Average Error: 38.4 → 0.2
Time: 7.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 8.861479379567599 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le 8.861479379567599 \cdot 10^{-05}:\\
\;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\

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

\end{array}
double f(double x) {
        double r2820314 = 1.0;
        double r2820315 = x;
        double r2820316 = r2820314 + r2820315;
        double r2820317 = log(r2820316);
        return r2820317;
}
↓
double f(double x) {
        double r2820318 = x;
        double r2820319 = 8.861479379567599e-05;
        bool r2820320 = r2820318 <= r2820319;
        double r2820321 = -0.5;
        double r2820322 = 0.3333333333333333;
        double r2820323 = r2820322 * r2820318;
        double r2820324 = r2820321 + r2820323;
        double r2820325 = r2820318 * r2820324;
        double r2820326 = r2820325 * r2820318;
        double r2820327 = r2820318 + r2820326;
        double r2820328 = 1.0;
        double r2820329 = r2820318 + r2820328;
        double r2820330 = log(r2820329);
        double r2820331 = r2820320 ? r2820327 : r2820330;
        return r2820331;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.4
Target0.3
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < 8.861479379567599e-05

    1. Initial program 58.9

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

    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

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

    if 8.861479379567599e-05 < x

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 8.861479379567599 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot \left(\frac{-1}{2} + \frac{1}{3} \cdot x\right)\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019133 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))