Average Error: 29.6 → 0.6
Time: 3.1s
Precision: binary64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \le -6.355315326815234 \cdot 10^{-4} \lor \neg \left(a \cdot x \le 5.429422779301956 \cdot 10^{-10}\right):\\ \;\;\;\;\left(\sqrt[3]{e^{a \cdot x} - 1} \cdot \sqrt[3]{e^{a \cdot x} - 1}\right) \cdot \sqrt[3]{e^{a \cdot x} - 1}\\ \mathbf{else}:\\ \;\;\;\;x \cdot a\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \le -6.355315326815234 \cdot 10^{-4} \lor \neg \left(a \cdot x \le 5.429422779301956 \cdot 10^{-10}\right):\\
\;\;\;\;\left(\sqrt[3]{e^{a \cdot x} - 1} \cdot \sqrt[3]{e^{a \cdot x} - 1}\right) \cdot \sqrt[3]{e^{a \cdot x} - 1}\\

\mathbf{else}:\\
\;\;\;\;x \cdot a\\

\end{array}
double code(double a, double x) {
	return ((double) (((double) exp(((double) (a * x)))) - 1.0));
}
↓
double code(double a, double x) {
	double VAR;
	if (((((double) (a * x)) <= -0.0006355315326815234) || !(((double) (a * x)) <= 5.429422779301956e-10))) {
		VAR = ((double) (((double) (((double) cbrt(((double) (((double) exp(((double) (a * x)))) - 1.0)))) * ((double) cbrt(((double) (((double) exp(((double) (a * x)))) - 1.0)))))) * ((double) cbrt(((double) (((double) exp(((double) (a * x)))) - 1.0))))));
	} else {
		VAR = ((double) (x * a));
	}
	return VAR;
}

Error

Bits error versus a

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original29.6
Target0.2
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt 0.10000000000000001:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a x) < -6.355315326815234e-4 or 5.429422779301956e-10 < (* a x)

    1. Initial program 0.5

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

    3. Applied add-cube-cbrt0.5

      \[\leadsto \color{blue}{\left(\sqrt[3]{e^{a \cdot x} - 1} \cdot \sqrt[3]{e^{a \cdot x} - 1}\right) \cdot \sqrt[3]{e^{a \cdot x} - 1}}\]

    if -6.355315326815234e-4 < (* a x) < 5.429422779301956e-10

    1. Initial program 45.1

      \[e^{a \cdot x} - 1\]

    2. Taylor expanded around 0 13.9

      \[\leadsto \color{blue}{\frac{1}{2} \cdot \left({a}^{2} \cdot {x}^{2}\right) + \left(\frac{1}{6} \cdot \left({a}^{3} \cdot {x}^{3}\right) + a \cdot x\right)}\]

    3. Simplified13.9

      \[\leadsto \color{blue}{x \cdot \left(a + \left(\frac{1}{2} \cdot {a}^{2}\right) \cdot x\right) + \frac{1}{6} \cdot \left({a}^{3} \cdot {x}^{3}\right)}\]

    4. Taylor expanded around 0 8.0

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

    5. Simplified4.3

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

    6. Taylor expanded around 0 0.6

      \[\leadsto \color{blue}{a \cdot x}\]

    7. Simplified0.6

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \le -6.355315326815234 \cdot 10^{-4} \lor \neg \left(a \cdot x \le 5.429422779301956 \cdot 10^{-10}\right):\\ \;\;\;\;\left(\sqrt[3]{e^{a \cdot x} - 1} \cdot \sqrt[3]{e^{a \cdot x} - 1}\right) \cdot \sqrt[3]{e^{a \cdot x} - 1}\\ \mathbf{else}:\\ \;\;\;\;x \cdot a\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(FPCore (a x)
  :name "expax (section 3.5)"
  :precision binary64
  :herbie-expected 14

  :herbie-target
  (if (< (fabs (* a x)) 0.1) (* (* a x) (+ 1.0 (+ (/ (* a x) 2.0) (/ (pow (* a x) 2.0) 6.0)))) (- (exp (* a x)) 1.0))

  (- (exp (* a x)) 1.0))