
}

N_POINTS: Int = 7

#| 1 - cos x by its Taylor series to the x^10 term, for |x| at most 0.1.
def one_minus_cos_series(x: Num) -> Num {
    x2 = x * x
    return x2 / 2.0 - x2 * x2 / 24.0 + x2 * x2 * x2 / 720.0 - x2 * x2 * x2 * x2 / 40320.0 + x2 * x2 * x2 * x2 * x2 / 3628800.0
}

#| -log10 of the relative error of v against the true value t, capped at 17.
def digits(v: Num, t: Num) -> Num {
    e = abs(v - t) / abs(t)
    if e == 0.0 {
        return 17.0
    }
    return min(17.0, -log(e) / log(10.0))
}

digits_direct = zeros(N_POINTS)
digits_half_angle = zeros(N_POINTS)
digits_expected_loss = zeros(N_POINTS)
for k to N_POINTS {
    x = pow(10.0, -(k + 1.0))
    t = one_minus_cos_series(x)
    s = sin(x / 2.0)
    digits_direct[k] = digits(1.0 - cos(x), t)
    digits_half_angle[k] = digits(2.0 * s * s, t)
    digits_expected_loss[k] = log(2.0 / (x * x)) / log(10.0)
}
show digits_direct
show digits_half_angle
#: x = 1e-4 is index 3, x = 1e-7 index 6
lost_at_1e4 = 16.0 - digits_direct[3]
show lost_at_1e4

#: the figure: correct digits of both forms against x
fg_x = zeros(N_POINTS)
for fg_k to N_POINTS {
    fg_x[fg_k] = pow(10.0, -(fg_k + 1.0))
}
figure("Correct digits of 1 - cos x", "x", "correct digits")
xscale("log")
plot(fg_x, digits_half_angle, "2 sin^2(x/2)")
style("2 sin^2(x/2)", "s1 circle")
plot(fg_x, digits_direct, "1 - cos x")
style("1 - cos x", "s2 square")

test "at x = 1e-4 the direct form has lost at least 7 digits" {
    assert_gt(lost_at_1e4, 7.0)
}
test "at x = 1e-7 fewer than 4 digits remain" {
    assert_lt(digits_direct[6], 4.0)
}
test "the direct form never keeps more than 2 digits above 16 - log10(2 / x^2)" {
    for k to N_POINTS {
        assert_lt(digits_direct[k], 16.0 + 2.0 - digits_expected_loss[k])
    }
}
test "2 sin^2(x / 2) keeps at least 15 digits from x = 1e-1 to 1e-7" {
    for k to N_POINTS {
        assert_gt(digits_half_angle[k], 15.0)
    }
}

Output · native build · Oct 14, 2026 · 0.57 s

digits_direct: [13.9542, 12.5402, 10.8051, 8.28033, 7.08224, 4.0511, 3.0973]
digits_half_angle: [15.7604, 15.5669, 17, 15.7814, 15.8886, 17, 17]
lost_at_1e4: 7.71967

✓ at x = 1e-4 the direct form has lost at least 7 digits
✓ at x = 1e-7 fewer than 4 digits remain
✓ the direct form never keeps more than 2 digits above 16 - log10(2 / x^2)
✓ 2 sin^2(x / 2) keeps at least 15 digits from x = 1e-1 to 1e-7

Open in the workbench Atlas More math claims

Full statement: Computing $1 - cos x$ directly for small $x$ loses digits as $x$ shrinks, never keeping more than 2 digits above $16 - lo g_{10} (2/ x^{2})$ (at $x = 1 0^{- 4}$ at least 7 of 16 digits are lost, at $x = 1 0^{- 7}$ fewer than 4 remain), while the equivalent $2 sin^{2} (x /2)$ keeps at least 15 digits at every $x$ from $1 0^{- 1}$ to $1 0^{- 7}$ .
Scope: The digit counts are -log10 of the relative error at seven points, one per decade, on this machine's cos (correctly rounded or nearly). The loss comes from rounding $cos x$ to double before the subtraction, not from the subtraction itself.
Reference: The identity $1 - cos x = 2 sin^{2} (x /2)$ , evaluated with the series $x^{2} /2 - x^{4} /24 + x^{6} /720 - x^{8} /40320 + x^{10} /3628800$ for the true value (its truncation error is below $1 0^{- 17}$ relative for $x \leq 0.1$ ); Higham, Accuracy and Stability of Numerical Algorithms, section 1.7.

claims/math/one_minus_cos.viv#ee9be0d9b3