Average Error: 30.0 → 3.7
Time: 6.0s
Precision: binary64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq -4499064321.424441:\\ \;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(x + a \cdot \left(\left(x \cdot x\right) \cdot \left(0.5 + \left(a \cdot x\right) \cdot 0.16666666666666666\right)\right)\right)\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \leq -4499064321.424441:\\
\;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\

\mathbf{else}:\\
\;\;\;\;a \cdot \left(x + a \cdot \left(\left(x \cdot x\right) \cdot \left(0.5 + \left(a \cdot x\right) \cdot 0.16666666666666666\right)\right)\right)\\

\end{array}
(FPCore (a x) :precision binary64 (- (exp (* a x)) 1.0))
↓
(FPCore (a x)
 :precision binary64
 (if (<= (* a x) -4499064321.424441)
   (log (exp (- (exp (* a x)) 1.0)))
   (* a (+ x (* a (* (* x x) (+ 0.5 (* (* a x) 0.16666666666666666))))))))
double code(double a, double x) {
	return exp(a * x) - 1.0;
}
↓
double code(double a, double x) {
	double tmp;
	if ((a * x) <= -4499064321.424441) {
		tmp = log(exp(exp(a * x) - 1.0));
	} else {
		tmp = a * (x + (a * ((x * x) * (0.5 + ((a * x) * 0.16666666666666666)))));
	}
	return tmp;
}

Error

Bits error versus a

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original30.0
Target0.2
Herbie3.7
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| < 0.1:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (*.f64 a x) < -4499064321.42444134

    1. Initial program 0

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

    3. Applied add-log-exp_binary64_14610

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

    4. Applied add-log-exp_binary64_14610

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

    5. Applied diff-log_binary64_15140

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

    6. Simplified0

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

    if -4499064321.42444134 < (*.f64 a x)

    1. Initial program 44.0

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

    2. Taylor expanded around 0 15.1

      \[\leadsto \color{blue}{0.16666666666666666 \cdot \left({a}^{3} \cdot {x}^{3}\right) + \left(a \cdot x + 0.5 \cdot \left({a}^{2} \cdot {x}^{2}\right)\right)}\]

    3. Simplified5.5

      \[\leadsto \color{blue}{a \cdot \left(x + a \cdot \left(\left(x \cdot x\right) \cdot \left(0.5 + 0.16666666666666666 \cdot \left(a \cdot x\right)\right)\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification3.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \leq -4499064321.424441:\\ \;\;\;\;\log \left(e^{e^{a \cdot x} - 1}\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(x + a \cdot \left(\left(x \cdot x\right) \cdot \left(0.5 + \left(a \cdot x\right) \cdot 0.16666666666666666\right)\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020280 
(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))