Average Error: 60.2 → 49.6
Time: 13.7s
Precision: binary64
\[-1 \lt \varepsilon \land \varepsilon \lt 1\]
\[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
↓
\[\frac{1}{a + \left(\log 1 + \varepsilon \cdot \left(\frac{1}{2} \cdot {\left(\log 1\right)}^{2}\right)\right)} \cdot \frac{{\left(e^{a + b}\right)}^{\varepsilon} - 1}{{\left(e^{b}\right)}^{\varepsilon} - 1}\]
\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}{a + \left(\log 1 + \varepsilon \cdot \left(\frac{1}{2} \cdot {\left(\log 1\right)}^{2}\right)\right)} \cdot \frac{{\left(e^{a + b}\right)}^{\varepsilon} - 1}{{\left(e^{b}\right)}^{\varepsilon} - 1}
double code(double a, double b, double eps) {
	return ((double) (((double) (eps * ((double) (((double) exp(((double) (((double) (a + b)) * eps)))) - 1.0)))) / ((double) (((double) (((double) exp(((double) (a * eps)))) - 1.0)) * ((double) (((double) exp(((double) (b * eps)))) - 1.0))))));
}
↓
double code(double a, double b, double eps) {
	return ((double) (((double) (1.0 / ((double) (a + ((double) (((double) log(1.0)) + ((double) (eps * ((double) (0.5 * ((double) pow(((double) log(1.0)), 2.0)))))))))))) * ((double) (((double) (((double) pow(((double) exp(((double) (a + b)))), eps)) - 1.0)) / ((double) (((double) pow(((double) exp(b)), eps)) - 1.0))))));
}

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
Target15.0
Herbie49.6
\[\frac{a + b}{a \cdot b}\]

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. Simplified60.4

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

  3. Taylor expanded around 0 54.5

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

  4. Simplified54.5

    \[\leadsto \varepsilon \cdot \frac{{\left(e^{a + b}\right)}^{\varepsilon} - 1}{\color{blue}{\left(\varepsilon \cdot \left(a + \left(\log 1 + \varepsilon \cdot \left(\frac{1}{2} \cdot {\left(\log 1\right)}^{2}\right)\right)\right)\right)} \cdot \left({\left(e^{b}\right)}^{\varepsilon} - 1\right)}\]
  5. Using strategy rm

  6. Applied *-un-lft-identity54.5

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

  7. Applied times-frac54.5

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

  8. Applied associate-*r*54.5

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

  9. Simplified49.6

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

  10. Final simplification49.6

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

Reproduce

herbie shell --seed 2020181 
(FPCore (a b eps)
  :name "expq3 (problem 3.4.2)"
  :precision binary64
  :pre (and (< -1.0 eps) (< eps 1.0))

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

  (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))))