ln(1 + x)

Average Error: 39.3 → 0.4

Time: 3.3s

Precision: 64

Math

\[\log \left(1 + x\right)\]

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000568305:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;{\left(\log \left(1 + x\right)\right)}^{1}\\ \end{array}\]

C

double code(double x) {
	return log((1.0 + x));
}

↓

double code(double x) {
	double temp;
	if (((1.0 + x) <= 1.0000000000568305)) {
		temp = fma(x, 1.0, (log(1.0) - (0.5 * (pow(x, 2.0) / pow(1.0, 2.0)))));
	} else {
		temp = pow(log((1.0 + x)), 1.0);
	}
	return temp;
}

Error

Bits error versus x

Target

Original	39.3
Target	0.2
Herbie	0.4

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

Derivation

1. Split input into 2 regimes

2. if (+ 1.0 x) < 1.0000000000568305

   1. Initial program 59.3

      \[\log \left(1 + x\right)\]

   2. Taylor expanded around 0 0.3

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

   3. Simplified 0.3

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

   if 1.0000000000568305 < (+ 1.0 x)

   1. Initial program 0.5

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

   3. Applied pow1 0.5

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

4. Final simplification 0.4

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000000000568305:\\ \;\;\;\;\mathsf{fma}\left(x, 1, \log 1 - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;{\left(\log \left(1 + x\right)\right)}^{1}\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 +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)))