expq3 (problem 3.4.2)

Average Error: 60.7 → 0.6

Time: 9.2s

Precision: binary64

\[-1 \lt \varepsilon \land \varepsilon \lt 1\]

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)} = -inf.0 \lor \neg \left(\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)} \le 2.37754048057683172 \cdot 10^{-65}\right):\\ \;\;\;\;\frac{1}{b} + \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)}\\ \end{array}\]

C

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) {
	double VAR;
	if (((((double) (((double) (eps * ((double) (((double) exp(((double) (eps * ((double) (a + b)))))) - 1.0)))) / ((double) (((double) (((double) exp(((double) (eps * a)))) - 1.0)) * ((double) (((double) exp(((double) (eps * b)))) - 1.0)))))) <= -inf.0) || !(((double) (((double) (eps * ((double) (((double) exp(((double) (eps * ((double) (a + b)))))) - 1.0)))) / ((double) (((double) (((double) exp(((double) (eps * a)))) - 1.0)) * ((double) (((double) exp(((double) (eps * b)))) - 1.0)))))) <= 2.3775404805768317e-65))) {
		VAR = ((double) (((double) (1.0 / b)) + ((double) (1.0 / a))));
	} else {
		VAR = ((double) (((double) (eps * ((double) (((double) exp(((double) (eps * ((double) (a + b)))))) - 1.0)))) / ((double) (((double) (((double) exp(((double) (eps * a)))) - 1.0)) * ((double) (((double) exp(((double) (eps * b)))) - 1.0))))));
	}
	return VAR;
}

Error

Bits error versus a

Bits error versus b

Bits error versus eps

Target

Original	60.7
Target	15.1
Herbie	0.6

\[\frac{a + b}{a \cdot b}\]

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 2.37754048057683172e-65 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0)))

   1. Initial program 63.5

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

   2. Taylor expanded around 0 58.3

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

   3. Simplified 58.3

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

   5. Applied associate-*l* 58.2

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

   6. Simplified 57.9

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

   7. Taylor expanded around 0 0.5

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

   if -inf.0 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))) < 2.37754048057683172e-65

   1. Initial program 3.9

      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.6

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)} = -inf.0 \lor \neg \left(\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)} \le 2.37754048057683172 \cdot 10^{-65}\right):\\ \;\;\;\;\frac{1}{b} + \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{\varepsilon \cdot \left(e^{\varepsilon \cdot \left(a + b\right)} - 1\right)}{\left(e^{\varepsilon \cdot a} - 1\right) \cdot \left(e^{\varepsilon \cdot b} - 1\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020179 
(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))))