Average Error: 40.6 → 0.3
Time: 8.9s
Precision: binary64
\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.6367972920606469 \cdot 10^{-4}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{x \cdot \left(1 + 0.5 \cdot x\right) + 2}\\ \end{array}\]
\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.6367972920606469 \cdot 10^{-4}:\\
\;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{x \cdot \left(1 + 0.5 \cdot x\right) + 2}\\

\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 <= -0.0001636797292060647)) {
		VAR = ((double) sqrt(((double) (((double) (((double) exp(((double) (2.0 * x)))) - 1.0)) / ((double) log(((double) exp(((double) (((double) exp(x)) - 1.0))))))))));
	} else {
		VAR = ((double) sqrt(((double) (((double) (x * ((double) (1.0 + ((double) (0.5 * x)))))) + 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 < -1.6367972920606469e-4

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.0

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

    5. Applied diff-log0.0

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

    6. Simplified0.0

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

    if -1.6367972920606469e-4 < x

    1. Initial program 61.5

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

    2. Taylor expanded around 0 0.5

      \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]

    3. Simplified0.5

      \[\leadsto \sqrt{\color{blue}{x \cdot \left(1 + 0.5 \cdot x\right) + 2}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.6367972920606469 \cdot 10^{-4}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{x \cdot \left(1 + 0.5 \cdot x\right) + 2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020164 
(FPCore (x)
  :name "sqrtexp (problem 3.4.4)"
  :precision binary64
  (sqrt (/ (- (exp (* 2.0 x)) 1.0) (- (exp x) 1.0))))