求4/π等于1-1/3+1/5-1/7+…1/n
时间: 2023-10-04 15:14:37 浏览: 97
求数列4*(1-1/3+1/5-1/7·····)的值,直到最后一项的绝对值小于10的负6次方为止
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根据莱布尼茨公式,有:
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots$$
将上式化简得:
$$\begin{aligned}\frac{4}{\pi}&=\frac{4}{\pi}\left( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots\right) \\ &=\frac{4}{\pi}-\frac{4}{3\pi}+\frac{4}{5\pi}-\frac{4}{7\pi}+\frac{4}{9\pi}-\frac{4}{11\pi}+\cdots\end{aligned}$$
移项得:
$$\frac{4}{3\pi}-\frac{4}{5\pi}+\frac{4}{7\pi}-\frac{4}{9\pi}+\frac{4}{11\pi}-\cdots=\frac{4}{\pi}-\frac{4}{\pi}=0$$
所以,有:
$$\begin{aligned} \frac{4}{\pi}&=\frac{4}{3\pi}-\frac{4}{5\pi}+\frac{4}{7\pi}-\frac{4}{9\pi}+\frac{4}{11\pi}-\cdots \\ &=\frac{1}{\frac{3}{4}\pi}-\frac{1}{\frac{5}{4}\pi}+\frac{1}{\frac{7}{4}\pi}-\frac{1}{\frac{9}{4}\pi}+\frac{1}{\frac{11}{4}\pi}-\cdots \\ &=\frac{1}{\tan^{-1}\frac{3}{4}}-\frac{1}{\tan^{-1}\frac{5}{4}}+\frac{1}{\tan^{-1}\frac{7}{4}}-\frac{1}{\tan^{-1}\frac{9}{4}}+\frac{1}{\tan^{-1}\frac{11}{4}}-\cdots \\ \end{aligned}$$
因此,
$$\begin{aligned} \frac{4}{\pi}&=\frac{1}{\tan^{-1}\frac{3}{4}}-\frac{1}{\tan^{-1}\frac{5}{4}}+\frac{1}{\tan^{-1}\frac{7}{4}}-\frac{1}{\tan^{-1}\frac{9}{4}}+\frac{1}{\tan^{-1}\frac{11}{4}}-\cdots \\ &=\frac{1}{\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{5}{4}+\tan^{-1}\frac{7}{4}-\tan^{-1}\frac{9}{4}+\tan^{-1}\frac{11}{4}-\cdots} \\ \end{aligned}$$
根据反正切函数的性质,有:
$$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$$
因此,
$$\begin{aligned} \frac{4}{\pi}&=\frac{1}{\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{5}{4}+\tan^{-1}\frac{7}{4}-\tan^{-1}\frac{9}{4}+\tan^{-1}\frac{11}{4}-\cdots} \\ &=\frac{1}{\tan^{-1}\frac{3}{4}\cdot \frac{4}{5}\cdot \frac{7}{8}\cdots-\tan^{-1}\frac{5}{4}\cdot \frac{4}{3}\cdot \frac{8}{7}\cdots+\tan^{-1}\frac{7}{4}\cdot \frac{4}{3}\cdot \frac{10}{9}\cdots-\tan^{-1}\frac{9}{4}\cdot \frac{4}{3}\cdot \frac{12}{11}\cdots+\tan^{-1}\frac{11}{4}\cdot \frac{4}{3}\cdot \frac{14}{13}\cdots-\cdots} \\ &=\frac{1}{\tan^{-1}\frac{3}{4}\cdot \frac{4}{5}\cdot \frac{7}{8}\cdots} \\ &=\frac{1}{\frac{3}{4}\cdot \frac{4}{5}\cdot \frac{7}{8}\cdots} \\ &=\frac{4}{3}\cdot \frac{5}{4}\cdot \frac{8}{7}\cdots \\ \end{aligned}$$
因此,
$$\frac{4}{\pi}=\frac{4}{3}\cdot \frac{5}{4}\cdot \frac{8}{7}\cdots=\frac{4\cdot 5\cdot 8\cdots}{3\cdot 4\cdot 7\cdots}$$
两边同时乘以$\frac{\pi}{4}$,得:
$$1=\frac{\pi}{4}\cdot \frac{4}{\pi}=\frac{4\cdot 5\cdot 8\cdots}{3\cdot 4\cdot 7\cdots}\cdot \frac{\pi}{4}=\frac{\pi}{3}\cdot \frac{5}{7}\cdot \frac{9}{11}\cdots$$
因此,
$$\frac{4}{\pi}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots=\frac{3}{4}\cdot \frac{5}{7}\cdot \frac{9}{11}\cdots$$
这就是所求的式子。
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