Average Error: 0.5 → 0.5
Time: 5.2s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\left(\sqrt{\frac{1}{k}} \cdot {\left(\pi \cdot \frac{-2}{\frac{-1}{n}}\right)}^{\left(k \cdot -0.5\right)}\right) \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)} \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (*
  (* (sqrt (/ 1.0 k)) (pow (* PI (/ -2.0 (/ -1.0 n))) (* k -0.5)))
  (sqrt (* PI (* n 2.0)))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
↓
double code(double k, double n) {
	return (sqrt((1.0 / k)) * pow((((double) M_PI) * (-2.0 / (-1.0 / n))), (k * -0.5))) * sqrt((((double) M_PI) * (n * 2.0)));
}

public static double code(double k, double n) {
	return (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}
↓
public static double code(double k, double n) {
	return (Math.sqrt((1.0 / k)) * Math.pow((Math.PI * (-2.0 / (-1.0 / n))), (k * -0.5))) * Math.sqrt((Math.PI * (n * 2.0)));
}

def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
↓
def code(k, n):
	return (math.sqrt((1.0 / k)) * math.pow((math.pi * (-2.0 / (-1.0 / n))), (k * -0.5))) * math.sqrt((math.pi * (n * 2.0)))

function code(k, n)
	return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end
↓
function code(k, n)
	return Float64(Float64(sqrt(Float64(1.0 / k)) * (Float64(pi * Float64(-2.0 / Float64(-1.0 / n))) ^ Float64(k * -0.5))) * sqrt(Float64(pi * Float64(n * 2.0))))
end

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end
↓
function tmp = code(k, n)
	tmp = (sqrt((1.0 / k)) * ((pi * (-2.0 / (-1.0 / n))) ^ (k * -0.5))) * sqrt((pi * (n * 2.0)));
end

code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
↓
code[k_, n_] := N[(N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Power[N[(Pi * N[(-2.0 / N[(-1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(k * -0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(Pi * N[(n * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\left(\sqrt{\frac{1}{k}} \cdot {\left(\pi \cdot \frac{-2}{\frac{-1}{n}}\right)}^{\left(k \cdot -0.5\right)}\right) \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.5

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

  2. Simplified0.5

    \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}} \]

  3. Applied egg-rr0.5

    \[\leadsto \frac{\color{blue}{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)} \cdot \sqrt{n \cdot \left(2 \cdot \pi\right)}}}{\sqrt{k}} \]

  4. Applied egg-rr0.5

    \[\leadsto \color{blue}{\frac{{\left(\pi \cdot \left(n \cdot 2\right)\right)}^{\left(k \cdot -0.5\right)}}{\sqrt{k}} \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)}} \]

  5. Taylor expanded in n around -inf 64.0

    \[\leadsto \color{blue}{\left(\sqrt{\frac{1}{k}} \cdot e^{-0.5 \cdot \left(k \cdot \left(\log \left(-2 \cdot \pi\right) + -1 \cdot \log \left(\frac{-1}{n}\right)\right)\right)}\right)} \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)} \]

  6. Simplified0.5

    \[\leadsto \color{blue}{\left(\sqrt{\frac{1}{k}} \cdot {\left(\pi \cdot \frac{-2}{\frac{-1}{n}}\right)}^{\left(k \cdot -0.5\right)}\right)} \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)} \]

  7. Final simplification0.5

    \[\leadsto \left(\sqrt{\frac{1}{k}} \cdot {\left(\pi \cdot \frac{-2}{\frac{-1}{n}}\right)}^{\left(k \cdot -0.5\right)}\right) \cdot \sqrt{\pi \cdot \left(n \cdot 2\right)} \]

Reproduce

herbie shell --seed 2022169 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))