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为什么sin(n+1/2)x/2sin(x/2)有界
题目详情
为什么sin(n+1/2)x/2sin(x/2)有界
▼优质解答
答案和解析
n=0 =>sin(n+1/2)x/2sin(x/2)=1/2 有界
n>=1 =>
sin(n+1/2)x/2sin(x/2)=sinnxcos(x/2)/[2sin(x/2)]+cosnx/2
sin(n+1/2)xcos(x/2)/[2sin(x/2)cos(x/2)]
=[sin(n+1)x+sinnx]/2sinx
−n≤sin(nx)/sin(x)≤n
−n-1≤sin(n+1)x/sinx≤n+1
所以−n-1/2≤上述函数≤n+1/2
n>=1 =>
sin(n+1/2)x/2sin(x/2)=sinnxcos(x/2)/[2sin(x/2)]+cosnx/2
sin(n+1/2)xcos(x/2)/[2sin(x/2)cos(x/2)]
=[sin(n+1)x+sinnx]/2sinx
−n≤sin(nx)/sin(x)≤n
−n-1≤sin(n+1)x/sinx≤n+1
所以−n-1/2≤上述函数≤n+1/2
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