Average Error: 39.5 → 0.5
Time: 11.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000000036:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0000000000000036:\\
\;\;\;\;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 r1253199 = 1.0;
        double r1253200 = x;
        double r1253201 = r1253199 + r1253200;
        double r1253202 = log(r1253201);
        return r1253202;
}
↓
double f(double x) {
        double r1253203 = x;
        double r1253204 = 1.0;
        double r1253205 = r1253203 + r1253204;
        double r1253206 = 1.0000000000000036;
        bool r1253207 = r1253205 <= r1253206;
        double r1253208 = 0.3333333333333333;
        double r1253209 = r1253203 * r1253208;
        double r1253210 = -0.5;
        double r1253211 = r1253209 + r1253210;
        double r1253212 = r1253203 * r1253203;
        double r1253213 = r1253211 * r1253212;
        double r1253214 = r1253203 + r1253213;
        double r1253215 = log(r1253205);
        double r1253216 = r1253207 ? r1253214 : r1253215;
        return r1253216;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.5
Target0.2
Herbie0.5
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1 x) < 1.0000000000000036

    1. Initial program 59.5

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

    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 1.0000000000000036 < (+ 1 x)

    1. Initial program 1.1

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

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0000000000000036:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019154 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))