Average Error: 38.8 → 0.5
Time: 9.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000397237798210881010163575411:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000000397237798210881010163575411:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(1 + x\right)\\

\end{array}
double f(double x) {
        double r62850 = 1.0;
        double r62851 = x;
        double r62852 = r62850 + r62851;
        double r62853 = log(r62852);
        return r62853;
}
↓
double f(double x) {
        double r62854 = 1.0;
        double r62855 = x;
        double r62856 = r62854 + r62855;
        double r62857 = 1.0000000000003972;
        bool r62858 = r62856 <= r62857;
        double r62859 = r62854 * r62855;
        double r62860 = log(r62854);
        double r62861 = r62859 + r62860;
        double r62862 = 0.5;
        double r62863 = 2.0;
        double r62864 = pow(r62855, r62863);
        double r62865 = pow(r62854, r62863);
        double r62866 = r62864 / r62865;
        double r62867 = r62862 * r62866;
        double r62868 = r62861 - r62867;
        double r62869 = log(r62856);
        double r62870 = r62858 ? r62868 : r62869;
        return r62870;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.8
Target0.2
Herbie0.5
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000003972

    1. Initial program 59.3

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

    2. Taylor expanded around 0 0.4

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

    if 1.0000000000003972 < (+ 1.0 x)

    1. Initial program 0.8

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

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000397237798210881010163575411:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019308 
(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)))