Falkner and Boettcher, Appendix A

Average Error: 2.2 → 0.2

Time: 5.1s

Precision: 64

Math

\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

↓

\[\begin{array}{l} \mathbf{if}\;k \le 1407855369986978.5:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)\\ \end{array}\]

C

double code(double a, double k, double m) {
	return ((a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k)));
}

↓

double code(double a, double k, double m) {
	double VAR;
	if ((k <= 1407855369986978.5)) {
		VAR = ((a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k)));
	} else {
		VAR = (((99.0 * ((pow(pow((1.0 / k), -0.6666666666666666), m) * a) * pow(pow((1.0 / k), -0.3333333333333333), m))) / pow(k, 4.0)) + ((((pow(pow((1.0 / k), -0.6666666666666666), m) * a) / k) * (pow(pow((1.0 / k), -0.3333333333333333), m) / k)) - ((10.0 * ((pow(pow((1.0 / k), -0.6666666666666666), m) * a) * pow(pow((1.0 / k), -0.3333333333333333), m))) / pow(k, 3.0))));
	}
	return VAR;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

1. Split input into 2 regimes

2. if k < 1407855369986978.5

   1. Initial program 0.0

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

   if 1407855369986978.5 < k

   1. Initial program 5.9

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

   2. Simplified 5.9

      \[\leadsto \color{blue}{\frac{{k}^{m}}{k \cdot \left(10 + k\right) + 1} \cdot a}\]
   3. Using strategy rm

   4. Applied *-un-lft-identity 5.9

      \[\leadsto \frac{{k}^{m}}{\color{blue}{1 \cdot \left(k \cdot \left(10 + k\right) + 1\right)}} \cdot a\]

   5. Applied add-cube-cbrt 5.9

      \[\leadsto \frac{{\color{blue}{\left(\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right) \cdot \sqrt[3]{k}\right)}}^{m}}{1 \cdot \left(k \cdot \left(10 + k\right) + 1\right)} \cdot a\]

   6. Applied unpow-prod-down 5.9

      \[\leadsto \frac{\color{blue}{{\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m} \cdot {\left(\sqrt[3]{k}\right)}^{m}}}{1 \cdot \left(k \cdot \left(10 + k\right) + 1\right)} \cdot a\]

   7. Applied times-frac 5.9

      \[\leadsto \color{blue}{\left(\frac{{\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}}{1} \cdot \frac{{\left(\sqrt[3]{k}\right)}^{m}}{k \cdot \left(10 + k\right) + 1}\right)} \cdot a\]

   8. Simplified 5.9

      \[\leadsto \left(\color{blue}{{\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}} \cdot \frac{{\left(\sqrt[3]{k}\right)}^{m}}{k \cdot \left(10 + k\right) + 1}\right) \cdot a\]

   9. Taylor expanded around inf 5.9

      \[\leadsto \color{blue}{\left(\frac{e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right) \cdot m} \cdot \left(a \cdot e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right) \cdot m}\right)}{{k}^{2}} + 99 \cdot \frac{e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right) \cdot m} \cdot \left(a \cdot e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right) \cdot m}\right)}{{k}^{4}}\right) - 10 \cdot \frac{e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right) \cdot m} \cdot \left(a \cdot e^{\log \left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right) \cdot m}\right)}{{k}^{3}}}\]

   10. Simplified 0.4

       \[\leadsto \color{blue}{\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1407855369986978.5:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020106 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))