Average Error: 39.0 → 0.4
Time: 3.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000077041040213998712715692818165:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000077041040213998712715692818165:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r70806 = 1.0;
        double r70807 = x;
        double r70808 = r70806 + r70807;
        double r70809 = log(r70808);
        return r70809;
}
↓
double f(double x) {
        double r70810 = 1.0;
        double r70811 = x;
        double r70812 = r70810 + r70811;
        double r70813 = 1.000000000077041;
        bool r70814 = r70812 <= r70813;
        double r70815 = r70810 * r70811;
        double r70816 = log(r70810);
        double r70817 = r70815 + r70816;
        double r70818 = 0.5;
        double r70819 = 2.0;
        double r70820 = pow(r70811, r70819);
        double r70821 = pow(r70810, r70819);
        double r70822 = r70820 / r70821;
        double r70823 = r70818 * r70822;
        double r70824 = r70817 - r70823;
        double r70825 = log(r70812);
        double r70826 = r70814 ? r70824 : r70825;
        return r70826;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.000000000077041

    1. Initial program 59.5

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

    2. Taylor expanded around 0 0.3

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

    if 1.000000000077041 < (+ 1.0 x)

    1. Initial program 0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000077041040213998712715692818165:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019353 
(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)))