Average Error: 38.8 → 0.3
Time: 15.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000053125128296471757494146004319191:\\ \;\;\;\;\frac{-1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) + \left(\log 1 + 1 \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000053125128296471757494146004319191:\\
\;\;\;\;\frac{-1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) + \left(\log 1 + 1 \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r9553022 = 1.0;
        double r9553023 = x;
        double r9553024 = r9553022 + r9553023;
        double r9553025 = log(r9553024);
        return r9553025;
}
↓
double f(double x) {
        double r9553026 = 1.0;
        double r9553027 = x;
        double r9553028 = r9553026 + r9553027;
        double r9553029 = 1.0000000531251283;
        bool r9553030 = r9553028 <= r9553029;
        double r9553031 = -0.5;
        double r9553032 = r9553027 / r9553026;
        double r9553033 = r9553032 * r9553032;
        double r9553034 = r9553031 * r9553033;
        double r9553035 = log(r9553026);
        double r9553036 = r9553026 * r9553027;
        double r9553037 = r9553035 + r9553036;
        double r9553038 = r9553034 + r9553037;
        double r9553039 = log(r9553028);
        double r9553040 = r9553030 ? r9553038 : r9553039;
        return r9553040;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.8
Target0.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000531251283

    1. Initial program 59.1

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

    if 1.0000000531251283 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000053125128296471757494146004319191:\\ \;\;\;\;\frac{-1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) + \left(\log 1 + 1 \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019173 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))