Average Error: 29.6 → 0.0
Time: 2.9s
Precision: 64
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \le -0.01231770057346326 \lor \neg \left(-2 \cdot x \le 1.54899453418180804 \cdot 10^{-4}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \end{array}\]
\frac{2}{1 + e^{-2 \cdot x}} - 1
\begin{array}{l}
\mathbf{if}\;-2 \cdot x \le -0.01231770057346326 \lor \neg \left(-2 \cdot x \le 1.54899453418180804 \cdot 10^{-4}\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 ((2.0 / (1.0 + exp((-2.0 * x)))) - 1.0);
}
double code(double x, double y) {
	double VAR;
	if ((((-2.0 * x) <= -0.01231770057346326) || !((-2.0 * x) <= 0.0001548994534181808))) {
		VAR = log(exp(((2.0 / (1.0 + exp((-2.0 * x)))) - 1.0)));
	} else {
		VAR = ((1.0 * x) - ((5.551115123125783e-17 * pow(x, 4.0)) + (0.33333333333333337 * 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) < -0.01231770057346326 or 0.0001548994534181808 < (* -2.0 x)

    1. Initial program 0.1

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
    2. Using strategy rm
    3. Applied add-log-exp0.1

      \[\leadsto \frac{2}{1 + e^{-2 \cdot x}} - \color{blue}{\log \left(e^{1}\right)}\]
    4. Applied add-log-exp0.1

      \[\leadsto \color{blue}{\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}}}\right)} - \log \left(e^{1}\right)\]
    5. Applied diff-log0.1

      \[\leadsto \color{blue}{\log \left(\frac{e^{\frac{2}{1 + e^{-2 \cdot x}}}}{e^{1}}\right)}\]
    6. Simplified0.1

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

    if -0.01231770057346326 < (* -2.0 x) < 0.0001548994534181808

    1. Initial program 59.1

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
    2. Taylor expanded around 0 0.0

      \[\leadsto \color{blue}{1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)}\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;-2 \cdot x \le -0.01231770057346326 \lor \neg \left(-2 \cdot x \le 1.54899453418180804 \cdot 10^{-4}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + e^{-2 \cdot x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot x - \left(5.55112 \cdot 10^{-17} \cdot {x}^{4} + 0.33333333333333337 \cdot {x}^{3}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020105 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2 (+ 1 (exp (* -2 x)))) 1))