Average Error: 38.9 → 0.6
Time: 12.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.0:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.0:\\
\;\;\;\;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 r3588938 = 1.0;
        double r3588939 = x;
        double r3588940 = r3588938 + r3588939;
        double r3588941 = log(r3588940);
        return r3588941;
}
↓
double f(double x) {
        double r3588942 = x;
        double r3588943 = 1.0;
        double r3588944 = r3588942 + r3588943;
        double r3588945 = 1.0;
        bool r3588946 = r3588944 <= r3588945;
        double r3588947 = 0.3333333333333333;
        double r3588948 = r3588942 * r3588947;
        double r3588949 = -0.5;
        double r3588950 = r3588948 + r3588949;
        double r3588951 = r3588942 * r3588942;
        double r3588952 = r3588950 * r3588951;
        double r3588953 = r3588942 + r3588952;
        double r3588954 = log(r3588944);
        double r3588955 = r3588946 ? r3588953 : r3588954;
        return r3588955;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
Target0.3
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 x) < 1.0

    1. Initial program 59.4

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    if 1.0 < (+ 1 x)

    1. Initial program 1.3

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.0:\\ \;\;\;\;x + \left(x \cdot \frac{1}{3} + \frac{-1}{2}\right) \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))