Average Error: 38.7 → 0.4
Time: 11.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000810462807976364274509251118:\\ \;\;\;\;\left(0.3333333333333333148296162562473909929395 \cdot {x}^{3} + 1 \cdot x\right) - 0.5 \cdot {x}^{2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000000810462807976364274509251118:\\
\;\;\;\;\left(0.3333333333333333148296162562473909929395 \cdot {x}^{3} + 1 \cdot x\right) - 0.5 \cdot {x}^{2}\\

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

\end{array}
double f(double x) {
        double r58501 = 1.0;
        double r58502 = x;
        double r58503 = r58501 + r58502;
        double r58504 = log(r58503);
        return r58504;
}
↓
double f(double x) {
        double r58505 = 1.0;
        double r58506 = x;
        double r58507 = r58505 + r58506;
        double r58508 = 1.0000000000008105;
        bool r58509 = r58507 <= r58508;
        double r58510 = 0.3333333333333333;
        double r58511 = 3.0;
        double r58512 = pow(r58506, r58511);
        double r58513 = r58510 * r58512;
        double r58514 = r58505 * r58506;
        double r58515 = r58513 + r58514;
        double r58516 = 0.5;
        double r58517 = 2.0;
        double r58518 = pow(r58506, r58517);
        double r58519 = r58516 * r58518;
        double r58520 = r58515 - r58519;
        double r58521 = log(r58507);
        double r58522 = r58509 ? r58520 : r58521;
        return r58522;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.7
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000008105

    1. Initial program 59.4

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

    3. Applied flip-+59.4

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

    4. Applied log-div59.4

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

    5. Taylor expanded around 0 0.2

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

    6. Taylor expanded around 0 0.2

      \[\leadsto \color{blue}{\left(0.3333333333333333148296162562473909929395 \cdot {x}^{3} + 1 \cdot x\right) - 0.5 \cdot {x}^{2}}\]

    if 1.0000000000008105 < (+ 1.0 x)

    1. Initial program 0.7

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000810462807976364274509251118:\\ \;\;\;\;\left(0.3333333333333333148296162562473909929395 \cdot {x}^{3} + 1 \cdot x\right) - 0.5 \cdot {x}^{2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019209 
(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)))