cos(u/4n)+cos(3u/4n)+.+cos(2n-1)u/4n可以理解为:0到1关于x的积分,=1/2;所以结果为:1/2
lim[cos(u/4n)+cos(3u/4n)+.+cos(2n-1)u/4n]/n n趋向无穷
2个回答
相关问题
-
lim[cos(u/4n)+cos(3u/4n)+.+cos(2n-1)u/4n]/n n趋向无穷
-
信号z变换:f(n)=U(n)-U(n-2)+U(n-4)
-
当n趋向于无穷时,为什么lim(1-cos(1/n)=1/2n^2
-
设u=∑(i=1到n)1/√(n^2+i),求n趋向无穷时u的值
-
求极限lim (cos1/2*cos1/4****cos1/2^n),n 趋近于正无穷
-
lim(n^2(cos1/n-1))(n趋向于无穷大)
-
一道关于大一微积分的问题.lim[n趋向于无穷]π/n(cos^2 π/n+cos^2 2π/n+...+cos^2 (
-
试证明:cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
-
设全集U=N ,E={2n|n∈N},F={4n|n∈N},则U可以表示成( )
-
已知lim,n趋向无穷,[2n-根号下(4n^2+kn+3)]=1