Average Error: 39.0 → 0.2
Time: 10.0s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 0.00013763371456035665:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 0.00013763371456035665:\\
\;\;\;\;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 r2914527 = 1.0;
        double r2914528 = x;
        double r2914529 = r2914527 + r2914528;
        double r2914530 = log(r2914529);
        return r2914530;
}
↓
double f(double x) {
        double r2914531 = x;
        double r2914532 = 0.00013763371456035665;
        bool r2914533 = r2914531 <= r2914532;
        double r2914534 = -0.5;
        double r2914535 = 0.3333333333333333;
        double r2914536 = r2914535 * r2914531;
        double r2914537 = r2914534 + r2914536;
        double r2914538 = r2914531 * r2914537;
        double r2914539 = r2914538 * r2914531;
        double r2914540 = r2914531 + r2914539;
        double r2914541 = 1.0;
        double r2914542 = r2914531 + r2914541;
        double r2914543 = log(r2914542);
        double r2914544 = r2914533 ? r2914540 : r2914543;
        return r2914544;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
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 < 0.00013763371456035665

    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

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

    if 0.00013763371456035665 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00013763371456035665:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019121 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))