Average Error: 39.3 → 0.2
Time: 12.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00014368607062820952:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00014368607062820952:\\
\;\;\;\;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 r2301463 = 1.0;
        double r2301464 = x;
        double r2301465 = r2301463 + r2301464;
        double r2301466 = log(r2301465);
        return r2301466;
}
↓
double f(double x) {
        double r2301467 = x;
        double r2301468 = 0.00014368607062820952;
        bool r2301469 = r2301467 <= r2301468;
        double r2301470 = r2301467 * r2301467;
        double r2301471 = 0.3333333333333333;
        double r2301472 = r2301471 * r2301467;
        double r2301473 = 0.5;
        double r2301474 = r2301472 - r2301473;
        double r2301475 = r2301470 * r2301474;
        double r2301476 = r2301467 + r2301475;
        double r2301477 = 1.0;
        double r2301478 = r2301467 + r2301477;
        double r2301479 = log(r2301478);
        double r2301480 = r2301469 ? r2301476 : r2301479;
        return r2301480;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
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 < 0.00014368607062820952

    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 0.00014368607062820952 < x

    1. Initial program 0.0

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00014368607062820952:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019134 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))