Average Error: 29.1 → 13.5
Time: 16.1s
Precision: 64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \le -5.0289586708105234 \cdot 10^{+104}:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\ \mathbf{elif}\;a \le 4.1836874807488166 \cdot 10^{+54}:\\ \;\;\;\;x \cdot a + \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \left(\left(x \cdot \frac{1}{6}\right) \cdot a + \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \le -5.0289586708105234 \cdot 10^{+104}:\\
\;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\

\mathbf{elif}\;a \le 4.1836874807488166 \cdot 10^{+54}:\\
\;\;\;\;x \cdot a + \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \left(\left(x \cdot \frac{1}{6}\right) \cdot a + \frac{1}{2}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\

\end{array}
double f(double a, double x) {
        double r4658688 = a;
        double r4658689 = x;
        double r4658690 = r4658688 * r4658689;
        double r4658691 = exp(r4658690);
        double r4658692 = 1.0;
        double r4658693 = r4658691 - r4658692;
        return r4658693;
}
↓
double f(double a, double x) {
        double r4658694 = a;
        double r4658695 = -5.0289586708105234e+104;
        bool r4658696 = r4658694 <= r4658695;
        double r4658697 = -1.0;
        double r4658698 = x;
        double r4658699 = r4658698 + r4658698;
        double r4658700 = r4658698 + r4658699;
        double r4658701 = 3.0;
        double r4658702 = r4658701 * r4658694;
        double r4658703 = r4658700 * r4658702;
        double r4658704 = exp(r4658703);
        double r4658705 = r4658697 + r4658704;
        double r4658706 = 1.0;
        double r4658707 = r4658698 * r4658694;
        double r4658708 = r4658701 * r4658707;
        double r4658709 = exp(r4658708);
        double r4658710 = r4658709 + r4658706;
        double r4658711 = r4658709 * r4658710;
        double r4658712 = r4658706 + r4658711;
        double r4658713 = r4658705 / r4658712;
        double r4658714 = exp(r4658707);
        double r4658715 = r4658714 + r4658706;
        double r4658716 = r4658714 * r4658715;
        double r4658717 = r4658716 + r4658706;
        double r4658718 = r4658713 / r4658717;
        double r4658719 = 4.1836874807488166e+54;
        bool r4658720 = r4658694 <= r4658719;
        double r4658721 = r4658707 * r4658707;
        double r4658722 = 0.16666666666666666;
        double r4658723 = r4658698 * r4658722;
        double r4658724 = r4658723 * r4658694;
        double r4658725 = 0.5;
        double r4658726 = r4658724 + r4658725;
        double r4658727 = r4658721 * r4658726;
        double r4658728 = r4658707 + r4658727;
        double r4658729 = r4658720 ? r4658728 : r4658718;
        double r4658730 = r4658696 ? r4658718 : r4658729;
        return r4658730;
}

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
Herbie13.5
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt \frac{1}{10}:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if a < -5.0289586708105234e+104 or 4.1836874807488166e+54 < a

    1. Initial program 16.7

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

    3. Applied flip3--16.8

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

    4. Simplified16.7

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

    5. Simplified16.7

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

    7. Applied flip3--16.7

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

    8. Simplified16.6

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

    9. Simplified16.6

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

    if -5.0289586708105234e+104 < a < 4.1836874807488166e+54

    1. Initial program 34.1

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

    2. Taylor expanded around 0 19.5

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

    3. Simplified12.3

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

  4. Final simplification13.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \le -5.0289586708105234 \cdot 10^{+104}:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\ \mathbf{elif}\;a \le 4.1836874807488166 \cdot 10^{+54}:\\ \;\;\;\;x \cdot a + \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \left(\left(x \cdot \frac{1}{6}\right) \cdot a + \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(x + \left(x + x\right)\right) \cdot \left(3 \cdot a\right)}}{1 + e^{3 \cdot \left(x \cdot a\right)} \cdot \left(e^{3 \cdot \left(x \cdot a\right)} + 1\right)}}{e^{x \cdot a} \cdot \left(e^{x \cdot a} + 1\right) + 1}\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(FPCore (a x)
  :name "expax (section 3.5)"
  :herbie-expected 14

  :herbie-target
  (if (< (fabs (* a x)) 1/10) (* (* a x) (+ 1 (+ (/ (* a x) 2) (/ (pow (* a x) 2) 6)))) (- (exp (* a x)) 1))

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