Average Error: 41.1 → 0.4
Time: 11.2s
Precision: binary64
\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.82753052962640882 \cdot 10^{-4}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;0.5 \cdot \log \left(e^{\frac{x}{\sqrt{2}}}\right) + \left(\sqrt{2} + \frac{{x}^{2}}{\sqrt{2}} \cdot \left(0.25 - \frac{0.125}{2}\right)\right)\\ \end{array}\]
\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.82753052962640882 \cdot 10^{-4}:\\
\;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\

\mathbf{else}:\\
\;\;\;\;0.5 \cdot \log \left(e^{\frac{x}{\sqrt{2}}}\right) + \left(\sqrt{2} + \frac{{x}^{2}}{\sqrt{2}} \cdot \left(0.25 - \frac{0.125}{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 <= -0.00018275305296264088)) {
		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) (((double) (0.5 * ((double) log(((double) exp(((double) (x / ((double) sqrt(2.0)))))))))) + ((double) (((double) sqrt(2.0)) + ((double) (((double) (((double) pow(x, 2.0)) / ((double) sqrt(2.0)))) * ((double) (0.25 - ((double) (0.125 / 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 < -1.82753052962640882e-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-log0.1

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

    6. Simplified0.0

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

    if -1.82753052962640882e-4 < x

    1. Initial program 61.8

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

    2. Taylor expanded around 0 0.5

      \[\leadsto \color{blue}{\left(0.25 \cdot \frac{{x}^{2}}{\sqrt{2}} + \left(\sqrt{2} + 0.5 \cdot \frac{x}{\sqrt{2}}\right)\right) - 0.125 \cdot \frac{{x}^{2}}{{\left(\sqrt{2}\right)}^{3}}}\]

    3. Simplified0.5

      \[\leadsto \color{blue}{0.5 \cdot \frac{x}{\sqrt{2}} + \left(\sqrt{2} + \frac{{x}^{2}}{\sqrt{2}} \cdot \left(0.25 - \frac{0.125}{2}\right)\right)}\]
    4. Using strategy rm

    5. Applied add-log-exp0.5

      \[\leadsto 0.5 \cdot \color{blue}{\log \left(e^{\frac{x}{\sqrt{2}}}\right)} + \left(\sqrt{2} + \frac{{x}^{2}}{\sqrt{2}} \cdot \left(0.25 - \frac{0.125}{2}\right)\right)\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.82753052962640882 \cdot 10^{-4}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\log \left(e^{e^{x} - 1}\right)}}\\ \mathbf{else}:\\ \;\;\;\;0.5 \cdot \log \left(e^{\frac{x}{\sqrt{2}}}\right) + \left(\sqrt{2} + \frac{{x}^{2}}{\sqrt{2}} \cdot \left(0.25 - \frac{0.125}{2}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(FPCore (x)
  :name "sqrtexp (problem 3.4.4)"
  :precision binary64
  (sqrt (/ (- (exp (* 2.0 x)) 1.0) (- (exp x) 1.0))))