Average Error: 39.2 → 0.3
Time: 12.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000005037382:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000000005037382:\\
\;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\

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

\end{array}
double f(double x) {
        double r1426831 = 1.0;
        double r1426832 = x;
        double r1426833 = r1426831 + r1426832;
        double r1426834 = log(r1426833);
        return r1426834;
}
↓
double f(double x) {
        double r1426835 = x;
        double r1426836 = 1.0;
        double r1426837 = r1426835 + r1426836;
        double r1426838 = 1.0000000005037382;
        bool r1426839 = r1426837 <= r1426838;
        double r1426840 = 0.3333333333333333;
        double r1426841 = r1426835 * r1426840;
        double r1426842 = -0.5;
        double r1426843 = r1426841 + r1426842;
        double r1426844 = r1426835 * r1426835;
        double r1426845 = r1426843 * r1426844;
        double r1426846 = r1426835 + r1426845;
        double r1426847 = log(r1426837);
        double r1426848 = r1426839 ? r1426846 : r1426847;
        return r1426848;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
Target0.3
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 x) < 1.0000000005037382

    1. Initial program 59.3

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

    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

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

    if 1.0000000005037382 < (+ 1 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000005037382:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019153 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))