Average Error: 29.0 → 0.1
Time: 3.1s
Precision: binary64
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \le -0.0075881199963341384:\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{elif}\;-2 \cdot x \le 1.9792632333110027 \cdot 10^{-10}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\frac{2}{1 + e^{-2 \cdot x}} - 1\right)}^{3}}\\ \end{array}\]
\frac{2}{1 + e^{-2 \cdot x}} - 1
↓
\begin{array}{l}
\mathbf{if}\;-2 \cdot x \le -0.0075881199963341384:\\
\;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\

\mathbf{elif}\;-2 \cdot x \le 1.9792632333110027 \cdot 10^{-10}:\\
\;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\

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

\end{array}
double code(double x, double y) {
	return ((double) (((double) (2.0 / ((double) (1.0 + ((double) exp(((double) (-2.0 * x)))))))) - 1.0));
}
↓
double code(double x, double y) {
	double VAR;
	if ((((double) (-2.0 * x)) <= -0.007588119996334138)) {
		VAR = ((double) log(((double) exp(((double) (((double) (2.0 / ((double) (1.0 + ((double) exp(((double) (-2.0 * x)))))))) - 1.0))))));
	} else {
		double VAR_1;
		if ((((double) (-2.0 * x)) <= 1.9792632333110027e-10)) {
			VAR_1 = ((double) (((double) (1.0 * x)) - ((double) (((double) (5.551115123125783e-17 * ((double) pow(x, 4.0)))) + ((double) (0.33333333333333337 * ((double) pow(x, 3.0))))))));
		} else {
			VAR_1 = ((double) cbrt(((double) pow(((double) (((double) (2.0 / ((double) (1.0 + ((double) exp(((double) (-2.0 * x)))))))) - 1.0)), 3.0))));
		}
		VAR = VAR_1;
	}
	return VAR;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 3 regimes

  2. if (* -2.0 x) < -0.0075881199963341384

    1. Initial program 0.0

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
    2. Using strategy rm

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.0

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

    5. Applied diff-log0.0

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

    6. Simplified0.0

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

    if -0.0075881199963341384 < (* -2.0 x) < 1.9792632333110027e-10

    1. Initial program 59.5

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1\]

    2. Taylor expanded around 0 0.0

      \[\leadsto \color{blue}{1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)}\]

    if 1.9792632333110027e-10 < (* -2.0 x)

    1. Initial program 0.5

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
    2. Using strategy rm

    3. Applied add-cbrt-cube0.5

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

    4. Simplified0.5

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;-2 \cdot x \le -0.0075881199963341384:\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{elif}\;-2 \cdot x \le 1.9792632333110027 \cdot 10^{-10}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\frac{2}{1 + e^{-2 \cdot x}} - 1\right)}^{3}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020163 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2.0 (+ 1.0 (exp (* -2.0 x)))) 1.0))