从几何意义出发【星形线】

青青子衿

已知$\theta_{i}\in R,\theta_{i}<\theta_{i+1}$
求证：\[\sum_{i=1}^n\sqrt{1-cos(\theta_{i+1}-\theta_{i})}<\frac{\theta_n-\theta_1}{\sqrt{2}}-\sqrt{1-cos(\theta_{n}-\theta_{1})}\]

青青子衿

已知$\sum_{i=1}^n\sqrt{1-cos(\theta_{i+1}-\theta_{i})}<\frac{\theta_n-\theta_1}{\sqrt{2}}-\sqrt{1-cos(\theta_{n}-\theta_{1})}$
青青子衿 发表于 2014-2-1 09:02

在单位圆上任意两$点A(\cos\theta_1，\sin\theta_1),点B(\cos\theta_2，\sin\theta_2)$,有$|AB|<弧AB$，则(翻译为代数语言为)
\[\sqrt{(\cos\theta_1-\cos\theta_2)^2+(\sin\theta_1-\sin\theta_2)^2}<\theta_2-\theta_1\]
推广为：
\[\sqrt{(\cos^3\theta_1-\cos^3\theta_2)^2+(\sin^3\theta_1-\sin^3\theta_2)^2}<星形线的弧长\]