Average Error: 39.1 → 0.6
Time: 15.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1:\\ \;\;\;\;x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1:\\
\;\;\;\;x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\

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

\end{array}
double f(double x) {
        double r41536 = 1.0;
        double r41537 = x;
        double r41538 = r41536 + r41537;
        double r41539 = log(r41538);
        return r41539;
}
↓
double f(double x) {
        double r41540 = 1.0;
        double r41541 = x;
        double r41542 = r41540 + r41541;
        bool r41543 = r41542 <= r41540;
        double r41544 = -0.5;
        double r41545 = r41540 * r41540;
        double r41546 = r41545 / r41541;
        double r41547 = r41544 / r41546;
        double r41548 = r41540 + r41547;
        double r41549 = r41541 * r41548;
        double r41550 = log(r41540);
        double r41551 = r41549 + r41550;
        double r41552 = log(r41542);
        double r41553 = r41543 ? r41551 : r41552;
        return r41553;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
Target0.2
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0

    1. Initial program 59.6

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    if 1.0 < (+ 1.0 x)

    1. Initial program 1.3

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

  4. Final simplification0.6

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

Reproduce

herbie shell --seed 2019326 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

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

  (log (+ 1 x)))