Average Error: 29.8 → 0.4
Time: 19.6s
Precision: 64
\[e^{a \cdot x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;a \cdot x \le -0.01289790256637027461572575504078486119397:\\ \;\;\;\;\frac{\frac{e^{9 \cdot \left(a \cdot x\right)} - \left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}{e^{3 \cdot \left(a \cdot x\right)} \cdot e^{3 \cdot \left(a \cdot x\right)} + \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right) + e^{3 \cdot \left(a \cdot x\right)} \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}}{e^{a \cdot x} \cdot e^{a \cdot x} + \left(1 \cdot 1 + e^{a \cdot x} \cdot 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \frac{1}{2} + x \cdot \left(\left(\frac{1}{6} \cdot a\right) \cdot \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) + a\right)\\ \end{array}\]
e^{a \cdot x} - 1
↓
\begin{array}{l}
\mathbf{if}\;a \cdot x \le -0.01289790256637027461572575504078486119397:\\
\;\;\;\;\frac{\frac{e^{9 \cdot \left(a \cdot x\right)} - \left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}{e^{3 \cdot \left(a \cdot x\right)} \cdot e^{3 \cdot \left(a \cdot x\right)} + \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right) + e^{3 \cdot \left(a \cdot x\right)} \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}}{e^{a \cdot x} \cdot e^{a \cdot x} + \left(1 \cdot 1 + e^{a \cdot x} \cdot 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \frac{1}{2} + x \cdot \left(\left(\frac{1}{6} \cdot a\right) \cdot \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) + a\right)\\

\end{array}
double f(double a, double x) {
        double r7746440 = a;
        double r7746441 = x;
        double r7746442 = r7746440 * r7746441;
        double r7746443 = exp(r7746442);
        double r7746444 = 1.0;
        double r7746445 = r7746443 - r7746444;
        return r7746445;
}
↓
double f(double a, double x) {
        double r7746446 = a;
        double r7746447 = x;
        double r7746448 = r7746446 * r7746447;
        double r7746449 = -0.012897902566370275;
        bool r7746450 = r7746448 <= r7746449;
        double r7746451 = 9.0;
        double r7746452 = r7746451 * r7746448;
        double r7746453 = exp(r7746452);
        double r7746454 = 1.0;
        double r7746455 = r7746454 * r7746454;
        double r7746456 = r7746454 * r7746455;
        double r7746457 = r7746456 * r7746456;
        double r7746458 = r7746456 * r7746457;
        double r7746459 = r7746453 - r7746458;
        double r7746460 = 3.0;
        double r7746461 = r7746460 * r7746448;
        double r7746462 = exp(r7746461);
        double r7746463 = r7746462 * r7746462;
        double r7746464 = r7746462 * r7746456;
        double r7746465 = r7746457 + r7746464;
        double r7746466 = r7746463 + r7746465;
        double r7746467 = r7746459 / r7746466;
        double r7746468 = exp(r7746448);
        double r7746469 = r7746468 * r7746468;
        double r7746470 = r7746468 * r7746454;
        double r7746471 = r7746455 + r7746470;
        double r7746472 = r7746469 + r7746471;
        double r7746473 = r7746467 / r7746472;
        double r7746474 = r7746447 * r7746446;
        double r7746475 = r7746474 * r7746474;
        double r7746476 = 0.5;
        double r7746477 = r7746475 * r7746476;
        double r7746478 = 0.16666666666666666;
        double r7746479 = r7746478 * r7746446;
        double r7746480 = r7746479 * r7746475;
        double r7746481 = r7746480 + r7746446;
        double r7746482 = r7746447 * r7746481;
        double r7746483 = r7746477 + r7746482;
        double r7746484 = r7746450 ? r7746473 : r7746483;
        return r7746484;
}

Error

Bits error versus a

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original29.8
Target0.1
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| \lt 0.1000000000000000055511151231257827021182:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a x) < -0.012897902566370275

    1. Initial program 0.0

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

    3. Applied flip3--0.0

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

    4. Simplified0.0

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

    6. Applied flip3--0.0

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

    7. Simplified0.0

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

    if -0.012897902566370275 < (* a x)

    1. Initial program 44.7

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

    2. Taylor expanded around 0 15.5

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

    3. Simplified0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot x \le -0.01289790256637027461572575504078486119397:\\ \;\;\;\;\frac{\frac{e^{9 \cdot \left(a \cdot x\right)} - \left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}{e^{3 \cdot \left(a \cdot x\right)} \cdot e^{3 \cdot \left(a \cdot x\right)} + \left(\left(1 \cdot \left(1 \cdot 1\right)\right) \cdot \left(1 \cdot \left(1 \cdot 1\right)\right) + e^{3 \cdot \left(a \cdot x\right)} \cdot \left(1 \cdot \left(1 \cdot 1\right)\right)\right)}}{e^{a \cdot x} \cdot e^{a \cdot x} + \left(1 \cdot 1 + e^{a \cdot x} \cdot 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) \cdot \frac{1}{2} + x \cdot \left(\left(\frac{1}{6} \cdot a\right) \cdot \left(\left(x \cdot a\right) \cdot \left(x \cdot a\right)\right) + a\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019174 
(FPCore (a x)
  :name "expax (section 3.5)"
  :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))