ln(1 + x)

Average Error: 39.4 → 0.3

Time: 3.7s

Precision: binary64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.0000001392773128:\\ \;\;\;\;1 \cdot x + \left(\log 1 + \left(x \cdot x\right) \cdot \frac{-0.5}{1 \cdot 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

C

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

↓

double code(double x) {
	double VAR;
	if ((((double) (1.0 + x)) <= 1.0000001392773128)) {
		VAR = ((double) (((double) (1.0 * x)) + ((double) (((double) log(1.0)) + ((double) (((double) (x * x)) * (-0.5 / ((double) (1.0 * 1.0)))))))));
	} else {
		VAR = ((double) log(((double) (1.0 + x))));
	}
	return VAR;
}

Error

Bits error versus x

Target

Original	39.4
Target	0.2
Herbie	0.3

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

Derivation

1. Split input into 2 regimes

2. if (+ 1.0 x) < 1.0000001392773128

   1. Initial program 59.1

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

   2. Taylor expanded around 0 0.4

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

   3. Simplified 0.4

      \[\leadsto \color{blue}{1 \cdot x + \left(\log 1 + \left(x \cdot x\right) \cdot \frac{-0.5}{1 \cdot 1}\right)}\]

   if 1.0000001392773128 < (+ 1.0 x)

   1. Initial program 0.2

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

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \leq 1.0000001392773128:\\ \;\;\;\;1 \cdot x + \left(\log 1 + \left(x \cdot x\right) \cdot \frac{-0.5}{1 \cdot 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020196 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))