Average Error: 39.1 → 0.2
Time: 3.5s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000025653698017:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, x, \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right) - 1 \cdot {x}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000025653698017:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, x, \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right) - 1 \cdot {x}^{2}\right)\\

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

\end{array}
double code(double x) {
	return log((1.0 + x));
}
↓
double code(double x) {
	double VAR;
	if (((1.0 + x) <= 1.0000025653698017)) {
		VAR = fma(0.3333333333333333, (pow(x, 3.0) / pow(1.0, 3.0)), (fma(1.0, x, (0.5 * (pow(x, 2.0) / pow(1.0, 2.0)))) - (1.0 * pow(x, 2.0))));
	} else {
		VAR = log((1.0 + x));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
Target0.3
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000025653698017

    1. Initial program 59.0

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

    3. Applied flip-+59.0

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

    4. Applied log-div59.0

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

    5. Taylor expanded around 0 0.3

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

    6. Simplified0.3

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

    if 1.0000025653698017 < (+ 1.0 x)

    1. Initial program 0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000025653698017:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{3}, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, x, \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right) - 1 \cdot {x}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020071 +o rules:numerics
(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)))