Average Error: 39.1 → 0.6
Time: 12.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1:\\ \;\;\;\;x \cdot \left(1 - 0.5 \cdot x\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{x + 1}\right) + \log \left(\sqrt{x + 1}\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1:\\
\;\;\;\;x \cdot \left(1 - 0.5 \cdot x\right) + \log 1\\

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

\end{array}
double f(double x) {
        double r3543494 = 1.0;
        double r3543495 = x;
        double r3543496 = r3543494 + r3543495;
        double r3543497 = log(r3543496);
        return r3543497;
}
↓
double f(double x) {
        double r3543498 = x;
        double r3543499 = 1.0;
        double r3543500 = r3543498 + r3543499;
        bool r3543501 = r3543500 <= r3543499;
        double r3543502 = 0.5;
        double r3543503 = r3543502 * r3543498;
        double r3543504 = r3543499 - r3543503;
        double r3543505 = r3543498 * r3543504;
        double r3543506 = log(r3543499);
        double r3543507 = r3543505 + r3543506;
        double r3543508 = sqrt(r3543500);
        double r3543509 = log(r3543508);
        double r3543510 = r3543509 + r3543509;
        double r3543511 = r3543501 ? r3543507 : r3543510;
        return r3543511;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
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 59.5

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    4. Taylor expanded around 0 0.3

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

    5. Simplified0.3

      \[\leadsto \color{blue}{\log 1 + x \cdot \left(1 - 0.5 \cdot x\right)}\]

    if 1.0 < (+ 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 simplification0.6

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

Reproduce

herbie shell --seed 2019171 
(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)))