Average Error: 60.2 → 3.5
Time: 12.2s
Precision: 64
\[-1 \lt \varepsilon \land \varepsilon \lt 1\]
\[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
↓
\[\frac{1}{b} + \frac{1}{a}\]
\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)}
↓
\frac{1}{b} + \frac{1}{a}
double f(double a, double b, double eps) {
        double r121988 = eps;
        double r121989 = a;
        double r121990 = b;
        double r121991 = r121989 + r121990;
        double r121992 = r121991 * r121988;
        double r121993 = exp(r121992);
        double r121994 = 1.0;
        double r121995 = r121993 - r121994;
        double r121996 = r121988 * r121995;
        double r121997 = r121989 * r121988;
        double r121998 = exp(r121997);
        double r121999 = r121998 - r121994;
        double r122000 = r121990 * r121988;
        double r122001 = exp(r122000);
        double r122002 = r122001 - r121994;
        double r122003 = r121999 * r122002;
        double r122004 = r121996 / r122003;
        return r122004;
}
↓
double f(double a, double b, double __attribute__((unused)) eps) {
        double r122005 = 1.0;
        double r122006 = b;
        double r122007 = r122005 / r122006;
        double r122008 = a;
        double r122009 = r122005 / r122008;
        double r122010 = r122007 + r122009;
        return r122010;
}

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
Target14.7
Herbie3.5
\[\frac{a + b}{a \cdot b}\]

Derivation

  1. Initial program 60.2

    \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]

  2. Taylor expanded around 0 57.9

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

  3. Simplified57.9

    \[\leadsto \frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{1}{6}, {\varepsilon}^{3} \cdot {b}^{3}, \mathsf{fma}\left(\frac{1}{2}, {\varepsilon}^{2} \cdot {b}^{2}, \varepsilon \cdot b\right)\right)}}\]

  4. Taylor expanded around 0 3.5

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

  5. Final simplification3.5

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

Reproduce

herbie shell --seed 2020035 +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))))