Average Error: 38.3 → 0.2
Time: 15.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.0001027760451352802:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.0001027760451352802:\\
\;\;\;\;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 r2917439 = 1.0;
        double r2917440 = x;
        double r2917441 = r2917439 + r2917440;
        double r2917442 = log(r2917441);
        return r2917442;
}
↓
double f(double x) {
        double r2917443 = x;
        double r2917444 = 0.0001027760451352802;
        bool r2917445 = r2917443 <= r2917444;
        double r2917446 = -0.5;
        double r2917447 = 0.3333333333333333;
        double r2917448 = r2917447 * r2917443;
        double r2917449 = r2917446 + r2917448;
        double r2917450 = r2917443 * r2917449;
        double r2917451 = r2917450 * r2917443;
        double r2917452 = r2917443 + r2917451;
        double r2917453 = 1.0;
        double r2917454 = r2917443 + r2917453;
        double r2917455 = log(r2917454);
        double r2917456 = r2917445 ? r2917452 : r2917455;
        return r2917456;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.3
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 < 0.0001027760451352802

    1. Initial program 58.9

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

    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

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

    if 0.0001027760451352802 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.0001027760451352802:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019129 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))