[再寄小读者之数学篇](2014-04-08 from 1297503521@qq.com $\sin x-x\cos x=0$ 的根的估计)

(2014-04-08 from 1297503521@qq.com) 设方程 $\sin x-x\cos x=0$ 在 $(0,+\infty)$ 中的第 $n$ 个解为 $x_n$. 证明: $$\bex n\pi+\cfrac{\pi}{2}-\cfrac{1}{n\pi} <x_n<n\pi+\cfrac{\pi}{2}. \eex$$

证明: 设 $f(x)=\sin x-x\cos x$, 则 $$\bex f'(x)=x\sin x\sedd{\ba{ll} >0,&x\in I_{2n},\\ <0,&x\in I_{2n+1}, \ea} \eex$$ 其中 $I_n=(n\pi,(n+1)\pi)$. 又 $$\beex \bea f(0)&=0,\\ f(2n\pi)&=-2n\pi<0\quad(n\geq 1),\\ f((2n+1)\pi)&=(2n+1)\pi>0， \eea \eeex$$ 我们知 $f(x)=0$ 在 $(0,+\infty)$ 内的第 $n$ 个解 $x_n\in I_n$. 再注意到 $$\beex \bea &\quad f\sex{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}}\\ &=\cos\cfrac{1}{2n\pi} -\sex{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}}\sin\cfrac{1}{2n\pi}\\ &=\sex{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}} \cos\cfrac{1}{2n\pi}\cdot\sex{ \cfrac{1}{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}} -\tan\cfrac{1}{2n\pi} }\\ &<\sex{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}} \cos\cfrac{1}{2n\pi}\cdot\sex{ \cfrac{1}{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi}} -\cfrac{1}{2n\pi}}\\ &<0,\\ f\sex{2n\pi+\cfrac{\pi}{2}}&=1>0, \eea \eeex$$ 我们有 $$\bex x_{2n}\in \sex{2n\pi+\cfrac{\pi}{2}-\cfrac{1}{2n\pi},2n\pi+\cfrac{\pi}{2}}. \eex$$ 同理, 由 $$\beex \bea &\quad f\sex{(2n+1)\pi+\cfrac{\pi}{2}-\cfrac{1}{(2n+1)\pi}}\\ &=-\cos\cfrac{1}{(2n+1)\pi} +\sez{(2n+1)\pi+\cfrac{\pi}{2}-\cfrac{1}{(2n+1)\pi}}\sin\cfrac{1}{(2n+1)\pi}\\ &=-\sez{(2n+1)\pi+\cfrac{\pi}{2}-\cfrac{1}{(2n+1)\pi}} \cos\cfrac{1}{(2n+1)\pi}\\ &\quad\cdot \sez{\cfrac{1}{(2n+1)\pi+\cfrac{\pi}{2}-\cfrac{1}{(2n+1)\pi}}-\tan\cfrac{1}{(2n+1)\pi}}\\ &>0,\\ &\quad f\sex{(2n+1)\pi+\cfrac{\pi}{2}}\\ &=-1<0 \eea \eeex$$ 我们知 $$\bex x_{2n+1}\in \sex{ (2n+1)\pi+\cfrac{\pi}{2}-\cfrac{1}{(2n+1)\pi}, (2n+1)\pi+\cfrac{\pi}{2} }. \eex$$

[再寄小读者之数学篇](2014-04-08 from 1297503521@qq.com $\sin x-x\cos x=0$ 的根的估计)的相关教程结束。