Average Error: 29.1 → 8.9
Time: 4.0s
Precision: binary64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \le -5.52449317808649714 \cdot 10^{-11}:\\ \;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(a + \left(\frac{1}{2} \cdot {a}^{2}\right) \cdot x\right) + \frac{1}{6} \cdot \left({a}^{3} \cdot {x}^{3}\right)\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \le -5.52449317808649714 \cdot 10^{-11}:\\
\;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\

\mathbf{else}:\\
\;\;\;\;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)\\

\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)) <= -5.524493178086497e-11)) {
		VAR = ((double) log(((double) exp(((double) (((double) exp(((double) (a * x)))) - 1.0))))));
	} else {
		VAR = ((double) (((double) (x * ((double) (a + ((double) (((double) (0.5 * ((double) pow(a, 2.0)))) * x)))))) + ((double) (0.16666666666666666 * ((double) (((double) pow(a, 3.0)) * ((double) pow(x, 3.0))))))));
	}
	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

Original29.1
Target0.2
Herbie8.9
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt 0.10000000000000001:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a x) < -5.52449317808649714e-11

    1. Initial program 0.4

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

    3. Applied add-log-exp0.4

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

    4. Applied add-log-exp0.5

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

    5. Applied diff-log0.5

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

    6. Simplified0.5

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

    if -5.52449317808649714e-11 < (* a x)

    1. Initial program 44.5

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

    2. Taylor expanded around 0 13.4

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

    3. Simplified13.4

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

  4. Final simplification8.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \le -5.52449317808649714 \cdot 10^{-11}:\\ \;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(a + \left(\frac{1}{2} \cdot {a}^{2}\right) \cdot x\right) + \frac{1}{6} \cdot \left({a}^{3} \cdot {x}^{3}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020147 
(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))