Average Error: 39.1 → 0.6
Time: 12.1s
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 r3804946 = 1.0;
        double r3804947 = x;
        double r3804948 = r3804946 + r3804947;
        double r3804949 = log(r3804948);
        return r3804949;
}
↓
double f(double x) {
        double r3804950 = x;
        double r3804951 = 1.0;
        double r3804952 = r3804950 + r3804951;
        bool r3804953 = r3804952 <= r3804951;
        double r3804954 = 0.5;
        double r3804955 = r3804954 * r3804950;
        double r3804956 = r3804951 - r3804955;
        double r3804957 = r3804950 * r3804956;
        double r3804958 = log(r3804951);
        double r3804959 = r3804957 + r3804958;
        double r3804960 = sqrt(r3804952);
        double r3804961 = log(r3804960);
        double r3804962 = r3804961 + r3804961;
        double r3804963 = r3804953 ? r3804959 : r3804962;
        return r3804963;
}

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

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

    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

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

    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)))