Average Error: 38.7 → 0.3
Time: 9.6s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000001309967689877566954237408936024:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000001309967689877566954237408936024:\\
\;\;\;\;\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 r41940 = 1.0;
        double r41941 = x;
        double r41942 = r41940 + r41941;
        double r41943 = log(r41942);
        return r41943;
}
↓
double f(double x) {
        double r41944 = 1.0;
        double r41945 = x;
        double r41946 = r41944 + r41945;
        double r41947 = 1.0000000013099677;
        bool r41948 = r41946 <= r41947;
        double r41949 = r41944 * r41945;
        double r41950 = log(r41944);
        double r41951 = r41949 + r41950;
        double r41952 = 0.5;
        double r41953 = 2.0;
        double r41954 = pow(r41945, r41953);
        double r41955 = pow(r41944, r41953);
        double r41956 = r41954 / r41955;
        double r41957 = r41952 * r41956;
        double r41958 = r41951 - r41957;
        double r41959 = sqrt(r41946);
        double r41960 = log(r41959);
        double r41961 = log(r41946);
        double r41962 = r41952 * r41961;
        double r41963 = r41960 + r41962;
        double r41964 = r41948 ? r41958 : r41963;
        return r41964;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.7
Target0.3
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000013099677

    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 1.0000000013099677 < (+ 1.0 x)

    1. Initial program 0.3

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

    3. Applied add-sqr-sqrt0.4

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

    4. Applied log-prod0.4

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

    6. Applied pow1/20.4

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

    7. Applied log-pow0.3

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000001309967689877566954237408936024:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{1 + x}\right) + \frac{1}{2} \cdot \log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019298 
(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)))