Average Error: 38.6 → 0.7
Time: 4.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000000000222044604925031308084726:\\
\;\;\;\;\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 r67169 = 1.0;
        double r67170 = x;
        double r67171 = r67169 + r67170;
        double r67172 = log(r67171);
        return r67172;
}
↓
double f(double x) {
        double r67173 = 1.0;
        double r67174 = x;
        double r67175 = r67173 + r67174;
        double r67176 = 1.0000000000000002;
        bool r67177 = r67175 <= r67176;
        double r67178 = r67173 * r67174;
        double r67179 = log(r67173);
        double r67180 = r67178 + r67179;
        double r67181 = 0.5;
        double r67182 = 2.0;
        double r67183 = pow(r67174, r67182);
        double r67184 = pow(r67173, r67182);
        double r67185 = r67183 / r67184;
        double r67186 = r67181 * r67185;
        double r67187 = r67180 - r67186;
        double r67188 = log(r67175);
        double r67189 = r67177 ? r67187 : r67188;
        return r67189;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.6
Target0.2
Herbie0.7
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000000002

    1. Initial program 59.4

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

    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 1.0000000000000002 < (+ 1.0 x)

    1. Initial program 1.2

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000000222044604925031308084726:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019318 
(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)))