Average Error: 0.5 → 0.4
Time: 7.4s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\frac{\sqrt{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(1 - k\right)}}}{\sqrt{k}} \]
(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 (pow (* (* 2.0 PI) n) (- 1.0 k))) (sqrt k)))
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(pow(((2.0 * ((double) M_PI)) * n), (1.0 - k))) / sqrt(k);
}

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(Math.pow(((2.0 * Math.PI) * n), (1.0 - k))) / Math.sqrt(k);
}

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(math.pow(((2.0 * math.pi) * n), (1.0 - k))) / math.sqrt(k)

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(sqrt((Float64(Float64(2.0 * pi) * n) ^ Float64(1.0 - k))) / sqrt(k))
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((((2.0 * pi) * n) ^ (1.0 - k))) / sqrt(k);
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[Sqrt[N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

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

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. Applied egg-rr0.5

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

  3. Applied egg-rr0.4

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

  4. Final simplification0.4

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

Reproduce

herbie shell --seed 2022131 
(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))))