Average Error: 39.0 → 0.7
Time: 4.1s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000000009:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000000000009:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r62509 = 1.0;
        double r62510 = x;
        double r62511 = r62509 + r62510;
        double r62512 = log(r62511);
        return r62512;
}
↓
double f(double x) {
        double r62513 = 1.0;
        double r62514 = x;
        double r62515 = r62513 + r62514;
        double r62516 = 1.0000000000000009;
        bool r62517 = r62515 <= r62516;
        double r62518 = r62513 * r62514;
        double r62519 = log(r62513);
        double r62520 = r62518 + r62519;
        double r62521 = 0.5;
        double r62522 = 2.0;
        double r62523 = pow(r62514, r62522);
        double r62524 = pow(r62513, r62522);
        double r62525 = r62523 / r62524;
        double r62526 = r62521 * r62525;
        double r62527 = r62520 - r62526;
        double r62528 = sqrt(r62515);
        double r62529 = log(r62528);
        double r62530 = r62529 + r62529;
        double r62531 = r62517 ? r62527 : r62530;
        return r62531;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.3
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) < 1.0000000000000009

    1. Initial program 59.3

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

    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 1.0000000000000009 < (+ 1.0 x)

    1. Initial program 1.1

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

    3. Applied add-sqr-sqrt1.2

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

    4. Applied log-prod1.1

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000000009:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020100 
(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)))