Average Error: 38.9 → 0.3
Time: 25.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000014348698229582623753231018781662:\\ \;\;\;\;\left(\left(\frac{x}{1} \cdot \frac{x}{1}\right) \cdot \frac{-1}{2} + \log 1\right) + x \cdot 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000014348698229582623753231018781662:\\
\;\;\;\;\left(\left(\frac{x}{1} \cdot \frac{x}{1}\right) \cdot \frac{-1}{2} + \log 1\right) + x \cdot 1\\

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

\end{array}
double f(double x) {
        double r4085150 = 1.0;
        double r4085151 = x;
        double r4085152 = r4085150 + r4085151;
        double r4085153 = log(r4085152);
        return r4085153;
}
↓
double f(double x) {
        double r4085154 = 1.0;
        double r4085155 = x;
        double r4085156 = r4085154 + r4085155;
        double r4085157 = 1.0000000143486982;
        bool r4085158 = r4085156 <= r4085157;
        double r4085159 = r4085155 / r4085154;
        double r4085160 = r4085159 * r4085159;
        double r4085161 = -0.5;
        double r4085162 = r4085160 * r4085161;
        double r4085163 = log(r4085154);
        double r4085164 = r4085162 + r4085163;
        double r4085165 = r4085155 * r4085154;
        double r4085166 = r4085164 + r4085165;
        double r4085167 = log(r4085156);
        double r4085168 = r4085158 ? r4085166 : r4085167;
        return r4085168;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
Target0.2
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 (+ 1.0 x) < 1.0000000143486982

    1. Initial program 59.1

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    if 1.0000000143486982 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000014348698229582623753231018781662:\\ \;\;\;\;\left(\left(\frac{x}{1} \cdot \frac{x}{1}\right) \cdot \frac{-1}{2} + \log 1\right) + x \cdot 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019200 
(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)))