Average Error: 38.8 → 0.5
Time: 3.7s
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 r69464 = 1.0;
        double r69465 = x;
        double r69466 = r69464 + r69465;
        double r69467 = log(r69466);
        return r69467;
}
↓
double f(double x) {
        double r69468 = 1.0;
        double r69469 = x;
        double r69470 = r69468 + r69469;
        double r69471 = 1.0000000000000033;
        bool r69472 = r69470 <= r69471;
        double r69473 = r69468 * r69469;
        double r69474 = log(r69468);
        double r69475 = r69473 + r69474;
        double r69476 = 0.5;
        double r69477 = 2.0;
        double r69478 = pow(r69469, r69477);
        double r69479 = pow(r69468, r69477);
        double r69480 = r69478 / r69479;
        double r69481 = r69476 * r69480;
        double r69482 = r69475 - r69481;
        double r69483 = sqrt(r69470);
        double r69484 = log(r69483);
        double r69485 = r69484 + r69484;
        double r69486 = r69472 ? r69482 : r69485;
        return r69486;
}

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)))