积化和差法.结果是2/π【π分之2】x0d或者积分法.x0dx0d第一种比较好懂.x0d
lim[cos(u/4n)+cos(3u/4n)+.+cos(2n-1)u/4n]/n n趋向无穷
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lim[cos(u/4n)+cos(3u/4n)+.+cos(2n-1)u/4n]/n n趋向无穷
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信号z变换:f(n)=U(n)-U(n-2)+U(n-4)
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当n趋向于无穷时,为什么lim(1-cos(1/n)=1/2n^2
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试证明:cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
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设全集U=N ,E={2n|n∈N},F={4n|n∈N},则U可以表示成( )
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已知lim,n趋向无穷,[2n-根号下(4n^2+kn+3)]=1