Average Error: 39.2 → 0.4
Time: 3.9s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000133302457960837728023761883378:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000000133302457960837728023761883378:\\
\;\;\;\;\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 r77553 = 1.0;
        double r77554 = x;
        double r77555 = r77553 + r77554;
        double r77556 = log(r77555);
        return r77556;
}
↓
double f(double x) {
        double r77557 = 1.0;
        double r77558 = x;
        double r77559 = r77557 + r77558;
        double r77560 = 1.0000000013330246;
        bool r77561 = r77559 <= r77560;
        double r77562 = r77557 * r77558;
        double r77563 = log(r77557);
        double r77564 = r77562 + r77563;
        double r77565 = 0.5;
        double r77566 = 2.0;
        double r77567 = pow(r77558, r77566);
        double r77568 = pow(r77557, r77566);
        double r77569 = r77567 / r77568;
        double r77570 = r77565 * r77569;
        double r77571 = r77564 - r77570;
        double r77572 = log(r77559);
        double r77573 = r77561 ? r77571 : r77572;
        return r77573;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000013330246

    1. Initial program 59.2

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

    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 1.0000000013330246 < (+ 1.0 x)

    1. Initial program 0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000133302457960837728023761883378:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019362 
(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)))