Average Error: 39.3 → 0.6
Time: 10.7s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000000000000222044604925031308084726:\\
\;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\

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

\end{array}
double f(double x) {
        double r3054010 = 1.0;
        double r3054011 = x;
        double r3054012 = r3054010 + r3054011;
        double r3054013 = log(r3054012);
        return r3054013;
}
↓
double f(double x) {
        double r3054014 = x;
        double r3054015 = 1.0;
        double r3054016 = r3054014 + r3054015;
        double r3054017 = 1.0000000000000002;
        bool r3054018 = r3054016 <= r3054017;
        double r3054019 = r3054015 * r3054014;
        double r3054020 = log(r3054015);
        double r3054021 = r3054014 / r3054015;
        double r3054022 = 0.5;
        double r3054023 = r3054022 * r3054021;
        double r3054024 = r3054021 * r3054023;
        double r3054025 = r3054020 - r3054024;
        double r3054026 = r3054019 + r3054025;
        double r3054027 = log(r3054016);
        double r3054028 = r3054018 ? r3054026 : r3054027;
        return r3054028;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
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 (+ 1.0 x) < 1.0000000000000002

    1. Initial program 59.5

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

    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

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

    if 1.0000000000000002 < (+ 1.0 x)

    1. Initial program 1.2

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019192 
(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)))