Average Error: 58.8 → 0.0
Time: 24.8s
Precision: 64
\[-1.700000000000000122124532708767219446599 \cdot 10^{-4} \lt x\]
\[e^{x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 1.000116207589260675092646124539896845818:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(-1\right) + e^{x}\\ \end{array}\]
e^{x} - 1
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 1.000116207589260675092646124539896845818:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right)\\

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

\end{array}
double f(double x) {
        double r3775162 = x;
        double r3775163 = exp(r3775162);
        double r3775164 = 1.0;
        double r3775165 = r3775163 - r3775164;
        return r3775165;
}
↓
double f(double x) {
        double r3775166 = x;
        double r3775167 = exp(r3775166);
        double r3775168 = 1.0001162075892607;
        bool r3775169 = r3775167 <= r3775168;
        double r3775170 = r3775166 * r3775166;
        double r3775171 = 0.5;
        double r3775172 = 0.16666666666666666;
        double r3775173 = r3775166 * r3775172;
        double r3775174 = r3775171 + r3775173;
        double r3775175 = r3775170 * r3775174;
        double r3775176 = r3775166 + r3775175;
        double r3775177 = 1.0;
        double r3775178 = -r3775177;
        double r3775179 = r3775178 + r3775167;
        double r3775180 = r3775169 ? r3775176 : r3775179;
        return r3775180;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original58.8
Target0.4
Herbie0.0
\[x \cdot \left(\left(1 + \frac{x}{2}\right) + \frac{x \cdot x}{6}\right)\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 1.0001162075892607

    1. Initial program 59.3

      \[e^{x} - 1\]

    2. Taylor expanded around 0 0.0

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

    3. Simplified0.0

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

    if 1.0001162075892607 < (exp x)

    1. Initial program 2.7

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

    3. Applied sub-neg2.7

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 1.000116207589260675092646124539896845818:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(-1\right) + e^{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2019168 
(FPCore (x)
  :name "expm1 (example 3.7)"
  :pre (< -0.00017 x)

  :herbie-target
  (* x (+ (+ 1.0 (/ x 2.0)) (/ (* x x) 6.0)))

  (- (exp x) 1.0))