Average Error: 39.1 → 0.6
Time: 15.0s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1:\\ \;\;\;\;x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1:\\
\;\;\;\;x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\

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

\end{array}
double f(double x) {
        double r40877 = 1.0;
        double r40878 = x;
        double r40879 = r40877 + r40878;
        double r40880 = log(r40879);
        return r40880;
}
↓
double f(double x) {
        double r40881 = 1.0;
        double r40882 = x;
        double r40883 = r40881 + r40882;
        bool r40884 = r40883 <= r40881;
        double r40885 = -0.5;
        double r40886 = r40881 * r40881;
        double r40887 = r40886 / r40882;
        double r40888 = r40885 / r40887;
        double r40889 = r40881 + r40888;
        double r40890 = r40882 * r40889;
        double r40891 = log(r40881);
        double r40892 = r40890 + r40891;
        double r40893 = log(r40883);
        double r40894 = r40884 ? r40892 : r40893;
        return r40894;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
Target0.2
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0

    1. Initial program 59.6

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

    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. Simplified0.3

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

    if 1.0 < (+ 1.0 x)

    1. Initial program 1.3

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

  4. Final simplification0.6

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

Reproduce

herbie shell --seed 2019326 
(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)))