Average Error: 39.1 → 0.2
Time: 17.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00015125854089797293:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{x + 1}\right) + \log \left(\sqrt{x + 1}\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00015125854089797293:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r6780709 = 1.0;
        double r6780710 = x;
        double r6780711 = r6780709 + r6780710;
        double r6780712 = log(r6780711);
        return r6780712;
}
↓
double f(double x) {
        double r6780713 = x;
        double r6780714 = 0.00015125854089797293;
        bool r6780715 = r6780713 <= r6780714;
        double r6780716 = r6780713 * r6780713;
        double r6780717 = -0.5;
        double r6780718 = 0.3333333333333333;
        double r6780719 = r6780718 * r6780713;
        double r6780720 = r6780717 + r6780719;
        double r6780721 = r6780716 * r6780720;
        double r6780722 = r6780713 + r6780721;
        double r6780723 = 1.0;
        double r6780724 = r6780713 + r6780723;
        double r6780725 = sqrt(r6780724);
        double r6780726 = log(r6780725);
        double r6780727 = r6780726 + r6780726;
        double r6780728 = r6780715 ? r6780722 : r6780727;
        return r6780728;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Split input into 2 regimes

  2. if x < 0.00015125854089797293

    1. Initial program 58.9

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    if 0.00015125854089797293 < x

    1. Initial program 0.1

      \[\log \left(1 + x\right)\]
    2. Using strategy rm

    3. Applied add-sqr-sqrt0.1

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

    4. Applied log-prod0.1

      \[\leadsto \color{blue}{\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00015125854089797293:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{x + 1}\right) + \log \left(\sqrt{x + 1}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019124 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))