Average Error: 29.1 → 1.1
Time: 5.7s
Precision: 64
\[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 130.404033395790208:\\ \;\;\;\;\frac{\left(0.66666666666666674 \cdot {x}^{3} + 2\right) - 1 \cdot {x}^{2}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{1 \cdot \left(\frac{e^{x \cdot \varepsilon - 1 \cdot x}}{\varepsilon} + e^{x \cdot \varepsilon - 1 \cdot x}\right) - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\\ \end{array}\]
\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 130.404033395790208:\\
\;\;\;\;\frac{\left(0.66666666666666674 \cdot {x}^{3} + 2\right) - 1 \cdot {x}^{2}}{2}\\

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

\end{array}
double f(double x, double eps) {
        double r36781 = 1.0;
        double r36782 = eps;
        double r36783 = r36781 / r36782;
        double r36784 = r36781 + r36783;
        double r36785 = r36781 - r36782;
        double r36786 = x;
        double r36787 = r36785 * r36786;
        double r36788 = -r36787;
        double r36789 = exp(r36788);
        double r36790 = r36784 * r36789;
        double r36791 = r36783 - r36781;
        double r36792 = r36781 + r36782;
        double r36793 = r36792 * r36786;
        double r36794 = -r36793;
        double r36795 = exp(r36794);
        double r36796 = r36791 * r36795;
        double r36797 = r36790 - r36796;
        double r36798 = 2.0;
        double r36799 = r36797 / r36798;
        return r36799;
}
↓
double f(double x, double eps) {
        double r36800 = x;
        double r36801 = 130.4040333957902;
        bool r36802 = r36800 <= r36801;
        double r36803 = 0.6666666666666667;
        double r36804 = 3.0;
        double r36805 = pow(r36800, r36804);
        double r36806 = r36803 * r36805;
        double r36807 = 2.0;
        double r36808 = r36806 + r36807;
        double r36809 = 1.0;
        double r36810 = 2.0;
        double r36811 = pow(r36800, r36810);
        double r36812 = r36809 * r36811;
        double r36813 = r36808 - r36812;
        double r36814 = r36813 / r36807;
        double r36815 = eps;
        double r36816 = r36800 * r36815;
        double r36817 = r36809 * r36800;
        double r36818 = r36816 - r36817;
        double r36819 = exp(r36818);
        double r36820 = r36819 / r36815;
        double r36821 = r36820 + r36819;
        double r36822 = r36809 * r36821;
        double r36823 = r36809 / r36815;
        double r36824 = r36823 - r36809;
        double r36825 = r36809 + r36815;
        double r36826 = r36825 * r36800;
        double r36827 = -r36826;
        double r36828 = exp(r36827);
        double r36829 = r36824 * r36828;
        double r36830 = r36822 - r36829;
        double r36831 = r36830 / r36807;
        double r36832 = r36802 ? r36814 : r36831;
        return r36832;
}

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 < 130.4040333957902

    1. Initial program 38.8

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

    2. Taylor expanded around 0 1.4

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

    if 130.4040333957902 < x

    1. Initial program 0.2

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

    2. Taylor expanded around inf 0.2

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

    3. Simplified0.2

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

  4. Final simplification1.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 130.404033395790208:\\ \;\;\;\;\frac{\left(0.66666666666666674 \cdot {x}^{3} + 2\right) - 1 \cdot {x}^{2}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{1 \cdot \left(\frac{e^{x \cdot \varepsilon - 1 \cdot x}}{\varepsilon} + e^{x \cdot \varepsilon - 1 \cdot x}\right) - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020060 
(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))