ln(1 + x)

Average Error: 39.1 → 0.3

Time: 21.6s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

C

double f(double x) {
        double r63343 = 1.0;
        double r63344 = x;
        double r63345 = r63343 + r63344;
        double r63346 = log(r63345);
        return r63346;
}

↓

double f(double x) {
        double r63347 = 1.0;
        double r63348 = x;
        double r63349 = r63347 + r63348;
        double r63350 = 1.000002024200259;
        bool r63351 = r63349 <= r63350;
        double r63352 = r63348 * r63347;
        double r63353 = -0.5;
        double r63354 = r63347 * r63347;
        double r63355 = r63354 / r63348;
        double r63356 = r63353 / r63355;
        double r63357 = r63348 * r63356;
        double r63358 = r63352 + r63357;
        double r63359 = log(r63347);
        double r63360 = r63358 + r63359;
        double r63361 = log(r63349);
        double r63362 = r63351 ? r63360 : r63361;
        return r63362;
}

Error

Bits error versus x

Target

Original	39.1
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) < 1.000002024200259

   1. Initial program 59.1

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

   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

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

   5. Applied distribute-lft-in 0.3

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

   if 1.000002024200259 < (+ 1.0 x)

   1. Initial program 0.1

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

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.00000202420025896543620547163300216198:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 
(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)))