Math FPCore C Java Python Julia MATLAB Wolfram TeX \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right)
\]
↓
\[\frac{0.5}{b - a} \cdot \frac{\frac{b - a}{b \cdot a}}{\frac{b + a}{\pi}}
\]
(FPCore (a b)
:precision binary64
(* (* (/ PI 2.0) (/ 1.0 (- (* b b) (* a a)))) (- (/ 1.0 a) (/ 1.0 b)))) ↓
(FPCore (a b)
:precision binary64
(* (/ 0.5 (- b a)) (/ (/ (- b a) (* b a)) (/ (+ b a) PI)))) double code(double a, double b) {
return ((((double) M_PI) / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
↓
double code(double a, double b) {
return (0.5 / (b - a)) * (((b - a) / (b * a)) / ((b + a) / ((double) M_PI)));
}
public static double code(double a, double b) {
return ((Math.PI / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
↓
public static double code(double a, double b) {
return (0.5 / (b - a)) * (((b - a) / (b * a)) / ((b + a) / Math.PI));
}
def code(a, b):
return ((math.pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b))
↓
def code(a, b):
return (0.5 / (b - a)) * (((b - a) / (b * a)) / ((b + a) / math.pi))
function code(a, b)
return Float64(Float64(Float64(pi / 2.0) * Float64(1.0 / Float64(Float64(b * b) - Float64(a * a)))) * Float64(Float64(1.0 / a) - Float64(1.0 / b)))
end
↓
function code(a, b)
return Float64(Float64(0.5 / Float64(b - a)) * Float64(Float64(Float64(b - a) / Float64(b * a)) / Float64(Float64(b + a) / pi)))
end
function tmp = code(a, b)
tmp = ((pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
end
↓
function tmp = code(a, b)
tmp = (0.5 / (b - a)) * (((b - a) / (b * a)) / ((b + a) / pi));
end
code[a_, b_] := N[(N[(N[(Pi / 2.0), $MachinePrecision] * N[(1.0 / N[(N[(b * b), $MachinePrecision] - N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 / a), $MachinePrecision] - N[(1.0 / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[a_, b_] := N[(N[(0.5 / N[(b - a), $MachinePrecision]), $MachinePrecision] * N[(N[(N[(b - a), $MachinePrecision] / N[(b * a), $MachinePrecision]), $MachinePrecision] / N[(N[(b + a), $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right)
↓
\frac{0.5}{b - a} \cdot \frac{\frac{b - a}{b \cdot a}}{\frac{b + a}{\pi}}
Alternatives Alternative 1 Accuracy 89.0% Cost 7177
\[\begin{array}{l}
\mathbf{if}\;a \leq -9.6 \cdot 10^{-12} \lor \neg \left(a \leq 9 \cdot 10^{-23}\right):\\
\;\;\;\;0.5 \cdot \frac{\pi}{a \cdot \left(b \cdot a\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{0.5}{b} \cdot \frac{\pi}{b \cdot a}\\
\end{array}
\]
Alternative 2 Accuracy 89.2% Cost 7177
\[\begin{array}{l}
\mathbf{if}\;a \leq -8.5 \cdot 10^{-12} \lor \neg \left(a \leq 7.5 \cdot 10^{-26}\right):\\
\;\;\;\;\frac{\frac{0.5}{b} \cdot \frac{\pi}{a}}{a}\\
\mathbf{else}:\\
\;\;\;\;\frac{0.5}{b} \cdot \frac{\pi}{b \cdot a}\\
\end{array}
\]
Alternative 3 Accuracy 89.2% Cost 7177
\[\begin{array}{l}
\mathbf{if}\;a \leq -1 \cdot 10^{-11} \lor \neg \left(a \leq 3.7 \cdot 10^{-23}\right):\\
\;\;\;\;\frac{\frac{0.5 \cdot \pi}{a}}{b \cdot a}\\
\mathbf{else}:\\
\;\;\;\;\frac{0.5}{b} \cdot \frac{\pi}{b \cdot a}\\
\end{array}
\]
Alternative 4 Accuracy 89.2% Cost 7176
\[\begin{array}{l}
\mathbf{if}\;a \leq -8.5 \cdot 10^{-12}:\\
\;\;\;\;\frac{\frac{0.5 \cdot \pi}{b \cdot a}}{a}\\
\mathbf{elif}\;a \leq 9 \cdot 10^{-23}:\\
\;\;\;\;\frac{0.5}{b} \cdot \frac{\pi}{b \cdot a}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{0.5 \cdot \pi}{a}}{b \cdot a}\\
\end{array}
\]
Alternative 5 Accuracy 99.6% Cost 7040
\[\frac{\pi}{b + a} \cdot \frac{0.5}{b \cdot a}
\]
Alternative 6 Accuracy 60.3% Cost 6912
\[\pi \cdot \frac{\frac{\frac{0.5}{a}}{b}}{b}
\]
Alternative 7 Accuracy 60.3% Cost 6912
\[\frac{0.5}{b} \cdot \frac{\pi}{b \cdot a}
\]