Average Error: 38.9 → 0.3
Time: 7.5s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000008116859207:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000008116859207:\\
\;\;\;\;\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 r80859 = 1.0;
        double r80860 = x;
        double r80861 = r80859 + r80860;
        double r80862 = log(r80861);
        return r80862;
}
↓
double f(double x) {
        double r80863 = 1.0;
        double r80864 = x;
        double r80865 = r80863 + r80864;
        double r80866 = 1.000000081168592;
        bool r80867 = r80865 <= r80866;
        double r80868 = r80863 * r80864;
        double r80869 = log(r80863);
        double r80870 = r80868 + r80869;
        double r80871 = 0.5;
        double r80872 = 2.0;
        double r80873 = pow(r80864, r80872);
        double r80874 = pow(r80863, r80872);
        double r80875 = r80873 / r80874;
        double r80876 = r80871 * r80875;
        double r80877 = r80870 - r80876;
        double r80878 = log(r80865);
        double r80879 = r80867 ? r80877 : r80878;
        return r80879;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
Target0.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.000000081168592

    1. Initial program 59.0

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

    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 1.000000081168592 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000008116859207:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020045 
(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)))