Average Error: 39.1 → 0.3
Time: 17.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000010117869209125274210236966609955:\\ \;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000010117869209125274210236966609955:\\
\;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\

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

\end{array}
double f(double x) {
        double r4660672 = 1.0;
        double r4660673 = x;
        double r4660674 = r4660672 + r4660673;
        double r4660675 = log(r4660674);
        return r4660675;
}
↓
double f(double x) {
        double r4660676 = x;
        double r4660677 = 1.0;
        double r4660678 = r4660676 + r4660677;
        double r4660679 = 1.0000000101178692;
        bool r4660680 = r4660678 <= r4660679;
        double r4660681 = r4660677 * r4660676;
        double r4660682 = 0.5;
        double r4660683 = r4660676 / r4660677;
        double r4660684 = r4660683 * r4660683;
        double r4660685 = r4660682 * r4660684;
        double r4660686 = log(r4660677);
        double r4660687 = r4660685 - r4660686;
        double r4660688 = r4660681 - r4660687;
        double r4660689 = log(r4660678);
        double r4660690 = r4660680 ? r4660688 : r4660689;
        return r4660690;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
Target0.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000101178692

    1. Initial program 59.2

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

    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

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

    if 1.0000000101178692 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000010117869209125274210236966609955:\\ \;\;\;\;1 \cdot x - \left(\frac{1}{2} \cdot \left(\frac{x}{1} \cdot \frac{x}{1}\right) - \log 1\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019169 
(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)))