Average Error: 38.9 → 0.4
Time: 5.0s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000004809:\\ \;\;\;\;\mathsf{fma}\left(-1, \mathsf{fma}\left(x, x, \frac{{x}^{3}}{{1}^{2}}\right), \mathsf{fma}\left(0.333333333333333315, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, {x}^{3}, \mathsf{fma}\left(0.5, \frac{{x}^{2}}{{1}^{2}}, 1 \cdot x\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000000000004809:\\
\;\;\;\;\mathsf{fma}\left(-1, \mathsf{fma}\left(x, x, \frac{{x}^{3}}{{1}^{2}}\right), \mathsf{fma}\left(0.333333333333333315, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, {x}^{3}, \mathsf{fma}\left(0.5, \frac{{x}^{2}}{{1}^{2}}, 1 \cdot x\right)\right)\right)\right)\\

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

\end{array}
double code(double x) {
	return log((1.0 + x));
}
↓
double code(double x) {
	double temp;
	if (((1.0 + x) <= 1.000000000000481)) {
		temp = fma(-1.0, fma(x, x, (pow(x, 3.0) / pow(1.0, 2.0))), fma(0.3333333333333333, (pow(x, 3.0) / pow(1.0, 3.0)), fma(1.0, pow(x, 3.0), fma(0.5, (pow(x, 2.0) / pow(1.0, 2.0)), (1.0 * x)))));
	} else {
		temp = log((1.0 + x));
	}
	return temp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.000000000000481

    1. Initial program 59.3

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

    3. Applied flip3-+59.4

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

    4. Applied log-div59.4

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

    5. Taylor expanded around 0 0.3

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + \left(0.5 \cdot \frac{{x}^{2}}{{1}^{2}} + 1 \cdot x\right)\right)\right) - \left(1 \cdot {x}^{2} + 1 \cdot \frac{{x}^{3}}{{1}^{2}}\right)}\]

    6. Simplified0.3

      \[\leadsto \color{blue}{\mathsf{fma}\left(-1, \mathsf{fma}\left(x, x, \frac{{x}^{3}}{{1}^{2}}\right), \mathsf{fma}\left(0.333333333333333315, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, {x}^{3}, \mathsf{fma}\left(0.5, \frac{{x}^{2}}{{1}^{2}}, 1 \cdot x\right)\right)\right)\right)}\]

    if 1.000000000000481 < (+ 1.0 x)

    1. Initial program 0.8

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000004809:\\ \;\;\;\;\mathsf{fma}\left(-1, \mathsf{fma}\left(x, x, \frac{{x}^{3}}{{1}^{2}}\right), \mathsf{fma}\left(0.333333333333333315, \frac{{x}^{3}}{{1}^{3}}, \mathsf{fma}\left(1, {x}^{3}, \mathsf{fma}\left(0.5, \frac{{x}^{2}}{{1}^{2}}, 1 \cdot x\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 +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)))