Average Error: 39.3 → 0.6
Time: 4.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r87560 = 1.0;
        double r87561 = x;
        double r87562 = r87560 + r87561;
        double r87563 = log(r87562);
        return r87563;
}
↓
double f(double x) {
        double r87564 = 1.0;
        double r87565 = x;
        double r87566 = r87564 + r87565;
        bool r87567 = r87566 <= r87564;
        double r87568 = r87564 * r87565;
        double r87569 = log(r87564);
        double r87570 = r87568 + r87569;
        double r87571 = 0.5;
        double r87572 = 2.0;
        double r87573 = pow(r87565, r87572);
        double r87574 = pow(r87564, r87572);
        double r87575 = r87573 / r87574;
        double r87576 = r87571 * r87575;
        double r87577 = r87570 - r87576;
        double r87578 = log(r87566);
        double r87579 = r87567 ? r87577 : r87578;
        return r87579;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
Target0.3
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0

    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 1.0 < (+ 1.0 x)

    1. Initial program 1.2

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

  4. Final simplification0.6

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

Reproduce

herbie shell --seed 2020046 
(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)))