Average Error: 29.6 → 1.8
Time: 6.6s
Precision: 64
\[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 1.0640289263225317 \cdot 10^{-25}:\\ \;\;\;\;\left(0.33333333333333337 \cdot {x}^{3} + 1\right) - 0.5 \cdot {x}^{2}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{1 + \frac{1}{\varepsilon}}{e^{\left(1 - \varepsilon\right) \cdot x}}}{2} - \frac{\frac{\frac{1}{\varepsilon}}{e^{\left(1 + \varepsilon\right) \cdot x}}}{2}\right) + \frac{\frac{1}{e^{\left(1 + \varepsilon\right) \cdot x}}}{2}\\ \end{array}\]
\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 1.0640289263225317 \cdot 10^{-25}:\\
\;\;\;\;\left(0.33333333333333337 \cdot {x}^{3} + 1\right) - 0.5 \cdot {x}^{2}\\

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

\end{array}
double f(double x, double eps) {
        double r40189 = 1.0;
        double r40190 = eps;
        double r40191 = r40189 / r40190;
        double r40192 = r40189 + r40191;
        double r40193 = r40189 - r40190;
        double r40194 = x;
        double r40195 = r40193 * r40194;
        double r40196 = -r40195;
        double r40197 = exp(r40196);
        double r40198 = r40192 * r40197;
        double r40199 = r40191 - r40189;
        double r40200 = r40189 + r40190;
        double r40201 = r40200 * r40194;
        double r40202 = -r40201;
        double r40203 = exp(r40202);
        double r40204 = r40199 * r40203;
        double r40205 = r40198 - r40204;
        double r40206 = 2.0;
        double r40207 = r40205 / r40206;
        return r40207;
}
↓
double f(double x, double eps) {
        double r40208 = x;
        double r40209 = 1.0640289263225317e-25;
        bool r40210 = r40208 <= r40209;
        double r40211 = 0.33333333333333337;
        double r40212 = 3.0;
        double r40213 = pow(r40208, r40212);
        double r40214 = r40211 * r40213;
        double r40215 = 1.0;
        double r40216 = r40214 + r40215;
        double r40217 = 0.5;
        double r40218 = 2.0;
        double r40219 = pow(r40208, r40218);
        double r40220 = r40217 * r40219;
        double r40221 = r40216 - r40220;
        double r40222 = eps;
        double r40223 = r40215 / r40222;
        double r40224 = r40215 + r40223;
        double r40225 = r40215 - r40222;
        double r40226 = r40225 * r40208;
        double r40227 = exp(r40226);
        double r40228 = r40224 / r40227;
        double r40229 = 2.0;
        double r40230 = r40228 / r40229;
        double r40231 = r40215 + r40222;
        double r40232 = r40231 * r40208;
        double r40233 = exp(r40232);
        double r40234 = r40223 / r40233;
        double r40235 = r40234 / r40229;
        double r40236 = r40230 - r40235;
        double r40237 = r40215 / r40233;
        double r40238 = r40237 / r40229;
        double r40239 = r40236 + r40238;
        double r40240 = r40210 ? r40221 : r40239;
        return r40240;
}

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 < 1.0640289263225317e-25

    1. Initial program 38.3

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

    2. Simplified38.3

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

    3. Taylor expanded around 0 1.1

      \[\leadsto \color{blue}{\left(0.33333333333333337 \cdot {x}^{3} + 1\right) - 0.5 \cdot {x}^{2}}\]

    if 1.0640289263225317e-25 < x

    1. Initial program 4.6

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

    2. Simplified4.6

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

    4. Applied div-sub4.6

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

    5. Applied div-sub4.6

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

    6. Applied associate--r-3.9

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

  4. Final simplification1.8

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 1.0640289263225317 \cdot 10^{-25}:\\ \;\;\;\;\left(0.33333333333333337 \cdot {x}^{3} + 1\right) - 0.5 \cdot {x}^{2}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{1 + \frac{1}{\varepsilon}}{e^{\left(1 - \varepsilon\right) \cdot x}}}{2} - \frac{\frac{\frac{1}{\varepsilon}}{e^{\left(1 + \varepsilon\right) \cdot x}}}{2}\right) + \frac{\frac{1}{e^{\left(1 + \varepsilon\right) \cdot x}}}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 
(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))