Kahan's exp quotient

Average Error: 39.8 → 0.4

Time: 2.4s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.69715010585821388 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)\right)\\ \end{array}\]

C

double f(double x) {
        double r76663 = x;
        double r76664 = exp(r76663);
        double r76665 = 1.0;
        double r76666 = r76664 - r76665;
        double r76667 = r76666 / r76663;
        return r76667;
}

↓

double f(double x) {
        double r76668 = x;
        double r76669 = -0.0001697150105858214;
        bool r76670 = r76668 <= r76669;
        double r76671 = 1.0;
        double r76672 = -r76671;
        double r76673 = r76668 + r76668;
        double r76674 = exp(r76673);
        double r76675 = fma(r76672, r76671, r76674);
        double r76676 = exp(r76668);
        double r76677 = r76676 + r76671;
        double r76678 = r76675 / r76677;
        double r76679 = r76678 / r76668;
        double r76680 = 0.052083333333333336;
        double r76681 = 2.0;
        double r76682 = pow(r76668, r76681);
        double r76683 = 0.25;
        double r76684 = log(r76681);
        double r76685 = fma(r76683, r76668, r76684);
        double r76686 = fma(r76680, r76682, r76685);
        double r76687 = expm1(r76686);
        double r76688 = r76670 ? r76679 : r76687;
        return r76688;
}

Error

Bits error versus x

Target

Original	39.8
Target	40.2
Herbie	0.4

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

Derivation

1. Split input into 2 regimes

2. if x < -0.0001697150105858214

   1. Initial program 0.1

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

   3. Applied flip-- 0.1

      \[\leadsto \frac{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}{x}\]

   4. Simplified 0.0

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

   if -0.0001697150105858214 < x

   1. Initial program 60.0

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

   2. Taylor expanded around 0 0.5

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

   3. Simplified 0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]
   4. Using strategy rm

   5. Applied expm1-log1p-u 0.5

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\right)\right)}\]

   6. Taylor expanded around 0 0.5

      \[\leadsto \mathsf{expm1}\left(\color{blue}{\frac{5}{96} \cdot {x}^{2} + \left(\frac{1}{4} \cdot x + \log 2\right)}\right)\]

   7. Simplified 0.5

      \[\leadsto \mathsf{expm1}\left(\color{blue}{\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)}\right)\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.4

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.69715010585821388 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 +o rules:numerics
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

  (/ (- (exp x) 1) x))