Average Error: 39.2 → 0.3
Time: 12.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000005236653777274113963358104228973:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000005236653777274113963358104228973:\\
\;\;\;\;\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 r46117 = 1.0;
        double r46118 = x;
        double r46119 = r46117 + r46118;
        double r46120 = log(r46119);
        return r46120;
}
↓
double f(double x) {
        double r46121 = 1.0;
        double r46122 = x;
        double r46123 = r46121 + r46122;
        double r46124 = 1.0000000052366538;
        bool r46125 = r46123 <= r46124;
        double r46126 = r46121 * r46122;
        double r46127 = log(r46121);
        double r46128 = r46126 + r46127;
        double r46129 = 0.5;
        double r46130 = 2.0;
        double r46131 = pow(r46122, r46130);
        double r46132 = pow(r46121, r46130);
        double r46133 = r46131 / r46132;
        double r46134 = r46129 * r46133;
        double r46135 = r46128 - r46134;
        double r46136 = log(r46123);
        double r46137 = r46125 ? r46135 : r46136;
        return r46137;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
Target0.3
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000052366538

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.4

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

    if 1.0000000052366538 < (+ 1.0 x)

    1. Initial program 0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000005236653777274113963358104228973:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019235 
(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)))