Toniolo and Linder, Equation (3a)

?

Percentage Accurate: 98.3% → 100.0%
Time: 25.0s
Precision: binary64
Cost: 32960

?

\[\sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}}\right)} \]
\[\sqrt{0.5 + 0.5 \cdot \frac{1}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}} \]
(FPCore (l Om kx ky)
 :precision binary64
 (sqrt
  (*
   (/ 1.0 2.0)
   (+
    1.0
    (/
     1.0
     (sqrt
      (+
       1.0
       (*
        (pow (/ (* 2.0 l) Om) 2.0)
        (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0))))))))))
(FPCore (l Om kx ky)
 :precision binary64
 (sqrt
  (+
   0.5
   (*
    0.5
    (/ 1.0 (hypot 1.0 (* (hypot (sin ky) (sin kx)) (* 2.0 (/ l Om)))))))))
double code(double l, double Om, double kx, double ky) {
	return sqrt(((1.0 / 2.0) * (1.0 + (1.0 / sqrt((1.0 + (pow(((2.0 * l) / Om), 2.0) * (pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))))))));
}
double code(double l, double Om, double kx, double ky) {
	return sqrt((0.5 + (0.5 * (1.0 / hypot(1.0, (hypot(sin(ky), sin(kx)) * (2.0 * (l / Om))))))));
}
public static double code(double l, double Om, double kx, double ky) {
	return Math.sqrt(((1.0 / 2.0) * (1.0 + (1.0 / Math.sqrt((1.0 + (Math.pow(((2.0 * l) / Om), 2.0) * (Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))))))));
}
public static double code(double l, double Om, double kx, double ky) {
	return Math.sqrt((0.5 + (0.5 * (1.0 / Math.hypot(1.0, (Math.hypot(Math.sin(ky), Math.sin(kx)) * (2.0 * (l / Om))))))));
}
def code(l, Om, kx, ky):
	return math.sqrt(((1.0 / 2.0) * (1.0 + (1.0 / math.sqrt((1.0 + (math.pow(((2.0 * l) / Om), 2.0) * (math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))))))))
def code(l, Om, kx, ky):
	return math.sqrt((0.5 + (0.5 * (1.0 / math.hypot(1.0, (math.hypot(math.sin(ky), math.sin(kx)) * (2.0 * (l / Om))))))))
function code(l, Om, kx, ky)
	return sqrt(Float64(Float64(1.0 / 2.0) * Float64(1.0 + Float64(1.0 / sqrt(Float64(1.0 + Float64((Float64(Float64(2.0 * l) / Om) ^ 2.0) * Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))))))))
end
function code(l, Om, kx, ky)
	return sqrt(Float64(0.5 + Float64(0.5 * Float64(1.0 / hypot(1.0, Float64(hypot(sin(ky), sin(kx)) * Float64(2.0 * Float64(l / Om))))))))
end
function tmp = code(l, Om, kx, ky)
	tmp = sqrt(((1.0 / 2.0) * (1.0 + (1.0 / sqrt((1.0 + ((((2.0 * l) / Om) ^ 2.0) * ((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))))))));
end
function tmp = code(l, Om, kx, ky)
	tmp = sqrt((0.5 + (0.5 * (1.0 / hypot(1.0, (hypot(sin(ky), sin(kx)) * (2.0 * (l / Om))))))));
end
code[l_, Om_, kx_, ky_] := N[Sqrt[N[(N[(1.0 / 2.0), $MachinePrecision] * N[(1.0 + N[(1.0 / N[Sqrt[N[(1.0 + N[(N[Power[N[(N[(2.0 * l), $MachinePrecision] / Om), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
code[l_, Om_, kx_, ky_] := N[Sqrt[N[(0.5 + N[(0.5 * N[(1.0 / N[Sqrt[1.0 ^ 2 + N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] * N[(2.0 * N[(l / Om), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}}\right)}
\sqrt{0.5 + 0.5 \cdot \frac{1}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}}

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 8 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Bogosity?

Bogosity

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 97.2%

    \[\sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}}\right)} \]
  2. Simplified97.2%

    \[\leadsto \color{blue}{\sqrt{0.5 + \frac{1}{\sqrt{1 + {\left(\frac{2}{\frac{Om}{\ell}}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot 0.5}} \]
    Step-by-step derivation

    [Start]97.2%

    \[ \sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}}\right)} \]

    distribute-rgt-in [=>]97.2%

    \[ \sqrt{\color{blue}{1 \cdot \frac{1}{2} + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot \frac{1}{2}}} \]

    metadata-eval [=>]97.2%

    \[ \sqrt{1 \cdot \color{blue}{0.5} + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot \frac{1}{2}} \]

    metadata-eval [=>]97.2%

    \[ \sqrt{\color{blue}{0.5} + \frac{1}{\sqrt{1 + {\left(\frac{2 \cdot \ell}{Om}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot \frac{1}{2}} \]

    associate-/l* [=>]97.2%

    \[ \sqrt{0.5 + \frac{1}{\sqrt{1 + {\color{blue}{\left(\frac{2}{\frac{Om}{\ell}}\right)}}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot \frac{1}{2}} \]

    metadata-eval [=>]97.2%

    \[ \sqrt{0.5 + \frac{1}{\sqrt{1 + {\left(\frac{2}{\frac{Om}{\ell}}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot \color{blue}{0.5}} \]
  3. Applied egg-rr100.0%

    \[\leadsto \sqrt{0.5 + \frac{1}{\color{blue}{e^{\mathsf{log1p}\left(\mathsf{hypot}\left(1, \left(2 \cdot \frac{\ell}{Om}\right) \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right)\right)\right)} - 1}} \cdot 0.5} \]
    Step-by-step derivation

    [Start]97.2%

    \[ \sqrt{0.5 + \frac{1}{\sqrt{1 + {\left(\frac{2}{\frac{Om}{\ell}}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}} \cdot 0.5} \]

    expm1-log1p-u [=>]97.2%

    \[ \sqrt{0.5 + \frac{1}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{1 + {\left(\frac{2}{\frac{Om}{\ell}}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}\right)\right)}} \cdot 0.5} \]

    expm1-udef [=>]97.2%

    \[ \sqrt{0.5 + \frac{1}{\color{blue}{e^{\mathsf{log1p}\left(\sqrt{1 + {\left(\frac{2}{\frac{Om}{\ell}}\right)}^{2} \cdot \left({\sin kx}^{2} + {\sin ky}^{2}\right)}\right)} - 1}} \cdot 0.5} \]
  4. Simplified100.0%

    \[\leadsto \sqrt{0.5 + \frac{1}{\color{blue}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}} \cdot 0.5} \]
    Step-by-step derivation

    [Start]100.0%

    \[ \sqrt{0.5 + \frac{1}{e^{\mathsf{log1p}\left(\mathsf{hypot}\left(1, \left(2 \cdot \frac{\ell}{Om}\right) \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right)\right)\right)} - 1} \cdot 0.5} \]

    expm1-def [=>]100.0%

    \[ \sqrt{0.5 + \frac{1}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{hypot}\left(1, \left(2 \cdot \frac{\ell}{Om}\right) \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right)\right)\right)\right)}} \cdot 0.5} \]

    expm1-log1p [=>]100.0%

    \[ \sqrt{0.5 + \frac{1}{\color{blue}{\mathsf{hypot}\left(1, \left(2 \cdot \frac{\ell}{Om}\right) \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right)\right)}} \cdot 0.5} \]

    *-commutative [=>]100.0%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)}\right)} \cdot 0.5} \]

    hypot-def [<=]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    unpow2 [<=]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    unpow2 [<=]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    +-commutative [<=]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    unpow2 [=>]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    unpow2 [=>]97.9%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]

    hypot-def [=>]100.0%

    \[ \sqrt{0.5 + \frac{1}{\mathsf{hypot}\left(1, \color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)} \cdot 0.5} \]
  5. Final simplification100.0%

    \[\leadsto \sqrt{0.5 + 0.5 \cdot \frac{1}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}} \]

Alternatives

Alternative 1
Accuracy100.0%
Cost32960
\[\sqrt{0.5 + 0.5 \cdot \frac{1}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}} \]
Alternative 2
Accuracy98.3%
Cost33092
\[\begin{array}{l} \mathbf{if}\;{\sin kx}^{2} \leq 5 \cdot 10^{-207}:\\ \;\;\;\;\sqrt{0.5 + \frac{0.5}{\mathsf{hypot}\left(1, \frac{\ell}{Om} \cdot \left(\sin ky \cdot 2\right)\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5 + 0.5 \cdot \frac{1}{\mathsf{hypot}\left(1, \sin kx \cdot \left(2 \cdot \frac{\ell}{Om}\right)\right)}}\\ \end{array} \]
Alternative 3
Accuracy93.6%
Cost19968
\[\sqrt{0.5 + \frac{0.5}{\mathsf{hypot}\left(1, \frac{\ell}{Om} \cdot \left(\sin ky \cdot 2\right)\right)}} \]
Alternative 4
Accuracy86.6%
Cost13833
\[\begin{array}{l} \mathbf{if}\;kx \leq -1.16 \cdot 10^{+104} \lor \neg \left(kx \leq -3.1 \cdot 10^{-164}\right):\\ \;\;\;\;\sqrt{0.5 + \frac{0.5}{\mathsf{hypot}\left(1, 2 \cdot \frac{ky \cdot \ell}{Om}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5 + \frac{0.5}{\mathsf{hypot}\left(1, 2 \cdot \frac{kx \cdot \ell}{Om}\right)}}\\ \end{array} \]
Alternative 5
Accuracy85.3%
Cost13700
\[\begin{array}{l} \mathbf{if}\;Om \leq -3.1 \cdot 10^{+89}:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5 + \frac{0.5}{\mathsf{hypot}\left(1, 2 \cdot \frac{kx \cdot \ell}{Om}\right)}}\\ \end{array} \]
Alternative 6
Accuracy80.4%
Cost8140
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5 \cdot 10^{-19}:\\ \;\;\;\;\sqrt{0.5}\\ \mathbf{elif}\;\ell \leq 2.6 \cdot 10^{-39}:\\ \;\;\;\;1\\ \mathbf{elif}\;\ell \leq 3.5 \cdot 10^{+140}:\\ \;\;\;\;\sqrt{0.5 + 0.5 \cdot \frac{1}{1 + 2 \cdot \frac{\ell \cdot \ell}{\frac{Om \cdot Om}{ky \cdot ky}}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5}\\ \end{array} \]
Alternative 7
Accuracy78.4%
Cost6728
\[\begin{array}{l} \mathbf{if}\;\ell \leq -6 \cdot 10^{-19}:\\ \;\;\;\;\sqrt{0.5}\\ \mathbf{elif}\;\ell \leq 4 \cdot 10^{+42}:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5}\\ \end{array} \]
Alternative 8
Accuracy62.2%
Cost64
\[1 \]

Reproduce?

herbie shell --seed 2023263 
(FPCore (l Om kx ky)
  :name "Toniolo and Linder, Equation (3a)"
  :precision binary64
  (sqrt (* (/ 1.0 2.0) (+ 1.0 (/ 1.0 (sqrt (+ 1.0 (* (pow (/ (* 2.0 l) Om) 2.0) (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0))))))))))