Average Error: 39.3 → 0.3
Time: 3.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000004755978571:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000004755978571:\\
\;\;\;\;\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 r68372 = 1.0;
        double r68373 = x;
        double r68374 = r68372 + r68373;
        double r68375 = log(r68374);
        return r68375;
}
↓
double f(double x) {
        double r68376 = 1.0;
        double r68377 = x;
        double r68378 = r68376 + r68377;
        double r68379 = 1.000000475597857;
        bool r68380 = r68378 <= r68379;
        double r68381 = r68376 * r68377;
        double r68382 = log(r68376);
        double r68383 = r68381 + r68382;
        double r68384 = 0.5;
        double r68385 = 2.0;
        double r68386 = pow(r68377, r68385);
        double r68387 = pow(r68376, r68385);
        double r68388 = r68386 / r68387;
        double r68389 = r68384 * r68388;
        double r68390 = r68383 - r68389;
        double r68391 = log(r68378);
        double r68392 = r68380 ? r68390 : r68391;
        return r68392;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.3
Target0.2
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) < 1.000000475597857

    1. Initial program 59.0

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

    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 1.000000475597857 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000004755978571:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020027 
(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)))