Average Error: 39.5 → 0.4
Time: 3.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000090492:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000000090492:\\
\;\;\;\;\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 r63011 = 1.0;
        double r63012 = x;
        double r63013 = r63011 + r63012;
        double r63014 = log(r63013);
        return r63014;
}
↓
double f(double x) {
        double r63015 = 1.0;
        double r63016 = x;
        double r63017 = r63015 + r63016;
        double r63018 = 1.0000000000090492;
        bool r63019 = r63017 <= r63018;
        double r63020 = r63015 * r63016;
        double r63021 = log(r63015);
        double r63022 = r63020 + r63021;
        double r63023 = 0.5;
        double r63024 = 2.0;
        double r63025 = pow(r63016, r63024);
        double r63026 = pow(r63015, r63024);
        double r63027 = r63025 / r63026;
        double r63028 = r63023 * r63027;
        double r63029 = r63022 - r63028;
        double r63030 = log(r63017);
        double r63031 = r63019 ? r63029 : r63030;
        return r63031;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.5
Target0.4
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000090492

    1. Initial program 59.4

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

    2. Taylor expanded around 0 0.3

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

    if 1.0000000000090492 < (+ 1.0 x)

    1. Initial program 0.7

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000090492:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020020 
(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)))