Average Error: 39.3 → 0.6
Time: 6.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1:\\ \;\;\;\;1 \cdot x + \left(\log 1 - \left(\frac{x}{1} \cdot \frac{1}{2}\right) \cdot \frac{x}{1}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1:\\
\;\;\;\;1 \cdot x + \left(\log 1 - \left(\frac{x}{1} \cdot \frac{1}{2}\right) \cdot \frac{x}{1}\right)\\

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

\end{array}
double f(double x) {
        double r3117999 = 1.0;
        double r3118000 = x;
        double r3118001 = r3117999 + r3118000;
        double r3118002 = log(r3118001);
        return r3118002;
}
↓
double f(double x) {
        double r3118003 = x;
        double r3118004 = 1.0;
        double r3118005 = r3118003 + r3118004;
        bool r3118006 = r3118005 <= r3118004;
        double r3118007 = r3118004 * r3118003;
        double r3118008 = log(r3118004);
        double r3118009 = r3118003 / r3118004;
        double r3118010 = 0.5;
        double r3118011 = r3118009 * r3118010;
        double r3118012 = r3118011 * r3118009;
        double r3118013 = r3118008 - r3118012;
        double r3118014 = r3118007 + r3118013;
        double r3118015 = log(r3118005);
        double r3118016 = r3118006 ? r3118014 : r3118015;
        return r3118016;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
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

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

    3. Simplified0.3

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

    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

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

Reproduce

herbie shell --seed 2019172 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))