Average Error: 39.6 → 0.2
Time: 13.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 9.93138264255617 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 9.93138264255617 \cdot 10^{-05}:\\
\;\;\;\;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 r1670434 = 1.0;
        double r1670435 = x;
        double r1670436 = r1670434 + r1670435;
        double r1670437 = log(r1670436);
        return r1670437;
}
↓
double f(double x) {
        double r1670438 = x;
        double r1670439 = 9.93138264255617e-05;
        bool r1670440 = r1670438 <= r1670439;
        double r1670441 = -0.5;
        double r1670442 = 0.3333333333333333;
        double r1670443 = r1670442 * r1670438;
        double r1670444 = r1670441 + r1670443;
        double r1670445 = r1670438 * r1670444;
        double r1670446 = r1670445 * r1670438;
        double r1670447 = r1670438 + r1670446;
        double r1670448 = 1.0;
        double r1670449 = r1670438 + r1670448;
        double r1670450 = log(r1670449);
        double r1670451 = r1670440 ? r1670447 : r1670450;
        return r1670451;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
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 < 9.93138264255617e-05

    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

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

    if 9.93138264255617e-05 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 9.93138264255617 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019130 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))