Average Error: 29.0 → 9.1
Time: 3.9s
Precision: 64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \le -1.1578740322654743 \cdot 10^{-29}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{2}, {a}^{2} \cdot {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, {a}^{3} \cdot {x}^{3}, a \cdot x\right)\right)\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \le -1.1578740322654743 \cdot 10^{-29}:\\
\;\;\;\;e^{a \cdot x} - 1\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{2}, {a}^{2} \cdot {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, {a}^{3} \cdot {x}^{3}, a \cdot x\right)\right)\\

\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)) <= -1.1578740322654743e-29)) {
		VAR = ((double) (((double) exp(((double) (a * x)))) - 1.0));
	} else {
		VAR = ((double) fma(0.5, ((double) (((double) pow(a, 2.0)) * ((double) pow(x, 2.0)))), ((double) fma(0.16666666666666666, ((double) (((double) pow(a, 3.0)) * ((double) pow(x, 3.0)))), ((double) (a * x))))));
	}
	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

Original29.0
Target0.2
Herbie9.1
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt 0.10000000000000001:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a x) < -1.1578740322654743e-29

    1. Initial program 3.0

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

    if -1.1578740322654743e-29 < (* a x)

    1. Initial program 44.4

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

    2. Taylor expanded around 0 12.8

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

    3. Simplified12.8

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{2}, {a}^{2} \cdot {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, {a}^{3} \cdot {x}^{3}, a \cdot x\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification9.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \le -1.1578740322654743 \cdot 10^{-29}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{2}, {a}^{2} \cdot {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, {a}^{3} \cdot {x}^{3}, a \cdot x\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020123 +o rules:numerics
(FPCore (a x)
  :name "expax (section 3.5)"
  :precision binary64
  :herbie-expected 14

  :herbie-target
  (if (< (fabs (* a x)) 0.1) (* (* a x) (+ 1 (+ (/ (* a x) 2) (/ (pow (* a x) 2) 6)))) (- (exp (* a x)) 1))

  (- (exp (* a x)) 1))