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正多边形中心到各个顶点的向量和为零向量

中心 向量 各个 顶点
2023-09-14 09:06:48 时间

引理

∑ k = 1 n cos ⁡ k x = sin ⁡ x ∑ k = 1 n cos ⁡ k x sin ⁡ x = sin ⁡ 2 x − sin ⁡ 0 x + sin ⁡ 3 x − sin ⁡ 1 x + ⋯ + sin ⁡ ( n + 1 ) x − sin ⁡ ( n − 1 ) x 2 sin ⁡ x = sin ⁡ ( n + 1 ) x + sin ⁡ n x − sin ⁡ x 2 sin ⁡ x \begin{aligned} &\quad \sum_{k=1}^{n} \cos kx \\ &= \frac{\sin x \sum_{k=1}^{n} \cos kx}{\sin x} \\ &= \frac{\sin 2x - \sin 0x +\sin 3x - \sin 1x+\dots + \sin (n+1)x -\sin (n-1)x}{2\sin x} \\ &=\frac{\sin (n+1)x +\sin nx -\sin x}{2\sin x} \end{aligned} ​k=1∑n​coskx=sinxsinx∑k=1n​coskx​=2sinxsin2x−sin0x+sin3x−sin1x+⋯+sin(n+1)x−sin(n−1)x​=2sinxsin(n+1)x+sinnx−sinx​​
∑ k = 1 n sin ⁡ k x = sin ⁡ x 2 ∑ k = 1 n sin ⁡ k x sin ⁡ x 2 = − cos ⁡ 3 x 2 − cos ⁡ x 2 + cos ⁡ 5 x 2 − cos ⁡ 3 x 2 + ⋯ + cos ⁡ ( 2 n + 1 ) x 2 − cos ⁡ ( 2 n − 1 ) x 2 2 sin ⁡ x 2 = cos ⁡ x 2 − cos ⁡ ( 2 n + 1 ) x 2 2 sin ⁡ x 2 \begin{aligned} &\quad \sum_{k=1}^{n} \sin kx \\ &= \frac{\sin \frac{x}{2} \sum_{k=1}^{n} \sin kx}{\sin \frac{x}{2}} \\ &= -\frac{\cos \frac{3x}{2} - \cos \frac{x}{2} +\cos \frac{5x}{2} - \cos \frac{3x}{2}+\dots + \cos \frac{(2n+1)x}{2} -\cos \frac{(2n-1)x}{2}}{2\sin \frac{x}{2}} \\ &=\frac{\cos \frac{x}{2}-\cos \frac{(2n+1)x}{2}}{2\sin \frac{x}{2}} \end{aligned} ​k=1∑n​sinkx=sin2x​sin2x​∑k=1n​sinkx​=−2sin2x​cos23x​−cos2x​+cos25x​−cos23x​+⋯+cos2(2n+1)x​−cos2(2n−1)x​​=2sin2x​cos2x​−cos2(2n+1)x​​​

∑ k = 0 n cos ⁡ k x = sin ⁡ ( n + 1 ) x + sin ⁡ n x + sin ⁡ x 2 sin ⁡ x \sum_{k=0}^{n} \cos kx =\frac{\sin (n+1)x +\sin nx + \sin x}{2\sin x} k=0∑n​coskx=2sinxsin(n+1)x+sinnx+sinx​
∑ k = 0 n sin ⁡ k x = cos ⁡ x 2 − cos ⁡ ( 2 n + 1 ) x 2 2 sin ⁡ x 2 \sum_{k=0}^{n}\sin kx = \frac{\cos \frac{x}{2}-\cos \frac{(2n+1)x}{2}}{2\sin \frac{x}{2}} k=0∑n​sinkx=2sin2x​cos2x​−cos2(2n+1)x​​

前置知识

  正 N N N边形中心角为 2 π N \frac{2\pi }{N} N2π​
   z = x + j y = ( x 2 + y 2 ) e j arg ⁡ z z=x+j y=(\sqrt{x^2+y^2})e^{j\arg z} z=x+jy=(x2+y2 ​)ejargz
  设 z 1 , z 2 z_1,z_2 z1​,z2​为两个非零复数，则
   z 1 z 2 = r 1 e j θ 1 r 2 e j θ 2 = ( r 1 r 2 ) e j ( θ 1 + θ 2 ) z_1 z_2=r_1 e^{j\theta_1} r_2e^{j\theta_2}= (r_1 r_2)e^{j(\theta_1 +\theta_2)} z1​z2​=r1​ejθ1​r2​ejθ2​=(r1​r2​)ej(θ1​+θ2​)
  所以复数相乘，幅角相加，模长相乘

原命题

设正 N N N边形 A 0 , A 1 , … , A N − 1 A_0,A_1,\dots,A_{N-1} A0​,A1​,…,AN−1​
考虑复平面里，中心在原点的正 N N N边形，并且有一个顶点在实轴的正半轴
在这里插入图片描述
相当于证明
S → = ∑ k = 0 N − 1 O A k → = ∑ k = 0 N − 1 e k 2 π j N = 0 \overrightarrow{S}=\sum_{k=0}^{N-1} \overrightarrow{OA_k}=\sum_{k=0}^{N-1} e^{k\frac{2\pi j}{N}}=0 S =k=0∑N−1​OAk​ ​=k=0∑N−1​ekN2πj​=0
证明：

方法1

∑ k = 0 N − 1 e k 2 π j N = ∑ k = 0 N − 1 ( cos ⁡ k 2 π N + j sin ⁡ k 2 π N ) = ∑ k = 0 N − 1 cos ⁡ k 2 π N + j ∑ k = 0 N − 1 sin ⁡ k 2 π N = sin ⁡ N 2 π N + sin ⁡ ( N − 1 ) 2 π N + sin ⁡ 2 π N 2 sin ⁡ 2 π N + j cos ⁡ 2 π 2 N − cos ⁡ 2 N − 1 2 2 π N 2 sin ⁡ 2 π 2 N = 0 \begin{aligned} &\quad \sum_{k=0}^{N-1} e^{k\frac{2\pi j}{N}} \\ &=\sum_{k=0}^{N-1}(\cos k\frac{2\pi}{N}+j\sin k\frac{2\pi}{N}) \\ &=\sum_{k=0}^{N-1}\cos k\frac{2\pi}{N}+j\sum_{k=0}^{N-1}\sin k\frac{2\pi}{N} \\ &=\frac{\sin N\frac{2\pi}{N}+\sin (N-1)\frac{2\pi}{N}+\sin\frac{2\pi}{N}}{2\sin\frac{2\pi}{N}} + j\frac{\cos \frac{2\pi}{2N}-\cos \frac{2N-1}{2}\frac{2\pi}{N}}{2\sin \frac{2\pi}{2N}} \\ &=0 \end{aligned} ​k=0∑N−1​ekN2πj​=k=0∑N−1​(coskN2π​+jsinkN2π​)=k=0∑N−1​coskN2π​+jk=0∑N−1​sinkN2π​=2sinN2π​sinNN2π​+sin(N−1)N2π​+sinN2π​​+j2sin2N2π​cos2N2π​−cos22N−1​N2π​​=0​

方法2

S → = ∑ k = 0 N − 1 O A k → \overrightarrow{S}=\sum_{k=0}^{N-1} \overrightarrow{OA_k} S =k=0∑N−1​OAk​ ​
两边同时旋转 2 π N \frac{2\pi}{N} N2π​(相当于乘 e 2 π N e^{\frac{2\pi}{N}} eN2π​）
O A k → \overrightarrow{OA_k} OAk​ ​就会旋转成 O A k + 1 → , k = 0 , 1 , 2 , … , N − 2 \overrightarrow{OA_{k+1}},k=0,1,2,\dots,N-2 OAk+1​ ​,k=0,1,2,…,N−2
O A N − 1 → \overrightarrow{OA_{N-1}} OAN−1​ ​就会旋转成 O A 0 → \overrightarrow{OA_0} OA0​ ​
因为等号右边旋转后相等，所以
S → = e 2 π N S → \overrightarrow{S}=e^{\frac{2\pi}{N}} \overrightarrow{S} S =eN2π​S
S → = 0 \overrightarrow{S}=0 S =0

方法3

∑ k = 0 N − 1 e k 2 π j N = 1 − e N 2 π j N 1 − e 2 π j N = 0 \sum_{k=0}^{N-1}e^{k\frac{2\pi j}{N}}=\frac{1-e^{N\frac{2\pi j}{N}}}{1-e^{\frac{2\pi j}{N}}}=0 k=0∑N−1​ekN2πj​=1−eN2πj​1−eNN2πj​​=0

推论

S → = ∑ k = 0 N − 1 V k → \overrightarrow{S}=\sum_{k=0}^{N-1} \overrightarrow{V_k} S =∑k=0N−1​Vk​ ​
e j θ V k → = V k + 1 → ( θ ≠ 2 n π , n ∈ Z ) e^{j\theta}\overrightarrow{V_k}=\overrightarrow{V_{k+1}} \left(\theta \ne 2n\pi ,n\in Z \right ) ejθVk​ ​=Vk+1​ ​(θ​=2nπ,n∈Z)
则 S → = 0 \overrightarrow{S}=0 S =0

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