Average Error: 29.9 → 1.1
Time: 33.6s
Precision: 64
\[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 2.224816812712469:\\ \;\;\;\;\frac{\left(2 - x \cdot x\right) + \left(\frac{2}{3} \cdot x\right) \cdot \left(x \cdot x\right)}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(\frac{1}{\varepsilon} + 1\right) \cdot e^{\left(1 - \varepsilon\right) \cdot \left(-x\right)} - \frac{\frac{1}{\varepsilon} - 1}{e^{\left(\varepsilon + 1\right) \cdot x}}}{2}\\ \end{array}\]
\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 2.224816812712469:\\
\;\;\;\;\frac{\left(2 - x \cdot x\right) + \left(\frac{2}{3} \cdot x\right) \cdot \left(x \cdot x\right)}{2}\\

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

\end{array}
double f(double x, double eps) {
        double r1998121 = 1.0;
        double r1998122 = eps;
        double r1998123 = r1998121 / r1998122;
        double r1998124 = r1998121 + r1998123;
        double r1998125 = r1998121 - r1998122;
        double r1998126 = x;
        double r1998127 = r1998125 * r1998126;
        double r1998128 = -r1998127;
        double r1998129 = exp(r1998128);
        double r1998130 = r1998124 * r1998129;
        double r1998131 = r1998123 - r1998121;
        double r1998132 = r1998121 + r1998122;
        double r1998133 = r1998132 * r1998126;
        double r1998134 = -r1998133;
        double r1998135 = exp(r1998134);
        double r1998136 = r1998131 * r1998135;
        double r1998137 = r1998130 - r1998136;
        double r1998138 = 2.0;
        double r1998139 = r1998137 / r1998138;
        return r1998139;
}
↓
double f(double x, double eps) {
        double r1998140 = x;
        double r1998141 = 2.224816812712469;
        bool r1998142 = r1998140 <= r1998141;
        double r1998143 = 2.0;
        double r1998144 = r1998140 * r1998140;
        double r1998145 = r1998143 - r1998144;
        double r1998146 = 0.6666666666666666;
        double r1998147 = r1998146 * r1998140;
        double r1998148 = r1998147 * r1998144;
        double r1998149 = r1998145 + r1998148;
        double r1998150 = r1998149 / r1998143;
        double r1998151 = 1.0;
        double r1998152 = eps;
        double r1998153 = r1998151 / r1998152;
        double r1998154 = r1998153 + r1998151;
        double r1998155 = r1998151 - r1998152;
        double r1998156 = -r1998140;
        double r1998157 = r1998155 * r1998156;
        double r1998158 = exp(r1998157);
        double r1998159 = r1998154 * r1998158;
        double r1998160 = r1998153 - r1998151;
        double r1998161 = r1998152 + r1998151;
        double r1998162 = r1998161 * r1998140;
        double r1998163 = exp(r1998162);
        double r1998164 = r1998160 / r1998163;
        double r1998165 = r1998159 - r1998164;
        double r1998166 = r1998165 / r1998143;
        double r1998167 = r1998142 ? r1998150 : r1998166;
        return r1998167;
}

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 < 2.224816812712469

    1. Initial program 39.3

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

    2. Taylor expanded around 0 1.3

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

    3. Simplified1.3

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

    if 2.224816812712469 < 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 exp-neg0.4

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

    4. Applied un-div-inv0.4

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

  4. Final simplification1.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 2.224816812712469:\\ \;\;\;\;\frac{\left(2 - x \cdot x\right) + \left(\frac{2}{3} \cdot x\right) \cdot \left(x \cdot x\right)}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(\frac{1}{\varepsilon} + 1\right) \cdot e^{\left(1 - \varepsilon\right) \cdot \left(-x\right)} - \frac{\frac{1}{\varepsilon} - 1}{e^{\left(\varepsilon + 1\right) \cdot x}}}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2019158 
(FPCore (x eps)
  :name "NMSE Section 6.1 mentioned, A"
  (/ (- (* (+ 1 (/ 1 eps)) (exp (- (* (- 1 eps) x)))) (* (- (/ 1 eps) 1) (exp (- (* (+ 1 eps) x))))) 2))