Average Error: 29.5 → 0.6
Time: 5.0s
Precision: 64
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \le -192566.99292489124 \lor \neg \left(-2 \cdot x \le 2.4638730624779373 \cdot 10^{-15}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \end{array}\]
\frac{2}{1 + e^{-2 \cdot x}} - 1
↓
\begin{array}{l}
\mathbf{if}\;-2 \cdot x \le -192566.99292489124 \lor \neg \left(-2 \cdot x \le 2.4638730624779373 \cdot 10^{-15}\right):\\
\;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\

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

\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)) <= -192566.99292489124) || !(((double) (-2.0 * x)) <= 2.4638730624779373e-15))) {
		VAR = ((double) log(((double) exp(((double) (((double) (2.0 / ((double) (1.0 + ((double) exp(((double) (-2.0 * x)))))))) - 1.0))))));
	} else {
		VAR = ((double) (((double) (1.0 * x)) - ((double) (((double) (5.551115123125783e-17 * ((double) pow(x, 4.0)))) + ((double) (0.33333333333333337 * ((double) pow(x, 3.0))))))));
	}
	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 2 regimes

  2. if (* -2.0 x) < -192566.99292489124 or 2.4638730624779373e-15 < (* -2.0 x)

    1. Initial program 0.5

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

    3. Applied add-log-exp0.5

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

    4. Applied add-log-exp0.5

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

    5. Applied diff-log0.5

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

    6. Simplified0.5

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

    if -192566.99292489124 < (* -2.0 x) < 2.4638730624779373e-15

    1. Initial program 59.0

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

    2. Taylor expanded around 0 0.6

      \[\leadsto \color{blue}{1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;-2 \cdot x \le -192566.99292489124 \lor \neg \left(-2 \cdot x \le 2.4638730624779373 \cdot 10^{-15}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020114 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2 (+ 1 (exp (* -2 x)))) 1))