Average Error: 39.2 → 0.3
Time: 16.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000003244205803554223166429437696933746:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000003244205803554223166429437696933746:\\
\;\;\;\;\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}
double f(double x) {
        double r56896 = 1.0;
        double r56897 = x;
        double r56898 = r56896 + r56897;
        double r56899 = log(r56898);
        return r56899;
}
↓
double f(double x) {
        double r56900 = 1.0;
        double r56901 = x;
        double r56902 = r56900 + r56901;
        double r56903 = 1.0000032442058036;
        bool r56904 = r56902 <= r56903;
        double r56905 = r56901 * r56900;
        double r56906 = -0.5;
        double r56907 = r56900 * r56900;
        double r56908 = r56907 / r56901;
        double r56909 = r56906 / r56908;
        double r56910 = r56901 * r56909;
        double r56911 = r56905 + r56910;
        double r56912 = log(r56900);
        double r56913 = r56911 + r56912;
        double r56914 = log(r56902);
        double r56915 = r56904 ? r56913 : r56914;
        return r56915;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
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) < 1.0000032442058036

    1. Initial program 59.0

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

    5. Applied distribute-lft-in0.4

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

    if 1.0000032442058036 < (+ 1.0 x)

    1. Initial program 0.1

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000003244205803554223166429437696933746:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019304 
(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)))