Average Error: 0.0 → 0.0
Time: 4.0s
Precision: binary64
Cost: 14528
\[\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}\]
\[\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \log \left(e^{t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}\right)}\]
\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}
\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \log \left(e^{t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}\right)}
(FPCore (t)
 :precision binary64
 (/
  (+ 1.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t))))
  (+ 2.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t))))))
(FPCore (t)
 :precision binary64
 (/
  (+ 1.0 (* t (/ 4.0 (+ 2.0 (+ t (/ 1.0 t))))))
  (+ 2.0 (log (exp (* t (/ 4.0 (+ 2.0 (+ t (/ 1.0 t))))))))))
double code(double t) {
	return (1.0 + (((2.0 * t) / (1.0 + t)) * ((2.0 * t) / (1.0 + t)))) / (2.0 + (((2.0 * t) / (1.0 + t)) * ((2.0 * t) / (1.0 + t))));
}
double code(double t) {
	return (1.0 + (t * (4.0 / (2.0 + (t + (1.0 / t)))))) / (2.0 + log(exp(t * (4.0 / (2.0 + (t + (1.0 / t)))))));
}

Error

Bits error versus t

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Alternatives

Alternative 1
Error30.6
Cost28288
\[\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \left(2 \cdot \log \left(\sqrt[3]{{\left(e^{t}\right)}^{\left(\frac{4}{2 + \left(t + \frac{1}{t}\right)}\right)}}\right) + \left(t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}\right) \cdot 0.3333333333333333\right)}\]
Alternative 2
Error30.6
Cost53696
\[\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \left(\log \left(\sqrt[3]{{\left(e^{t}\right)}^{\left(\frac{4}{2 + \left(t + \frac{1}{t}\right)}\right)}}\right) + 2 \cdot \log \left(\sqrt[3]{{\left(e^{t}\right)}^{\left(\frac{4}{2 + \left(t + \frac{1}{t}\right)}\right)}}\right)\right)}\]

Error

Derivation

  1. Initial program 0.0

    \[\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}\]
  2. Simplified0.0

    \[\leadsto \color{blue}{\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}}\]
  3. Using strategy rm
  4. Applied add-log-exp_binary64_7990.0

    \[\leadsto \frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \color{blue}{\log \left(e^{t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}\right)}}\]
  5. Simplified0.0

    \[\leadsto \color{blue}{\frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \log \left(e^{t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}\right)}}\]
  6. Final simplification0.0

    \[\leadsto \frac{1 + t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}{2 + \log \left(e^{t \cdot \frac{4}{2 + \left(t + \frac{1}{t}\right)}}\right)}\]

Reproduce

herbie shell --seed 2021043 
(FPCore (t)
  :name "Kahan p13 Example 1"
  :precision binary64
  (/ (+ 1.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t)))) (+ 2.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t))))))