Average Error: 60.2 → 0.3
Time: 9.4s
Precision: 64
\[-1 \lt \varepsilon \land \varepsilon \lt 1\]
\[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
\[\begin{array}{l} \mathbf{if}\;\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} = -\infty \lor \neg \left(\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} \le 0.0087632414871769045\right):\\ \;\;\;\;\frac{1}{b} + \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{\varepsilon}{\frac{e^{\mathsf{fma}\left(a, \varepsilon, \varepsilon \cdot b\right)} + \left(1 - \mathsf{fma}\left(1, e^{\varepsilon \cdot b}, 1 \cdot e^{a \cdot \varepsilon}\right)\right)}{e^{\left(a + b\right) \cdot \varepsilon} - 1}}\\ \end{array}\]
\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}
\begin{array}{l}
\mathbf{if}\;\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} = -\infty \lor \neg \left(\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} \le 0.0087632414871769045\right):\\
\;\;\;\;\frac{1}{b} + \frac{1}{a}\\

\mathbf{else}:\\
\;\;\;\;\frac{\varepsilon}{\frac{e^{\mathsf{fma}\left(a, \varepsilon, \varepsilon \cdot b\right)} + \left(1 - \mathsf{fma}\left(1, e^{\varepsilon \cdot b}, 1 \cdot e^{a \cdot \varepsilon}\right)\right)}{e^{\left(a + b\right) \cdot \varepsilon} - 1}}\\

\end{array}
double code(double a, double b, double eps) {
	return ((eps * (exp(((a + b) * eps)) - 1.0)) / ((exp((a * eps)) - 1.0) * (exp((b * eps)) - 1.0)));
}
double code(double a, double b, double eps) {
	double temp;
	if (((((eps * (exp(((a + b) * eps)) - 1.0)) / ((exp((a * eps)) - 1.0) * (exp((b * eps)) - 1.0))) <= -inf.0) || !(((eps * (exp(((a + b) * eps)) - 1.0)) / ((exp((a * eps)) - 1.0) * (exp((b * eps)) - 1.0))) <= 0.008763241487176904))) {
		temp = ((1.0 / b) + (1.0 / a));
	} else {
		temp = (eps / ((exp(fma(a, eps, (eps * b))) + (1.0 - fma(1.0, exp((eps * b)), (1.0 * exp((a * eps)))))) / (exp(((a + b) * eps)) - 1.0)));
	}
	return temp;
}

Error

Bits error versus a

Bits error versus b

Bits error versus eps

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original60.2
Target15.2
Herbie0.3
\[\frac{a + b}{a \cdot b}\]

Derivation

  1. Split input into 2 regimes
  2. if (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))) < -inf.0 or 0.008763241487176904 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0)))

    1. Initial program 64.0

      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
    2. Taylor expanded around 0 0.0

      \[\leadsto \color{blue}{\frac{1}{b} + \frac{1}{a}}\]

    if -inf.0 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))) < 0.008763241487176904

    1. Initial program 4.1

      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
    2. Taylor expanded around inf 29.9

      \[\leadsto \color{blue}{\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} \cdot e^{\varepsilon \cdot b} + 1\right) - \left(1 \cdot e^{\varepsilon \cdot b} + 1 \cdot e^{a \cdot \varepsilon}\right)}}\]
    3. Simplified4.4

      \[\leadsto \color{blue}{\frac{\varepsilon}{\frac{e^{\mathsf{fma}\left(a, \varepsilon, \varepsilon \cdot b\right)} + \left(1 - \mathsf{fma}\left(1, e^{\varepsilon \cdot b}, 1 \cdot e^{a \cdot \varepsilon}\right)\right)}{e^{\left(a + b\right) \cdot \varepsilon} - 1}}}\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} = -\infty \lor \neg \left(\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)} \le 0.0087632414871769045\right):\\ \;\;\;\;\frac{1}{b} + \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{\varepsilon}{\frac{e^{\mathsf{fma}\left(a, \varepsilon, \varepsilon \cdot b\right)} + \left(1 - \mathsf{fma}\left(1, e^{\varepsilon \cdot b}, 1 \cdot e^{a \cdot \varepsilon}\right)\right)}{e^{\left(a + b\right) \cdot \varepsilon} - 1}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 +o rules:numerics
(FPCore (a b eps)
  :name "expq3 (problem 3.4.2)"
  :precision binary64
  :pre (and (< -1 eps) (< eps 1))

  :herbie-target
  (/ (+ a b) (* a b))

  (/ (* eps (- (exp (* (+ a b) eps)) 1)) (* (- (exp (* a eps)) 1) (- (exp (* b eps)) 1))))