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Average Accuracy: 48.6% → 88.8%
Time: 3.0min
Precision: binary64

?

\[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
\[\begin{array}{l} t_1 := \left(\frac{\cos k}{t} \cdot \frac{{\left(\frac{-1}{k} \cdot \ell\right)}^{2}}{{\sin k}^{2}}\right) \cdot 2\\ \mathbf{if}\;k \leq -3.7 \cdot 10^{+101}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;k \leq 1.16 \cdot 10^{+77}:\\ \;\;\;\;{\left(\frac{\sqrt[3]{\frac{-2}{-2 - {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt[3]{\tan k} \cdot \left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \left(\sqrt[3]{\sin k} \cdot t\right)\right)}\right)}^{3}\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]
(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
         (*
          (* (/ (cos k) t) (/ (pow (* (/ -1.0 k) l) 2.0) (pow (sin k) 2.0)))
          2.0)))
   (if (<= k -3.7e+101)
     t_1
     (if (<= k 1.16e+77)
       (pow
        (/
         (cbrt (/ -2.0 (- -2.0 (pow (/ k t) 2.0))))
         (* (cbrt (tan k)) (* (pow (cbrt l) -2.0) (* (cbrt (sin k)) t))))
        3.0)
       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 = ((cos(k) / t) * (pow(((-1.0 / k) * l), 2.0) / pow(sin(k), 2.0))) * 2.0;
	double tmp;
	if (k <= -3.7e+101) {
		tmp = t_1;
	} else if (k <= 1.16e+77) {
		tmp = pow((cbrt((-2.0 / (-2.0 - pow((k / t), 2.0)))) / (cbrt(tan(k)) * (pow(cbrt(l), -2.0) * (cbrt(sin(k)) * t)))), 3.0);
	} else {
		tmp = 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.cos(k) / t) * (Math.pow(((-1.0 / k) * l), 2.0) / Math.pow(Math.sin(k), 2.0))) * 2.0;
	double tmp;
	if (k <= -3.7e+101) {
		tmp = t_1;
	} else if (k <= 1.16e+77) {
		tmp = Math.pow((Math.cbrt((-2.0 / (-2.0 - Math.pow((k / t), 2.0)))) / (Math.cbrt(Math.tan(k)) * (Math.pow(Math.cbrt(l), -2.0) * (Math.cbrt(Math.sin(k)) * t)))), 3.0);
	} else {
		tmp = 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 = Float64(Float64(Float64(cos(k) / t) * Float64((Float64(Float64(-1.0 / k) * l) ^ 2.0) / (sin(k) ^ 2.0))) * 2.0)
	tmp = 0.0
	if (k <= -3.7e+101)
		tmp = t_1;
	elseif (k <= 1.16e+77)
		tmp = Float64(cbrt(Float64(-2.0 / Float64(-2.0 - (Float64(k / t) ^ 2.0)))) / Float64(cbrt(tan(k)) * Float64((cbrt(l) ^ -2.0) * Float64(cbrt(sin(k)) * t)))) ^ 3.0;
	else
		tmp = 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[(N[(N[(N[Cos[k], $MachinePrecision] / t), $MachinePrecision] * N[(N[Power[N[(N[(-1.0 / k), $MachinePrecision] * l), $MachinePrecision], 2.0], $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * 2.0), $MachinePrecision]}, If[LessEqual[k, -3.7e+101], t$95$1, If[LessEqual[k, 1.16e+77], N[Power[N[(N[Power[N[(-2.0 / N[(-2.0 - N[Power[N[(k / t), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] / N[(N[Power[N[Tan[k], $MachinePrecision], 1/3], $MachinePrecision] * N[(N[Power[N[Power[l, 1/3], $MachinePrecision], -2.0], $MachinePrecision] * N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision], t$95$1]]]
\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 := \left(\frac{\cos k}{t} \cdot \frac{{\left(\frac{-1}{k} \cdot \ell\right)}^{2}}{{\sin k}^{2}}\right) \cdot 2\\
\mathbf{if}\;k \leq -3.7 \cdot 10^{+101}:\\
\;\;\;\;t_1\\

\mathbf{elif}\;k \leq 1.16 \cdot 10^{+77}:\\
\;\;\;\;{\left(\frac{\sqrt[3]{\frac{-2}{-2 - {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt[3]{\tan k} \cdot \left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \left(\sqrt[3]{\sin k} \cdot t\right)\right)}\right)}^{3}\\

\mathbf{else}:\\
\;\;\;\;t_1\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes
  2. if k < -3.6999999999999997e101 or 1.1600000000000001e77 < k

    1. Initial program 46.8%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
    2. Simplified47.2%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{{t}^{3} \cdot \left({\ell}^{-2} \cdot \sin k\right)}}{\tan k}}{{\left(\frac{k}{t}\right)}^{2} - -2}} \]
      Proof
    3. Taylor expanded in k around -inf 64.0%

      \[\leadsto \color{blue}{2 \cdot \frac{\cos k \cdot \left({\left(\frac{-1}{k}\right)}^{2} \cdot {\ell}^{2}\right)}{t \cdot {\sin k}^{2}}} \]
    4. Simplified88.1%

      \[\leadsto \color{blue}{\left(\frac{\cos k}{t} \cdot \frac{{\left(\frac{-1}{k} \cdot \ell\right)}^{2}}{{\sin k}^{2}}\right) \cdot 2} \]
      Proof

    if -3.6999999999999997e101 < k < 1.1600000000000001e77

    1. Initial program 50.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
    2. Simplified51.1%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{{t}^{3} \cdot \left({\ell}^{-2} \cdot \sin k\right)}}{\tan k}}{{\left(\frac{k}{t}\right)}^{2} - -2}} \]
      Proof
    3. Applied egg-rr60.6%

      \[\leadsto \frac{\color{blue}{\frac{\frac{2}{t \cdot \sqrt[3]{{\ell}^{-2} \cdot \sin k}}}{\tan k \cdot {\left(t \cdot \sqrt[3]{{\ell}^{-2} \cdot \sin k}\right)}^{2}}}}{{\left(\frac{k}{t}\right)}^{2} - -2} \]
    4. Applied egg-rr60.6%

      \[\leadsto \frac{\frac{\frac{2}{t \cdot \color{blue}{\left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \sqrt[3]{\sin k}\right)}}}{\tan k \cdot {\left(t \cdot \sqrt[3]{{\ell}^{-2} \cdot \sin k}\right)}^{2}}}{{\left(\frac{k}{t}\right)}^{2} - -2} \]
    5. Applied egg-rr84.6%

      \[\leadsto \frac{\frac{\frac{2}{t \cdot \left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \sqrt[3]{\sin k}\right)}}{\tan k \cdot {\left(t \cdot \color{blue}{\left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \sqrt[3]{\sin k}\right)}\right)}^{2}}}{{\left(\frac{k}{t}\right)}^{2} - -2} \]
    6. Applied egg-rr89.5%

      \[\leadsto \color{blue}{{\left(\frac{\sqrt[3]{\frac{-2}{-2 - {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt[3]{\tan k} \cdot \left(\left({\left(\sqrt[3]{\ell}\right)}^{-2} \cdot \sqrt[3]{\sin k}\right) \cdot t\right)}\right)}^{3}} \]
    7. Simplified89.4%

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

Reproduce?

herbie shell --seed 2023237 
(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))))