Average Error: 39.0 → 0.4
Time: 4.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000019218915:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000000019218915:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\

\end{array}
double f(double x) {
        double r78562 = 1.0;
        double r78563 = x;
        double r78564 = r78562 + r78563;
        double r78565 = log(r78564);
        return r78565;
}
↓
double f(double x) {
        double r78566 = 1.0;
        double r78567 = x;
        double r78568 = r78566 + r78567;
        double r78569 = 1.0000000001921892;
        bool r78570 = r78568 <= r78569;
        double r78571 = r78566 * r78567;
        double r78572 = log(r78566);
        double r78573 = r78571 + r78572;
        double r78574 = 0.5;
        double r78575 = 2.0;
        double r78576 = pow(r78567, r78575);
        double r78577 = pow(r78566, r78575);
        double r78578 = r78576 / r78577;
        double r78579 = r78574 * r78578;
        double r78580 = r78573 - r78579;
        double r78581 = sqrt(r78568);
        double r78582 = log(r78581);
        double r78583 = log(r78568);
        double r78584 = r78574 * r78583;
        double r78585 = r78582 + r78584;
        double r78586 = r78570 ? r78580 : r78585;
        return r78586;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
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) < 1.0000000001921892

    1. Initial program 59.3

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

    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 1.0000000001921892 < (+ 1.0 x)

    1. Initial program 0.5

      \[\log \left(1 + x\right)\]
    2. Using strategy rm

    3. Applied add-sqr-sqrt0.6

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

    4. Applied log-prod0.5

      \[\leadsto \color{blue}{\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)}\]
    5. Using strategy rm

    6. Applied pow1/20.5

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

    7. Applied log-pow0.5

      \[\leadsto \log \left(\sqrt{1 + x}\right) + \color{blue}{\frac{1}{2} \cdot \log \left(1 + x\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000019218915:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020034 
(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)))