Average Error: 32.6 → 10.8
Time: 12.7s
Precision: binary64
\[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
\[\begin{array}{l} t_1 := \sqrt[3]{\sin k}\\ t_2 := 2 + {\left(\frac{k}{t}\right)}^{2}\\ t_3 := \sqrt[3]{\tan k \cdot t_2}\\ t_4 := \ell \cdot \sqrt{2}\\ t_5 := \frac{t_4}{t \cdot \left(t_1 \cdot t_3\right)}\\ t_6 := \frac{t_5}{t \cdot t_3}\\ \mathbf{if}\;t \leq 2.1 \cdot 10^{-274}:\\ \;\;\;\;t_6 \cdot \frac{\frac{t_4}{t \cdot \left(t_1 \cdot \left(\sqrt[3]{\tan k} \cdot \sqrt[3]{t_2}\right)\right)}}{t_1}\\ \mathbf{elif}\;t \leq 10^{-96}:\\ \;\;\;\;\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\sin k \cdot \left(t \cdot \left(k \cdot k\right) + 2 \cdot {t}^{3}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;t_6 \cdot \frac{t_5}{t_1}\\ \end{array} \]
(FPCore (t l k)
 :precision binary64
 (/
  2.0
  (*
   (* (* (/ (pow t 3.0) (* l l)) (sin k)) (tan k))
   (+ (+ 1.0 (pow (/ k t) 2.0)) 1.0))))
(FPCore (t l k)
 :precision binary64
 (let* ((t_1 (cbrt (sin k)))
        (t_2 (+ 2.0 (pow (/ k t) 2.0)))
        (t_3 (cbrt (* (tan k) t_2)))
        (t_4 (* l (sqrt 2.0)))
        (t_5 (/ t_4 (* t (* t_1 t_3))))
        (t_6 (/ t_5 (* t t_3))))
   (if (<= t 2.1e-274)
     (* t_6 (/ (/ t_4 (* t (* t_1 (* (cbrt (tan k)) (cbrt t_2))))) t_1))
     (if (<= t 1e-96)
       (*
        (* l l)
        (/ 2.0 (* (tan k) (* (sin k) (+ (* t (* k k)) (* 2.0 (pow t 3.0)))))))
       (* t_6 (/ t_5 t_1))))))
double code(double t, double l, double k) {
	return 2.0 / ((((pow(t, 3.0) / (l * l)) * sin(k)) * tan(k)) * ((1.0 + pow((k / t), 2.0)) + 1.0));
}
double code(double t, double l, double k) {
	double t_1 = cbrt(sin(k));
	double t_2 = 2.0 + pow((k / t), 2.0);
	double t_3 = cbrt((tan(k) * t_2));
	double t_4 = l * sqrt(2.0);
	double t_5 = t_4 / (t * (t_1 * t_3));
	double t_6 = t_5 / (t * t_3);
	double tmp;
	if (t <= 2.1e-274) {
		tmp = t_6 * ((t_4 / (t * (t_1 * (cbrt(tan(k)) * cbrt(t_2))))) / t_1);
	} else if (t <= 1e-96) {
		tmp = (l * l) * (2.0 / (tan(k) * (sin(k) * ((t * (k * k)) + (2.0 * pow(t, 3.0))))));
	} else {
		tmp = t_6 * (t_5 / t_1);
	}
	return tmp;
}
public static double code(double t, double l, double k) {
	return 2.0 / ((((Math.pow(t, 3.0) / (l * l)) * Math.sin(k)) * Math.tan(k)) * ((1.0 + Math.pow((k / t), 2.0)) + 1.0));
}
public static double code(double t, double l, double k) {
	double t_1 = Math.cbrt(Math.sin(k));
	double t_2 = 2.0 + Math.pow((k / t), 2.0);
	double t_3 = Math.cbrt((Math.tan(k) * t_2));
	double t_4 = l * Math.sqrt(2.0);
	double t_5 = t_4 / (t * (t_1 * t_3));
	double t_6 = t_5 / (t * t_3);
	double tmp;
	if (t <= 2.1e-274) {
		tmp = t_6 * ((t_4 / (t * (t_1 * (Math.cbrt(Math.tan(k)) * Math.cbrt(t_2))))) / t_1);
	} else if (t <= 1e-96) {
		tmp = (l * l) * (2.0 / (Math.tan(k) * (Math.sin(k) * ((t * (k * k)) + (2.0 * Math.pow(t, 3.0))))));
	} else {
		tmp = t_6 * (t_5 / t_1);
	}
	return tmp;
}
function code(t, l, k)
	return Float64(2.0 / Float64(Float64(Float64(Float64((t ^ 3.0) / Float64(l * l)) * sin(k)) * tan(k)) * Float64(Float64(1.0 + (Float64(k / t) ^ 2.0)) + 1.0)))
end
function code(t, l, k)
	t_1 = cbrt(sin(k))
	t_2 = Float64(2.0 + (Float64(k / t) ^ 2.0))
	t_3 = cbrt(Float64(tan(k) * t_2))
	t_4 = Float64(l * sqrt(2.0))
	t_5 = Float64(t_4 / Float64(t * Float64(t_1 * t_3)))
	t_6 = Float64(t_5 / Float64(t * t_3))
	tmp = 0.0
	if (t <= 2.1e-274)
		tmp = Float64(t_6 * Float64(Float64(t_4 / Float64(t * Float64(t_1 * Float64(cbrt(tan(k)) * cbrt(t_2))))) / t_1));
	elseif (t <= 1e-96)
		tmp = Float64(Float64(l * l) * Float64(2.0 / Float64(tan(k) * Float64(sin(k) * Float64(Float64(t * Float64(k * k)) + Float64(2.0 * (t ^ 3.0)))))));
	else
		tmp = Float64(t_6 * Float64(t_5 / t_1));
	end
	return tmp
end
code[t_, l_, k_] := N[(2.0 / N[(N[(N[(N[(N[Power[t, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision] * N[Sin[k], $MachinePrecision]), $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 + N[Power[N[(k / t), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
code[t_, l_, k_] := Block[{t$95$1 = N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$2 = N[(2.0 + N[Power[N[(k / t), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[Power[N[(N[Tan[k], $MachinePrecision] * t$95$2), $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$4 = N[(l * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$5 = N[(t$95$4 / N[(t * N[(t$95$1 * t$95$3), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$6 = N[(t$95$5 / N[(t * t$95$3), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t, 2.1e-274], N[(t$95$6 * N[(N[(t$95$4 / N[(t * N[(t$95$1 * N[(N[Power[N[Tan[k], $MachinePrecision], 1/3], $MachinePrecision] * N[Power[t$95$2, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t, 1e-96], N[(N[(l * l), $MachinePrecision] * N[(2.0 / N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(t * N[(k * k), $MachinePrecision]), $MachinePrecision] + N[(2.0 * N[Power[t, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$6 * N[(t$95$5 / t$95$1), $MachinePrecision]), $MachinePrecision]]]]]]]]]
\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}
\begin{array}{l}
t_1 := \sqrt[3]{\sin k}\\
t_2 := 2 + {\left(\frac{k}{t}\right)}^{2}\\
t_3 := \sqrt[3]{\tan k \cdot t_2}\\
t_4 := \ell \cdot \sqrt{2}\\
t_5 := \frac{t_4}{t \cdot \left(t_1 \cdot t_3\right)}\\
t_6 := \frac{t_5}{t \cdot t_3}\\
\mathbf{if}\;t \leq 2.1 \cdot 10^{-274}:\\
\;\;\;\;t_6 \cdot \frac{\frac{t_4}{t \cdot \left(t_1 \cdot \left(\sqrt[3]{\tan k} \cdot \sqrt[3]{t_2}\right)\right)}}{t_1}\\

\mathbf{elif}\;t \leq 10^{-96}:\\
\;\;\;\;\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\sin k \cdot \left(t \cdot \left(k \cdot k\right) + 2 \cdot {t}^{3}\right)\right)}\\

\mathbf{else}:\\
\;\;\;\;t_6 \cdot \frac{t_5}{t_1}\\


\end{array}

Error

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 3 regimes
  2. if t < 2.09999999999999994e-274

    1. Initial program 34.0

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
    2. Simplified34.0

      \[\leadsto \color{blue}{\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\left(2 + {\left(\frac{k}{t}\right)}^{2}\right) \cdot \left({t}^{3} \cdot \sin k\right)\right)}} \]
    3. Applied egg-rr23.6

      \[\leadsto \color{blue}{\frac{\frac{\ell \cdot \left(\ell \cdot 2\right)}{{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \left(t \cdot \sqrt[3]{\sin k}\right)\right)}^{2}}}{\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \left(t \cdot \sqrt[3]{\sin k}\right)}} \]
    4. Applied egg-rr12.2

      \[\leadsto \color{blue}{\frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{t \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \cdot \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{\sqrt[3]{\sin k}}} \]
    5. Applied egg-rr12.2

      \[\leadsto \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{t \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \cdot \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \color{blue}{\left(\sqrt[3]{\tan k} \cdot \sqrt[3]{2 + {\left(\frac{k}{t}\right)}^{2}}\right)}\right)}}{\sqrt[3]{\sin k}} \]

    if 2.09999999999999994e-274 < t < 9.9999999999999991e-97

    1. Initial program 61.6

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
    2. Simplified61.7

      \[\leadsto \color{blue}{\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\left(2 + {\left(\frac{k}{t}\right)}^{2}\right) \cdot \left({t}^{3} \cdot \sin k\right)\right)}} \]
    3. Taylor expanded in k around inf 26.7

      \[\leadsto \left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \color{blue}{\left(2 \cdot \left(\sin k \cdot {t}^{3}\right) + {k}^{2} \cdot \left(\sin k \cdot t\right)\right)}} \]
    4. Simplified26.7

      \[\leadsto \left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \color{blue}{\left(\sin k \cdot \left(t \cdot \left(k \cdot k\right) + 2 \cdot {t}^{3}\right)\right)}} \]

    if 9.9999999999999991e-97 < t

    1. Initial program 22.9

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
    2. Simplified22.9

      \[\leadsto \color{blue}{\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\left(2 + {\left(\frac{k}{t}\right)}^{2}\right) \cdot \left({t}^{3} \cdot \sin k\right)\right)}} \]
    3. Applied egg-rr17.7

      \[\leadsto \color{blue}{\frac{\frac{\ell \cdot \left(\ell \cdot 2\right)}{{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \left(t \cdot \sqrt[3]{\sin k}\right)\right)}^{2}}}{\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \left(t \cdot \sqrt[3]{\sin k}\right)}} \]
    4. Applied egg-rr4.5

      \[\leadsto \color{blue}{\frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{t \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \cdot \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{\sqrt[3]{\sin k}}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification10.8

    \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 2.1 \cdot 10^{-274}:\\ \;\;\;\;\frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{t \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \cdot \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \left(\sqrt[3]{\tan k} \cdot \sqrt[3]{2 + {\left(\frac{k}{t}\right)}^{2}}\right)\right)}}{\sqrt[3]{\sin k}}\\ \mathbf{elif}\;t \leq 10^{-96}:\\ \;\;\;\;\left(\ell \cdot \ell\right) \cdot \frac{2}{\tan k \cdot \left(\sin k \cdot \left(t \cdot \left(k \cdot k\right) + 2 \cdot {t}^{3}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{t \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \cdot \frac{\frac{\ell \cdot \sqrt{2}}{t \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}\right)}}{\sqrt[3]{\sin k}}\\ \end{array} \]

Reproduce

herbie shell --seed 2022210 
(FPCore (t l k)
  :name "Toniolo and Linder, Equation (10+)"
  :precision binary64
  (/ 2.0 (* (* (* (/ (pow t 3.0) (* l l)) (sin k)) (tan k)) (+ (+ 1.0 (pow (/ k t) 2.0)) 1.0))))