Average Error: 29.1 → 0.0
Time: 2.9s
Precision: binary64
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -9.6621564271044073 \cdot 10^{-4}:\\ \;\;\;\;\frac{2 \cdot \frac{2}{\left(1 + {\left(e^{-2}\right)}^{x}\right) \cdot \left(1 + {\left(e^{-2}\right)}^{x}\right)} - 1 \cdot 1}{1 + \frac{2}{1 + {\left(e^{-2}\right)}^{x}}}\\ \mathbf{elif}\;x \le 9.27427744022323226 \cdot 10^{-4}:\\ \;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\right)\\ \end{array}\]
\frac{2}{1 + e^{-2 \cdot x}} - 1
↓
\begin{array}{l}
\mathbf{if}\;x \le -9.6621564271044073 \cdot 10^{-4}:\\
\;\;\;\;\frac{2 \cdot \frac{2}{\left(1 + {\left(e^{-2}\right)}^{x}\right) \cdot \left(1 + {\left(e^{-2}\right)}^{x}\right)} - 1 \cdot 1}{1 + \frac{2}{1 + {\left(e^{-2}\right)}^{x}}}\\

\mathbf{elif}\;x \le 9.27427744022323226 \cdot 10^{-4}:\\
\;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\right)\\

\mathbf{else}:\\
\;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\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 <= -0.0009662156427104407)) {
		VAR = ((double) (((double) (((double) (2.0 * ((double) (2.0 / ((double) (((double) (1.0 + ((double) pow(((double) exp(-2.0)), x)))) * ((double) (1.0 + ((double) pow(((double) exp(-2.0)), x)))))))))) - ((double) (1.0 * 1.0)))) / ((double) (1.0 + ((double) (2.0 / ((double) (1.0 + ((double) pow(((double) exp(-2.0)), x))))))))));
	} else {
		double VAR_1;
		if ((x <= 0.0009274277440223232)) {
			VAR_1 = ((double) (((double) (x * 1.0)) - ((double) (((double) pow(x, 3.0)) * ((double) (((double) (x * 5.551115123125783e-17)) + 0.33333333333333337))))));
		} else {
			VAR_1 = ((double) log(((double) exp(((double) (((double) (2.0 / ((double) (1.0 + ((double) pow(((double) exp(-2.0)), x)))))) - 1.0))))));
		}
		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 x < -9.6621564271044073e-4

    1. Initial program 0.0

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

    3. Applied flip--0.0

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

    4. Simplified0.0

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

    5. Simplified0.0

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

    if -9.6621564271044073e-4 < x < 9.27427744022323226e-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

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

    if 9.27427744022323226e-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

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -9.6621564271044073 \cdot 10^{-4}:\\ \;\;\;\;\frac{2 \cdot \frac{2}{\left(1 + {\left(e^{-2}\right)}^{x}\right) \cdot \left(1 + {\left(e^{-2}\right)}^{x}\right)} - 1 \cdot 1}{1 + \frac{2}{1 + {\left(e^{-2}\right)}^{x}}}\\ \mathbf{elif}\;x \le 9.27427744022323226 \cdot 10^{-4}:\\ \;\;\;\;x \cdot 1 - {x}^{3} \cdot \left(x \cdot 5.55112 \cdot 10^{-17} + 0.33333333333333337\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{\frac{2}{1 + {\left(e^{-2}\right)}^{x}} - 1}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020179 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2.0 (+ 1.0 (exp (* -2.0 x)))) 1.0))