Average Error: 4.6 → 1.0
Time: 5.3s
Precision: 64
\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.26321628786970431 \cdot 10^{-5}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]
\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.26321628786970431 \cdot 10^{-5}:\\
\;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\right)\right)\\

\mathbf{else}:\\
\;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\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 <= -1.2632162878697043e-05)) {
		VAR = ((double) expm1(((double) log1p(((double) sqrt(((double) (((double) (((double) (((double) exp(((double) (2.0 * x)))) - 1.0)) / ((double) fma(((double) -(1.0)), 1.0, ((double) exp(((double) (x + x)))))))) * ((double) (((double) exp(x)) + 1.0))))))))));
	} else {
		VAR = ((double) sqrt(((double) fma(0.5, ((double) pow(x, 2.0)), ((double) fma(1.0, 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 < -1.2632162878697043e-05

    1. Initial program 0.1

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

    3. Applied flip--0.1

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

    4. Applied associate-/r/0.1

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

    5. Simplified0.0

      \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}} \cdot \left(e^{x} + 1\right)}\]
    6. Using strategy rm

    7. Applied expm1-log1p-u0.0

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\right)\right)}\]

    if -1.2632162878697043e-05 < x

    1. Initial program 34.6

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

    2. Taylor expanded around 0 7.3

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

    3. Simplified7.3

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

  4. Final simplification1.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.26321628786970431 \cdot 10^{-5}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020123 +o rules:numerics
(FPCore (x)
  :name "sqrtexp (problem 3.4.4)"
  :precision binary64
  (sqrt (/ (- (exp (* 2 x)) 1) (- (exp x) 1))))