Average Error: 38.8 → 0.5
Time: 7.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000000000333:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000000000000333:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r68341 = 1.0;
        double r68342 = x;
        double r68343 = r68341 + r68342;
        double r68344 = log(r68343);
        return r68344;
}
↓
double f(double x) {
        double r68345 = 1.0;
        double r68346 = x;
        double r68347 = r68345 + r68346;
        double r68348 = 1.0000000000000033;
        bool r68349 = r68347 <= r68348;
        double r68350 = r68345 * r68346;
        double r68351 = log(r68345);
        double r68352 = r68350 + r68351;
        double r68353 = 0.5;
        double r68354 = 2.0;
        double r68355 = pow(r68346, r68354);
        double r68356 = pow(r68345, r68354);
        double r68357 = r68355 / r68356;
        double r68358 = r68353 * r68357;
        double r68359 = r68352 - r68358;
        double r68360 = sqrt(r68347);
        double r68361 = log(r68360);
        double r68362 = r68361 + r68361;
        double r68363 = r68349 ? r68359 : r68362;
        return r68363;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.8
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 (+ 1.0 x) < 1.0000000000000033

    1. Initial program 59.6

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

    2. Taylor expanded around 0 0.3

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

    if 1.0000000000000033 < (+ 1.0 x)

    1. Initial program 0.9

      \[\log \left(1 + x\right)\]
    2. Using strategy rm

    3. Applied add-sqr-sqrt1.0

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

    4. Applied log-prod1.0

      \[\leadsto \color{blue}{\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000000000333:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020047 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

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

  (log (+ 1 x)))