Average Error: 29.1 → 0.0
Time: 2.6s
Precision: binary64
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -7.83981168693725981 \cdot 10^{-4} \lor \neg \left(x \le 6.4231160015316654 \cdot 10^{-4}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\right)\\ \end{array}\]
\frac{2}{1 + e^{-2 \cdot x}} - 1
↓
\begin{array}{l}
\mathbf{if}\;x \le -7.83981168693725981 \cdot 10^{-4} \lor \neg \left(x \le 6.4231160015316654 \cdot 10^{-4}\right):\\
\;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\right)\\

\mathbf{else}:\\
\;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\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 (((x <= -0.000783981168693726) || !(x <= 0.0006423116001531665))) {
		VAR = ((double) log(((double) exp(((double) (((double) (2.0 / ((double) (1.0 + ((double) pow(((double) exp(-2.0)), x)))))) - 1.0))))));
	} else {
		VAR = ((double) (((double) (x * 1.0)) - ((double) (((double) pow(x, 3.0)) * ((double) (((double) (x * 5.551115123125783e-17)) + 0.33333333333333337))))));
	}
	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 x < -7.83981168693725981e-4 or 6.4231160015316654e-4 < x

    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

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

    if -7.83981168693725981e-4 < x < 6.4231160015316654e-4

    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. Simplified0.0

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -7.83981168693725981 \cdot 10^{-4} \lor \neg \left(x \le 6.4231160015316654 \cdot 10^{-4}\right):\\ \;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020184 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2.0 (+ 1.0 (exp (* -2.0 x)))) 1.0))