Average Error: 38.8 → 0.1
Time: 13.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.000133440903912734:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.000133440903912734:\\
\;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\

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

\end{array}
double f(double x) {
        double r2156172 = 1.0;
        double r2156173 = x;
        double r2156174 = r2156172 + r2156173;
        double r2156175 = log(r2156174);
        return r2156175;
}
↓
double f(double x) {
        double r2156176 = x;
        double r2156177 = 0.000133440903912734;
        bool r2156178 = r2156176 <= r2156177;
        double r2156179 = -0.5;
        double r2156180 = 0.3333333333333333;
        double r2156181 = r2156180 * r2156176;
        double r2156182 = r2156179 + r2156181;
        double r2156183 = r2156176 * r2156182;
        double r2156184 = r2156183 * r2156176;
        double r2156185 = r2156176 + r2156184;
        double r2156186 = 1.0;
        double r2156187 = r2156176 + r2156186;
        double r2156188 = log(r2156187);
        double r2156189 = r2156178 ? r2156185 : r2156188;
        return r2156189;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Split input into 2 regimes

  2. if x < 0.000133440903912734

    1. Initial program 58.8

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

    2. Taylor expanded around 0 0.2

      \[\leadsto \color{blue}{\left(x + \frac{1}{3} \cdot {x}^{3}\right) - \frac{1}{2} \cdot {x}^{2}}\]

    3. Simplified0.2

      \[\leadsto \color{blue}{x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x}\]

    if 0.000133440903912734 < x

    1. Initial program 0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.000133440903912734:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019135 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1 x) 1) x (/ (* x (log (+ 1 x))) (- (+ 1 x) 1)))

  (log (+ 1 x)))