ln(1 + x)

Average Error: 39.0 → 0.3

Time: 19.8s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000061006203200264508268446661531925:\\ \;\;\;\;\frac{x \cdot \left({1}^{3} + {\left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}^{3}\right)}{1 \cdot 1 + \left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} - 1 \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)} + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

C

double f(double x) {
        double r75878 = 1.0;
        double r75879 = x;
        double r75880 = r75878 + r75879;
        double r75881 = log(r75880);
        return r75881;
}

↓

double f(double x) {
        double r75882 = 1.0;
        double r75883 = x;
        double r75884 = r75882 + r75883;
        double r75885 = 1.0000000610062032;
        bool r75886 = r75884 <= r75885;
        double r75887 = 3.0;
        double r75888 = pow(r75882, r75887);
        double r75889 = -0.5;
        double r75890 = r75882 * r75882;
        double r75891 = r75890 / r75883;
        double r75892 = r75889 / r75891;
        double r75893 = pow(r75892, r75887);
        double r75894 = r75888 + r75893;
        double r75895 = r75883 * r75894;
        double r75896 = r75892 * r75892;
        double r75897 = r75882 * r75892;
        double r75898 = r75896 - r75897;
        double r75899 = r75890 + r75898;
        double r75900 = r75895 / r75899;
        double r75901 = log(r75882);
        double r75902 = r75900 + r75901;
        double r75903 = log(r75884);
        double r75904 = r75886 ? r75902 : r75903;
        return r75904;
}

Error

Bits error versus x

Target

Original	39.0
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) < 1.0000000610062032

   1. Initial program 59.1

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

   2. Taylor expanded around 0 0.4

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

   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 flip3-+ 0.4

      \[\leadsto x \cdot \color{blue}{\frac{{1}^{3} + {\left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}^{3}}{1 \cdot 1 + \left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} - 1 \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}} + \log 1\]

   6. Applied associate-*r/ 0.4

      \[\leadsto \color{blue}{\frac{x \cdot \left({1}^{3} + {\left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}^{3}\right)}{1 \cdot 1 + \left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} - 1 \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}} + \log 1\]

   if 1.0000000610062032 < (+ 1.0 x)

   1. Initial program 0.2

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

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000061006203200264508268446661531925:\\ \;\;\;\;\frac{x \cdot \left({1}^{3} + {\left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)}^{3}\right)}{1 \cdot 1 + \left(\frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}} - 1 \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right)} + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019325 
(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)))