Average Error: 31.1 → 15.7
Time: 4.5s
Precision: binary64
\[\sqrt{2 \cdot {x}^{2}}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 6.5285658990945 \cdot 10^{-312}:\\ \;\;\;\;\sqrt{2 \cdot {x}^{2}}\\ \mathbf{else}:\\ \;\;\;\;{x}^{1} \cdot \sqrt{2}\\ \end{array}\]
\sqrt{2 \cdot {x}^{2}}
↓
\begin{array}{l}
\mathbf{if}\;x \le 6.5285658990945 \cdot 10^{-312}:\\
\;\;\;\;\sqrt{2 \cdot {x}^{2}}\\

\mathbf{else}:\\
\;\;\;\;{x}^{1} \cdot \sqrt{2}\\

\end{array}
double code(double x) {
	return ((double) sqrt(((double) (2.0 * ((double) pow(x, 2.0))))));
}
↓
double code(double x) {
	double VAR;
	if ((x <= 6.5285658990945e-312)) {
		VAR = ((double) sqrt(((double) (2.0 * ((double) pow(x, 2.0))))));
	} else {
		VAR = ((double) (((double) pow(x, 1.0)) * ((double) sqrt(2.0))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if x < 6.5285658990945e-312

    1. Initial program 31.3

      \[\sqrt{2 \cdot {x}^{2}}\]

    if 6.5285658990945e-312 < x

    1. Initial program 30.8

      \[\sqrt{2 \cdot {x}^{2}}\]

    2. Taylor expanded around 0 5.7

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

    3. Simplified0.4

      \[\leadsto \color{blue}{{x}^{1} \cdot \sqrt{2}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification15.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 6.5285658990945 \cdot 10^{-312}:\\ \;\;\;\;\sqrt{2 \cdot {x}^{2}}\\ \mathbf{else}:\\ \;\;\;\;{x}^{1} \cdot \sqrt{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020171 
(FPCore (x)
  :name "sqrt D"
  :precision binary64
  (sqrt (* 2.0 (pow x 2.0))))