Average Error: 39.4 → 0.2
Time: 15.0s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00012930592350443132:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00012930592350443132:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\

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

\end{array}
double f(double x) {
        double r1592458 = 1.0;
        double r1592459 = x;
        double r1592460 = r1592458 + r1592459;
        double r1592461 = log(r1592460);
        return r1592461;
}
↓
double f(double x) {
        double r1592462 = x;
        double r1592463 = 0.00012930592350443132;
        bool r1592464 = r1592462 <= r1592463;
        double r1592465 = r1592462 * r1592462;
        double r1592466 = 0.3333333333333333;
        double r1592467 = r1592466 * r1592462;
        double r1592468 = 0.5;
        double r1592469 = r1592467 - r1592468;
        double r1592470 = r1592465 * r1592469;
        double r1592471 = r1592462 + r1592470;
        double r1592472 = 1.0;
        double r1592473 = r1592462 + r1592472;
        double r1592474 = log(r1592473);
        double r1592475 = r1592464 ? r1592471 : r1592474;
        return r1592475;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
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 < 0.00012930592350443132

    1. Initial program 59.0

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

    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

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

    if 0.00012930592350443132 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00012930592350443132:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019138 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))