作者Honor1984 (希望愿望成真)
看板trans_math
标题Re: [积分] 黎曼和
时间Fri Aug 21 13:21:08 2009
※ 引述《Honor1984 (希望愿望成真)》之铭言：
※ 引述《midarmyman (midarmyman)》之铭言：
= π/n [f(π/n)+f(2π/n)+....f(π)]
= π/(-2)nsin(π/2n) {-cos[π/2n]+cos[3π/2n] ...... +cos[(2n+1)π/2n]}
= π/(-2)nsin(π/2n) {cos[(2n+1)π/2n]-cos[π/2n]}
= π/(-2)nsin(π/2n) * (-2)sin[(n+1)π/2n]sin[nπ/2n]
πsin[(n+1)π/2n]
= -----------------
n sin[π/2n]
π cos[π/2n]
= --------------
n sin[π/2n]
接下来取极限
2 1
=> -------------- = 2
1
1F：推 midarmyman:为啥偶数项都不见了? 还有为啥会用两 114.38.156.190 08/20 22:35
2F：→ midarmyman:倍角? 114.38.156.190 08/20 22:35
积化和差(第二个等号)後从来就没有偶数项
也都没有用到两倍角公式
上面使用的只有积化和差
广义角的三角函数性质
再加上一些三角函数取极限的知识
3F：推 midarmyman:站内信 114.38.156.190 08/20 23:19
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4F：推 midarmyman:thanks 114.41.248.180 08/21 13:38