单位圆周上任取n个点, 构成的n边形面积是多少

王守恩

单位圆周上任取n个点, 构成的n边形面积是多少

王守恩

谢谢 northwolves！！！

\(S(3)=\frac{3}{2 \pi}=0.47746483,\)

\(S(4)=\frac{3}{\pi}= 0.95492966, \)

\(S(5)=\frac{5 (2\pi^2 - 3)}{2 \pi^3}=1.3496629,\)

\(S(6)=\frac{15 (\pi^2 - 3)}{2 \pi^3}=1.6616646, \)

\(S(7)=\frac{21 (15 + 2\pi^2 (\pi^2 - 5))}{4 \pi^5}=1.9063846,\)

\(S(8)=\frac{ 7 (2 \pi^4 - 15\pi^2 + 45)}{\pi^5}=2.0992728,\)

\(S(9)=\frac{9 (4 \pi^6 - 42 \pi^4 + 210\pi^2 - 315)}{2\pi^7)}=2.2527493,\)

\(\cdots\cdots\)


{0.47746483, 0.95492966, 1.3496629, 1.6616646, 1.9063846, 2.0992728, 2.2527493, 2.3762042, 2.4766132, 2.5591520, 2.6276838, 2.6851178, 2.7336674, 2.7750341, 2.8105396,
2.8412209, 2.8678994, 2.8912316, 2.9117466, 2.9298743, 2.9459668, 2.9603140, 2.9731569, 2.9846965, 2.9951016, 3.0045149, 3.0130574, 3.0208324, 3.0279285, 3.0344219}

1. J[0] = 0; J[1] = 1/(2 Pi); J[k_] := J[k] = (2 Pi - (k (k - 1)) J[k - 2])/(4 Pi^2); S[n_] := (n (n - 1) J[n - 2])/2; Table[S[n], {n, 3, 9}] // FullSimplify

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王守恩

谢谢 northwolves！！！

\(S(3)=\frac{3}{2 \pi}=0.47746483,\)

\(S(4)=\frac{3}{\pi}= 0.95492966, \)

\(S(5)=\frac{5 (2\pi^2 - 3)}{2 \pi^3}=1.3496629,\)

\(S(6)=\frac{15 (\pi^2 - 3)}{2 \pi^3}=1.6616646, \)

\(S(7)=\frac{21 (15 + 2\pi^2 (\pi^2 - 5))}{4 \pi^5}=1.9063846,\)

\(S(8)=\frac{ 7 (2 \pi^4 - 15\pi^2 + 45)}{\pi^5}=2.0992728,\)

\(S(9)=\frac{9 (4 \pi^6 - 42 \pi^4 + 210\pi^2 - 315)}{2\pi^7}=2.2527493,\)

\(\cdots\cdots\)

1. Table[{n,Pi HypergeometricPFQ[{1},{(n+1)/2,(n+2)/2},-Pi^2]//FullSimplify},{n,3,10}]

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王守恩

备忘——圆周上任取 n 个点,  求 n 个点落在同一半圆周内的概率。

Table[{n, n/2^(n - 1)}, {n, 10}]

{{1, 1}, {2, 1}, {3, 3/4}, {4, 1/2}, {5, 5/16}, {6, 3/16}, {7, 7/64}, {8, 1/16}, {9, 9/256}, {10, 5/256}}