Average Error: 29.5 → 1.0
Time: 6.3s
Precision: 64
\[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 3.591957276208114:\\ \;\;\;\;\frac{\mathsf{fma}\left({x}^{3}, 0.66666666666666674, 2 - 1 \cdot {x}^{2}\right)}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{fma}\left(1 + \frac{1}{\varepsilon}, e^{-\left(1 - \varepsilon\right) \cdot x}, -\left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}\right)}{2}\\ \end{array}\]
\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}
↓
\begin{array}{l}
\mathbf{if}\;x \le 3.591957276208114:\\
\;\;\;\;\frac{\mathsf{fma}\left({x}^{3}, 0.66666666666666674, 2 - 1 \cdot {x}^{2}\right)}{2}\\

\mathbf{else}:\\
\;\;\;\;\frac{\mathsf{fma}\left(1 + \frac{1}{\varepsilon}, e^{-\left(1 - \varepsilon\right) \cdot x}, -\left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}\right)}{2}\\

\end{array}
double code(double x, double eps) {
	return ((((1.0 + (1.0 / eps)) * exp(-((1.0 - eps) * x))) - (((1.0 / eps) - 1.0) * exp(-((1.0 + eps) * x)))) / 2.0);
}
↓
double code(double x, double eps) {
	double temp;
	if ((x <= 3.591957276208114)) {
		temp = (fma(pow(x, 3.0), 0.6666666666666667, (2.0 - (1.0 * pow(x, 2.0)))) / 2.0);
	} else {
		temp = (fma((1.0 + (1.0 / eps)), exp(-((1.0 - eps) * x)), -(((1.0 / eps) - 1.0) * exp(-((1.0 + eps) * x)))) / 2.0);
	}
	return temp;
}

Error

Bits error versus x

Bits error versus eps

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if x < 3.591957276208114

    1. Initial program 39.0

      \[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]

    2. Taylor expanded around 0 1.2

      \[\leadsto \frac{\color{blue}{\left(0.66666666666666674 \cdot {x}^{3} + 2\right) - 1 \cdot {x}^{2}}}{2}\]

    3. Simplified1.2

      \[\leadsto \frac{\color{blue}{\mathsf{fma}\left({x}^{3}, 0.66666666666666674, 2 - 1 \cdot {x}^{2}\right)}}{2}\]

    if 3.591957276208114 < x

    1. Initial program 0.4

      \[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
    2. Using strategy rm

    3. Applied fma-neg0.4

      \[\leadsto \frac{\color{blue}{\mathsf{fma}\left(1 + \frac{1}{\varepsilon}, e^{-\left(1 - \varepsilon\right) \cdot x}, -\left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}\right)}}{2}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification1.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 3.591957276208114:\\ \;\;\;\;\frac{\mathsf{fma}\left({x}^{3}, 0.66666666666666674, 2 - 1 \cdot {x}^{2}\right)}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{fma}\left(1 + \frac{1}{\varepsilon}, e^{-\left(1 - \varepsilon\right) \cdot x}, -\left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}\right)}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 +o rules:numerics
(FPCore (x eps)
  :name "NMSE Section 6.1 mentioned, A"
  :precision binary64
  (/ (- (* (+ 1 (/ 1 eps)) (exp (- (* (- 1 eps) x)))) (* (- (/ 1 eps) 1) (exp (- (* (+ 1 eps) x))))) 2))