ln(1 + x)

Average Error: 39.3 → 0.6

Time: 10.1s

Precision: 64

Math

\[\log \left(1 + x\right)\]

↓

\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;\mathsf{fma}\left(\frac{-1}{2} \cdot \frac{x}{1}, \frac{x}{1}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

C

double f(double x) {
        double r3392549 = 1.0;
        double r3392550 = x;
        double r3392551 = r3392549 + r3392550;
        double r3392552 = log(r3392551);
        return r3392552;
}

↓

double f(double x) {
        double r3392553 = x;
        double r3392554 = 1.0;
        double r3392555 = r3392553 + r3392554;
        double r3392556 = 1.0000000000000002;
        bool r3392557 = r3392555 <= r3392556;
        double r3392558 = -0.5;
        double r3392559 = r3392553 / r3392554;
        double r3392560 = r3392558 * r3392559;
        double r3392561 = log(r3392554);
        double r3392562 = fma(r3392554, r3392553, r3392561);
        double r3392563 = fma(r3392560, r3392559, r3392562);
        double r3392564 = log(r3392555);
        double r3392565 = r3392557 ? r3392563 : r3392564;
        return r3392565;
}

Error

Bits error versus x

Target

Original	39.3
Target	0.3
Herbie	0.6

\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

1. Split input into 2 regimes

2. if (+ 1.0 x) < 1.0000000000000002

   1. Initial program 59.5

      \[\log \left(1 + x\right)\]

   2. Taylor expanded around 0 0.3

      \[\leadsto \color{blue}{\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}}\]

   3. Simplified 0.3

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{-1}{2} \cdot \frac{x}{1}, \frac{x}{1}, \mathsf{fma}\left(1, x, \log 1\right)\right)}\]

   if 1.0000000000000002 < (+ 1.0 x)

   1. Initial program 1.2

      \[\log \left(1 + x\right)\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.6

   \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;\mathsf{fma}\left(\frac{-1}{2} \cdot \frac{x}{1}, \frac{x}{1}, \mathsf{fma}\left(1, x, \log 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019192 +o rules:numerics
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))