Average Error: 40.7 → 0.3
Time: 5.6s
Precision: binary64
\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -2.03176788136327477 \cdot 10^{-8}:\\ \;\;\;\;\sqrt{\frac{{\left(e^{x}\right)}^{2} - 1}{\frac{{\left(e^{x}\right)}^{2} - 1 \cdot 1}{e^{x} + 1}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2} + \left(\left(x \cdot \frac{x}{\sqrt{2}}\right) \cdot \frac{3}{16} + \frac{x}{\sqrt{2}} \cdot \frac{1}{2}\right)\\ \end{array}\]
\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}
↓
\begin{array}{l}
\mathbf{if}\;x \le -2.03176788136327477 \cdot 10^{-8}:\\
\;\;\;\;\sqrt{\frac{{\left(e^{x}\right)}^{2} - 1}{\frac{{\left(e^{x}\right)}^{2} - 1 \cdot 1}{e^{x} + 1}}}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{2} + \left(\left(x \cdot \frac{x}{\sqrt{2}}\right) \cdot \frac{3}{16} + \frac{x}{\sqrt{2}} \cdot \frac{1}{2}\right)\\

\end{array}
double code(double x) {
	return ((double) sqrt(((double) (((double) (((double) exp(((double) (2.0 * x)))) - 1.0)) / ((double) (((double) exp(x)) - 1.0))))));
}
↓
double code(double x) {
	double VAR;
	if ((x <= -2.0317678813632748e-08)) {
		VAR = ((double) sqrt(((double) (((double) (((double) pow(((double) exp(x)), 2.0)) - 1.0)) / ((double) (((double) (((double) pow(((double) exp(x)), 2.0)) - ((double) (1.0 * 1.0)))) / ((double) (((double) exp(x)) + 1.0))))))));
	} else {
		VAR = ((double) (((double) sqrt(2.0)) + ((double) (((double) (((double) (x * ((double) (x / ((double) sqrt(2.0)))))) * 0.1875)) + ((double) (((double) (x / ((double) sqrt(2.0)))) * 0.5))))));
	}
	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 < -2.03176788136327477e-8

    1. Initial program 0.2

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

    2. Simplified0.2

      \[\leadsto \color{blue}{\sqrt{\frac{{\left(e^{x}\right)}^{2} - 1}{e^{x} - 1}}}\]
    3. Using strategy rm

    4. Applied flip--0.0

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

    5. Simplified0.0

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

    if -2.03176788136327477e-8 < x

    1. Initial program 61.9

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

    2. Simplified61.4

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

    3. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\left(\frac{1}{2} \cdot \frac{x}{\sqrt{2}} + \left(\frac{1}{4} \cdot \frac{{x}^{2}}{\sqrt{2}} + \sqrt{2}\right)\right) - \frac{1}{8} \cdot \frac{{x}^{2}}{{\left(\sqrt{2}\right)}^{3}}}\]

    4. Simplified0.4

      \[\leadsto \color{blue}{\sqrt{2} + \left(\left(\frac{x}{\sqrt{2}} \cdot x\right) \cdot \frac{3}{16} + \frac{1}{2} \cdot \frac{x}{\sqrt{2}}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -2.03176788136327477 \cdot 10^{-8}:\\ \;\;\;\;\sqrt{\frac{{\left(e^{x}\right)}^{2} - 1}{\frac{{\left(e^{x}\right)}^{2} - 1 \cdot 1}{e^{x} + 1}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2} + \left(\left(x \cdot \frac{x}{\sqrt{2}}\right) \cdot \frac{3}{16} + \frac{x}{\sqrt{2}} \cdot \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020181 
(FPCore (x)
  :name "sqrtexp (problem 3.4.4)"
  :precision binary64
  (sqrt (/ (- (exp (* 2.0 x)) 1.0) (- (exp x) 1.0))))