Average Error: 39.2 → 0.6
Time: 3.6s
Precision: 64
\[\log \left(1 + x\right)\]
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000000666133814775093924254179:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \log \left(1 + x\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000000000000666133814775093924254179:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r67268 = 1.0;
        double r67269 = x;
        double r67270 = r67268 + r67269;
        double r67271 = log(r67270);
        return r67271;
}

double f(double x) {
        double r67272 = 1.0;
        double r67273 = x;
        double r67274 = r67272 + r67273;
        double r67275 = 1.0000000000000007;
        bool r67276 = r67274 <= r67275;
        double r67277 = r67272 * r67273;
        double r67278 = log(r67272);
        double r67279 = r67277 + r67278;
        double r67280 = 0.5;
        double r67281 = 2.0;
        double r67282 = pow(r67273, r67281);
        double r67283 = pow(r67272, r67281);
        double r67284 = r67282 / r67283;
        double r67285 = r67280 * r67284;
        double r67286 = r67279 - r67285;
        double r67287 = 1.0;
        double r67288 = r67287 / r67281;
        double r67289 = log(r67274);
        double r67290 = r67288 * r67289;
        double r67291 = r67290 + r67290;
        double r67292 = r67276 ? r67286 : r67291;
        return r67292;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.2
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 59.5

      \[\log \left(1 + x\right)\]
    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 1.0000000000000007 < (+ 1.0 x)

    1. Initial program 1.0

      \[\log \left(1 + x\right)\]
    2. Using strategy rm
    3. Applied add-sqr-sqrt1.1

      \[\leadsto \log \color{blue}{\left(\sqrt{1 + x} \cdot \sqrt{1 + x}\right)}\]
    4. Applied log-prod1.1

      \[\leadsto \color{blue}{\log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{1 + x}\right)}\]
    5. Using strategy rm
    6. Applied pow11.1

      \[\leadsto \log \left(\sqrt{1 + x}\right) + \log \left(\sqrt{\color{blue}{{\left(1 + x\right)}^{1}}}\right)\]
    7. Applied sqrt-pow11.1

      \[\leadsto \log \left(\sqrt{1 + x}\right) + \log \color{blue}{\left({\left(1 + x\right)}^{\left(\frac{1}{2}\right)}\right)}\]
    8. Applied log-pow1.0

      \[\leadsto \log \left(\sqrt{1 + x}\right) + \color{blue}{\frac{1}{2} \cdot \log \left(1 + x\right)}\]
    9. Using strategy rm
    10. Applied pow11.0

      \[\leadsto \log \left(\sqrt{\color{blue}{{\left(1 + x\right)}^{1}}}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\]
    11. Applied sqrt-pow11.0

      \[\leadsto \log \color{blue}{\left({\left(1 + x\right)}^{\left(\frac{1}{2}\right)}\right)} + \frac{1}{2} \cdot \log \left(1 + x\right)\]
    12. Applied log-pow1.0

      \[\leadsto \color{blue}{\frac{1}{2} \cdot \log \left(1 + x\right)} + \frac{1}{2} \cdot \log \left(1 + x\right)\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000000000000666133814775093924254179:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \log \left(1 + x\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019354 
(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)))