ln(1 + x)

Average Error: 38.7 → 0.3

Time: 22.6s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000005155942339740704483119770884514:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

C

double f(double x) {
        double r51721 = 1.0;
        double r51722 = x;
        double r51723 = r51721 + r51722;
        double r51724 = log(r51723);
        return r51724;
}

↓

double f(double x) {
        double r51725 = 1.0;
        double r51726 = x;
        double r51727 = r51725 + r51726;
        double r51728 = 1.0000000051559423;
        bool r51729 = r51727 <= r51728;
        double r51730 = r51726 * r51725;
        double r51731 = -0.5;
        double r51732 = r51725 * r51725;
        double r51733 = r51732 / r51726;
        double r51734 = r51731 / r51733;
        double r51735 = r51726 * r51734;
        double r51736 = r51730 + r51735;
        double r51737 = log(r51725);
        double r51738 = r51736 + r51737;
        double r51739 = log(r51727);
        double r51740 = r51729 ? r51738 : r51739;
        return r51740;
}

Error

Bits error versus x

Target

Original	38.7
Target	0.3
Herbie	0.3

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

Derivation

1. Split input into 2 regimes

2. if (+ 1.0 x) < 1.0000000051559423

   1. Initial program 59.2

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

   2. Taylor expanded around 0 0.3

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

   3. Simplified 0.4

      \[\leadsto \color{blue}{x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1}\]
   4. Using strategy rm

   5. Applied distribute-lft-in 0.3

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

   if 1.0000000051559423 < (+ 1.0 x)

   1. Initial program 0.3

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

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000005155942339740704483119770884514:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019322 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

  :herbie-target
  (if (== (+ 1 x) 1) x (/ (* x (log (+ 1 x))) (- (+ 1 x) 1)))

  (log (+ 1 x)))