Average Error: 29.6 → 0.4
Time: 7.4s
Precision: binary64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \le -1.3848736434257956 \cdot 10^{-5}:\\ \;\;\;\;\frac{{\left(e^{a \cdot x}\right)}^{3} - {1}^{3}}{e^{2 \cdot \left(a \cdot x\right)} + 1 \cdot \left(e^{a \cdot x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot x + \frac{1}{2} \cdot {\left(x \cdot a\right)}^{2}\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \le -1.3848736434257956 \cdot 10^{-5}:\\
\;\;\;\;\frac{{\left(e^{a \cdot x}\right)}^{3} - {1}^{3}}{e^{2 \cdot \left(a \cdot x\right)} + 1 \cdot \left(e^{a \cdot x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;a \cdot x + \frac{1}{2} \cdot {\left(x \cdot a\right)}^{2}\\

\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)) <= -1.3848736434257956e-05)) {
		VAR = (((double) (((double) pow(((double) exp(((double) (a * x)))), 3.0)) - ((double) pow(1.0, 3.0)))) / ((double) (((double) exp(((double) (2.0 * ((double) (a * x)))))) + ((double) (1.0 * ((double) (((double) exp(((double) (a * x)))) + 1.0)))))));
	} else {
		VAR = ((double) (((double) (a * x)) + ((double) (0.5 * ((double) pow(((double) (x * a)), 2.0))))));
	}
	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

Original29.6
Target0.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt 0.10000000000000001:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a x) < -1.3848736434257956e-5

    1. Initial program 0.1

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

    3. Applied flip3--0.1

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

    4. Simplified0.1

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

    if -1.3848736434257956e-5 < (* a x)

    1. Initial program 45.1

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

    2. Taylor expanded around 0 15.4

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

    3. Simplified8.1

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

    4. Taylor expanded around 0 8.9

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

    6. Applied pow-prod-down0.6

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

    7. Simplified0.6

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \le -1.3848736434257956 \cdot 10^{-5}:\\ \;\;\;\;\frac{{\left(e^{a \cdot x}\right)}^{3} - {1}^{3}}{e^{2 \cdot \left(a \cdot x\right)} + 1 \cdot \left(e^{a \cdot x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot x + \frac{1}{2} \cdot {\left(x \cdot a\right)}^{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020182 
(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))