Kahan's exp quotient

Average Error: 39.6 → 0.3

Time: 15.1s

Precision: 64

Math

\[\frac{e^{x} - 1}{x}\]

↓

\[\begin{array}{l} \mathbf{if}\;x \le -9.248950464724178875501603425135499492171 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{\frac{e^{\left(3 \cdot x + 3 \cdot x\right) + 3 \cdot x} - \left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(1 \cdot 1\right) \cdot 1\right)\right)}{e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(1 \cdot 1\right) \cdot 1 + e^{3 \cdot x}\right)}}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\ \end{array}\]

C

double f(double x) {
        double r6309382 = x;
        double r6309383 = exp(r6309382);
        double r6309384 = 1.0;
        double r6309385 = r6309383 - r6309384;
        double r6309386 = r6309385 / r6309382;
        return r6309386;
}

↓

double f(double x) {
        double r6309387 = x;
        double r6309388 = -9.248950464724179e-05;
        bool r6309389 = r6309387 <= r6309388;
        double r6309390 = 3.0;
        double r6309391 = r6309390 * r6309387;
        double r6309392 = r6309391 + r6309391;
        double r6309393 = r6309392 + r6309391;
        double r6309394 = exp(r6309393);
        double r6309395 = 1.0;
        double r6309396 = r6309395 * r6309395;
        double r6309397 = r6309396 * r6309395;
        double r6309398 = r6309397 * r6309397;
        double r6309399 = r6309397 * r6309398;
        double r6309400 = r6309394 - r6309399;
        double r6309401 = exp(r6309391);
        double r6309402 = r6309401 * r6309401;
        double r6309403 = r6309397 + r6309401;
        double r6309404 = r6309397 * r6309403;
        double r6309405 = r6309402 + r6309404;
        double r6309406 = r6309400 / r6309405;
        double r6309407 = exp(r6309387);
        double r6309408 = r6309407 * r6309407;
        double r6309409 = r6309407 * r6309395;
        double r6309410 = r6309396 + r6309409;
        double r6309411 = r6309408 + r6309410;
        double r6309412 = r6309406 / r6309411;
        double r6309413 = r6309412 / r6309387;
        double r6309414 = 1.0;
        double r6309415 = 0.5;
        double r6309416 = 0.16666666666666666;
        double r6309417 = r6309416 * r6309387;
        double r6309418 = r6309415 + r6309417;
        double r6309419 = r6309387 * r6309418;
        double r6309420 = r6309414 + r6309419;
        double r6309421 = r6309389 ? r6309413 : r6309420;
        return r6309421;
}

Error

Bits error versus x

Target

Original	39.6
Target	40.0
Herbie	0.3

\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

1. Split input into 2 regimes

2. if x < -9.248950464724179e-05

   1. Initial program 0.1

      \[\frac{e^{x} - 1}{x}\]
   2. Using strategy rm

   3. Applied flip3-- 0.1

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

   4. Simplified 0.1

      \[\leadsto \frac{\frac{\color{blue}{e^{\left(x + x\right) + x} - \left(1 \cdot 1\right) \cdot 1}}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\]
   5. Using strategy rm

   6. Applied flip3-- 0.1

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

   7. Simplified 0.1

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

   8. Simplified 0.1

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

   if -9.248950464724179e-05 < x

   1. Initial program 60.2

      \[\frac{e^{x} - 1}{x}\]

   2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\frac{1}{2} \cdot x + \left(\frac{1}{6} \cdot {x}^{2} + 1\right)}\]

   3. Simplified 0.4

      \[\leadsto \color{blue}{x \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + 1}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -9.248950464724178875501603425135499492171 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{\frac{e^{\left(3 \cdot x + 3 \cdot x\right) + 3 \cdot x} - \left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(1 \cdot 1\right) \cdot 1\right)\right)}{e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(\left(1 \cdot 1\right) \cdot 1\right) \cdot \left(\left(1 \cdot 1\right) \cdot 1 + e^{3 \cdot x}\right)}}{e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019169 
(FPCore (x)
  :name "Kahan's exp quotient"

  :herbie-target
  (if (and (< x 1.0) (> x -1.0)) (/ (- (exp x) 1.0) (log (exp x))) (/ (- (exp x) 1.0) x))

  (/ (- (exp x) 1.0) x))