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

northwolves

三角形和四边形的面积期望值分别是多少呢？

LLPikaPika

我找的一些信息
1. https://math.stackexchange.com/q ... ces-lie-on-a-circle
2. https://www.luogu.com/article/oq6dj4c4

LLPikaPika

LLPikaPika 发表于 2026-5-6 21:47
我找的一些信息
1. https://math.stackexchange.com/questions/160380/what-is-the-expected-area-of-a-pol ...

3. https://zhuanlan.zhihu.com/p/31050060

LLPikaPika

问题说得准确一些，不要少字

northwolves

https://www.mathpages.com/home/kmath516/kmath516.htm 这里有个公式，但三角形的值似乎是错的

northwolves

我计算的是$A_3=\frac{3\sqrt3}{4\pi}$。刚才随机生成了100000个三角形，模拟平均面积: 0.477866，看来$A_3=\frac{3}{2 \pi }$是对的

northwolves

1. f[n_]:=\[Pi] HypergeometricPFQ[{1},{(n+1)/2,(n+2)/2},-\[Pi]^2]

复制代码

$f(n)=\pi  * _1F_2\left(1;\frac{n+1}{2},\frac{n+2}{2};-\pi ^2\right)$

王守恩

northwolves 发表于 2026-5-6 21:53
https://www.mathpages.com/home/kmath516/kmath516.htm 这里有个公式，但三角形的值似乎是错的 ...

根据6#资料——是这串数？？？

{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}

Table[N[Sum[(n! (2 Pi)^(2 k - Mod[n, 2]) Cos[Pi*Floor[(n + 1)/2]])/((2 k - Mod[n, 2])! 2^n Pi^(n - 1) Cos[Pi*(k + 1)]), {k, (n - 1)/2}], 8], {n, 3, 32}]

\(\D \sum_{k=1}^{(n-1)/2}\frac{n! (2\pi)^{2 k - Mod[n, 2]}\cos[\pi(n + 1)/2]}{(2 k - Mod[n, 2])! 2^n\pi^{n - 1}\cos[\pi(k + 1)]}\)

{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}

Table[N[Pi*n!/I^(n(n + 1)) Sum[(2pi)^(2k - Mod[n, 2] - n)/((2k - Mod[n, 2])! I^(2(k + 1))), {k, (n - 1)/2}], 8], {n, 3, 32}]——这样也可以。

\(\D\frac{\pi*n!}{I^{n(n + 1)}}\sum_{k=1}^{(n-1)/2}\frac{(2\pi)^{2k - Mod[n, 2] - n}}{(2k - Mod[n, 2])! I^{2(k + 1)}}\)——还是不好看！

王守恩

northwolves 发表于 2026-5-6 21:53
https://www.mathpages.com/home/kmath516/kmath516.htm 这里有个公式，但三角形的值似乎是错的 ...

{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}

如果是这串数——可以用这个代码。

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[N[S[n], 8], {n, 3, 32}]

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

{3/(2 Pi), 3/Pi, 5 (2 Pi^2 - 3)/(2 Pi^3), 15 (Pi^2 - 3)/(2 Pi^3), 21 (15 + 2 Pi^2 (Pi^2 - 5))/(4 Pi^5), 7 (2 Pi^4 - 15Pi^2 + 45)/Pi^5, 9 (4 Pi^6 - 42 Pi^4 + 210 Pi^2 - 315)/(2 Pi^7)}