Average Error: 58.7 → 0.2
Time: 6.6s
Precision: 64
\[\frac{1}{2} \cdot \log \left(\frac{1 + x}{1 - x}\right)\]
↓
\[\frac{1}{2} \cdot \mathsf{fma}\left(\frac{2}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(2, x, \frac{2}{5} \cdot \frac{{x}^{5}}{{1}^{5}}\right)\right)\]
\frac{1}{2} \cdot \log \left(\frac{1 + x}{1 - x}\right)
↓
\frac{1}{2} \cdot \mathsf{fma}\left(\frac{2}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(2, x, \frac{2}{5} \cdot \frac{{x}^{5}}{{1}^{5}}\right)\right)
double f(double x) {
        double r107699 = 1.0;
        double r107700 = 2.0;
        double r107701 = r107699 / r107700;
        double r107702 = x;
        double r107703 = r107699 + r107702;
        double r107704 = r107699 - r107702;
        double r107705 = r107703 / r107704;
        double r107706 = log(r107705);
        double r107707 = r107701 * r107706;
        return r107707;
}
↓
double f(double x) {
        double r107708 = 1.0;
        double r107709 = 2.0;
        double r107710 = r107708 / r107709;
        double r107711 = 0.6666666666666666;
        double r107712 = x;
        double r107713 = 3.0;
        double r107714 = pow(r107712, r107713);
        double r107715 = pow(r107708, r107713);
        double r107716 = r107714 / r107715;
        double r107717 = 0.4;
        double r107718 = 5.0;
        double r107719 = pow(r107712, r107718);
        double r107720 = pow(r107708, r107718);
        double r107721 = r107719 / r107720;
        double r107722 = r107717 * r107721;
        double r107723 = fma(r107709, r107712, r107722);
        double r107724 = fma(r107711, r107716, r107723);
        double r107725 = r107710 * r107724;
        return r107725;
}

Error

Bits error versus x

Derivation

  1. Initial program 58.7

    \[\frac{1}{2} \cdot \log \left(\frac{1 + x}{1 - x}\right)\]
  2. Using strategy rm

  3. Applied log-div58.7

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

  4. Taylor expanded around 0 0.2

    \[\leadsto \frac{1}{2} \cdot \color{blue}{\left(\frac{2}{3} \cdot \frac{{x}^{3}}{{1}^{3}} + \left(2 \cdot x + \frac{2}{5} \cdot \frac{{x}^{5}}{{1}^{5}}\right)\right)}\]

  5. Simplified0.2

    \[\leadsto \frac{1}{2} \cdot \color{blue}{\mathsf{fma}\left(\frac{2}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(2, x, \frac{2}{5} \cdot \frac{{x}^{5}}{{1}^{5}}\right)\right)}\]

  6. Final simplification0.2

    \[\leadsto \frac{1}{2} \cdot \mathsf{fma}\left(\frac{2}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(2, x, \frac{2}{5} \cdot \frac{{x}^{5}}{{1}^{5}}\right)\right)\]

Reproduce

herbie shell --seed 2020081 +o rules:numerics
(FPCore (x)
  :name "Hyperbolic arc-(co)tangent"
  :precision binary64
  (* (/ 1 2) (log (/ (+ 1 x) (- 1 x)))))