Average Error: 29.3 → 1.1
Time: 7.0s
Precision: 64
\[\frac{\left(1 + \frac{1}{\varepsilon}\right) \cdot e^{-\left(1 - \varepsilon\right) \cdot x} - \left(\frac{1}{\varepsilon} - 1\right) \cdot e^{-\left(1 + \varepsilon\right) \cdot x}}{2}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 9.6260562945900503:\\ \;\;\;\;1 + {x}^{2} \cdot \left(x \cdot 0.33333333333333337 - 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\log \left(\frac{\frac{1 + \frac{1}{\varepsilon}}{e^{x \cdot \left(1 - \varepsilon\right)}}}{2} - \frac{\frac{\frac{\frac{1}{\varepsilon} - 1}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}} \cdot \sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{2}\right)}\\ \end{array}\]
\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 9.6260562945900503:\\
\;\;\;\;1 + {x}^{2} \cdot \left(x \cdot 0.33333333333333337 - 0.5\right)\\

\mathbf{else}:\\
\;\;\;\;e^{\log \left(\frac{\frac{1 + \frac{1}{\varepsilon}}{e^{x \cdot \left(1 - \varepsilon\right)}}}{2} - \frac{\frac{\frac{\frac{1}{\varepsilon} - 1}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}} \cdot \sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{2}\right)}\\

\end{array}
double f(double x, double eps) {
        double r41224 = 1.0;
        double r41225 = eps;
        double r41226 = r41224 / r41225;
        double r41227 = r41224 + r41226;
        double r41228 = r41224 - r41225;
        double r41229 = x;
        double r41230 = r41228 * r41229;
        double r41231 = -r41230;
        double r41232 = exp(r41231);
        double r41233 = r41227 * r41232;
        double r41234 = r41226 - r41224;
        double r41235 = r41224 + r41225;
        double r41236 = r41235 * r41229;
        double r41237 = -r41236;
        double r41238 = exp(r41237);
        double r41239 = r41234 * r41238;
        double r41240 = r41233 - r41239;
        double r41241 = 2.0;
        double r41242 = r41240 / r41241;
        return r41242;
}
↓
double f(double x, double eps) {
        double r41243 = x;
        double r41244 = 9.62605629459005;
        bool r41245 = r41243 <= r41244;
        double r41246 = 1.0;
        double r41247 = 2.0;
        double r41248 = pow(r41243, r41247);
        double r41249 = 0.33333333333333337;
        double r41250 = r41243 * r41249;
        double r41251 = 0.5;
        double r41252 = r41250 - r41251;
        double r41253 = r41248 * r41252;
        double r41254 = r41246 + r41253;
        double r41255 = eps;
        double r41256 = r41246 / r41255;
        double r41257 = r41246 + r41256;
        double r41258 = r41246 - r41255;
        double r41259 = r41243 * r41258;
        double r41260 = exp(r41259);
        double r41261 = r41257 / r41260;
        double r41262 = 2.0;
        double r41263 = r41261 / r41262;
        double r41264 = r41256 - r41246;
        double r41265 = r41246 + r41255;
        double r41266 = r41265 * r41243;
        double r41267 = exp(r41266);
        double r41268 = cbrt(r41267);
        double r41269 = r41268 * r41268;
        double r41270 = r41264 / r41269;
        double r41271 = r41270 / r41268;
        double r41272 = r41271 / r41262;
        double r41273 = r41263 - r41272;
        double r41274 = log(r41273);
        double r41275 = exp(r41274);
        double r41276 = r41245 ? r41254 : r41275;
        return r41276;
}

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 < 9.62605629459005

    1. Initial program 38.9

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

    2. Simplified38.9

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

    3. Taylor expanded around 0 1.3

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

    4. Simplified1.3

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

    if 9.62605629459005 < 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. Simplified0.4

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

    3. Taylor expanded around inf 0.4

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

    5. Applied add-cube-cbrt0.4

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

    6. Applied associate-/r*0.4

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

    8. Applied add-exp-log0.4

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

  4. Final simplification1.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 9.6260562945900503:\\ \;\;\;\;1 + {x}^{2} \cdot \left(x \cdot 0.33333333333333337 - 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\log \left(\frac{\frac{1 + \frac{1}{\varepsilon}}{e^{x \cdot \left(1 - \varepsilon\right)}}}{2} - \frac{\frac{\frac{\frac{1}{\varepsilon} - 1}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}} \cdot \sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{\sqrt[3]{e^{\left(1 + \varepsilon\right) \cdot x}}}}{2}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020021 
(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))