Average Error: 38.4 → 0.6
Time: 10.1s
Precision: 64
\[\log \left(x + 1\right)\]
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000006439293542825907934457064:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(x + 1\right)
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000000000006439293542825907934457064:\\
\;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\

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

\end{array}
double f(double x) {
        double r1084141 = x;
        double r1084142 = 1.0;
        double r1084143 = r1084141 + r1084142;
        double r1084144 = log(r1084143);
        return r1084144;
}

double f(double x) {
        double r1084145 = x;
        double r1084146 = 1.0;
        double r1084147 = r1084145 + r1084146;
        double r1084148 = 1.0000000000000064;
        bool r1084149 = r1084147 <= r1084148;
        double r1084150 = r1084146 * r1084145;
        double r1084151 = log(r1084146);
        double r1084152 = r1084150 + r1084151;
        double r1084153 = 0.5;
        double r1084154 = 2.0;
        double r1084155 = pow(r1084145, r1084154);
        double r1084156 = pow(r1084146, r1084154);
        double r1084157 = r1084155 / r1084156;
        double r1084158 = r1084153 * r1084157;
        double r1084159 = r1084152 - r1084158;
        double r1084160 = log(r1084147);
        double r1084161 = r1084149 ? r1084159 : r1084160;
        return r1084161;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes
  2. if (+ x 1.0) < 1.0000000000000064

    1. Initial program 59.4

      \[\log \left(x + 1\right)\]
    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 1.0000000000000064 < (+ x 1.0)

    1. Initial program 0.9

      \[\log \left(x + 1\right)\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000000006439293542825907934457064:\\ \;\;\;\;\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 1 
(FPCore (x)
  :name "log(x+1)"
  :precision binary64
  (log (+ x 1)))