Average Error: 38.5 → 0.6
Time: 3.4s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000000000067:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.00000000000000067:\\
\;\;\;\;\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 r56317 = 1.0;
        double r56318 = x;
        double r56319 = r56317 + r56318;
        double r56320 = log(r56319);
        return r56320;
}
↓
double f(double x) {
        double r56321 = 1.0;
        double r56322 = x;
        double r56323 = r56321 + r56322;
        double r56324 = 1.0000000000000007;
        bool r56325 = r56323 <= r56324;
        double r56326 = r56321 * r56322;
        double r56327 = log(r56321);
        double r56328 = r56326 + r56327;
        double r56329 = 0.5;
        double r56330 = 2.0;
        double r56331 = pow(r56322, r56330);
        double r56332 = pow(r56321, r56330);
        double r56333 = r56331 / r56332;
        double r56334 = r56329 * r56333;
        double r56335 = r56328 - r56334;
        double r56336 = sqrt(r56323);
        double r56337 = log(r56336);
        double r56338 = log(r56323);
        double r56339 = r56329 * r56338;
        double r56340 = r56337 + r56339;
        double r56341 = r56325 ? r56335 : r56340;
        return r56341;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.5
Target0.3
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000000007

    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 1.0000000000000007 < (+ 1.0 x)

    1. Initial program 1.0

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

    3. Applied add-sqr-sqrt1.0

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

    4. Applied log-prod1.0

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

    6. Applied pow1/21.0

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

    7. Applied log-pow1.0

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000000000000067:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020025 
(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)))