Average Error: 39.0 → 0.2
Time: 25.8s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00011713184139444529:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00011713184139444529:\\
\;\;\;\;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 r11144241 = 1.0;
        double r11144242 = x;
        double r11144243 = r11144241 + r11144242;
        double r11144244 = log(r11144243);
        return r11144244;
}
↓
double f(double x) {
        double r11144245 = x;
        double r11144246 = 0.00011713184139444529;
        bool r11144247 = r11144245 <= r11144246;
        double r11144248 = -0.5;
        double r11144249 = 0.3333333333333333;
        double r11144250 = r11144249 * r11144245;
        double r11144251 = r11144248 + r11144250;
        double r11144252 = r11144245 * r11144251;
        double r11144253 = r11144252 * r11144245;
        double r11144254 = r11144245 + r11144253;
        double r11144255 = 1.0;
        double r11144256 = r11144245 + r11144255;
        double r11144257 = log(r11144256);
        double r11144258 = r11144247 ? r11144254 : r11144257;
        return r11144258;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.3
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 < 0.00011713184139444529

    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

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

    if 0.00011713184139444529 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00011713184139444529:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019120 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))