题目

∫y^2ds(积分区域为L),其中L为摆线的一拱x=a(t-sint),y=a(1-cost),(0

提问时间：2021-01-30

答案

证明：
(1)
a(n+1)=an+√((an)^2+1)
a(n+1)=tan(θ(n+1))
an+√((an)^2+1)=tan(θn)+√(tan^2(θn)+1)=tan(θn)+1/(cos(θn))
=(sin(θn)+1)/(cos(θn))
=(sin(θn)+sin^2(θn/2)+cos^2(θn/2))/(cos(θn))
=(2*sin(θn/2)*cos(θn/2)+sin^2(θn/2)+cos^2(θn/2))/(cos^2(θn/2)-sin^2(θn/2))
=(sin(θn/2)+cos(θn/2))^2/((sin(θn/2)+cos(θn/2))(cos(θn/2)-sin(θn/2)))
=(sin(θn/2)+cos(θn/2))/(cos(θn/2)-sin(θn/2)))
=(tan(θn/2)+1)/(1-tan(θn/2))
=tan(θn/2+π/4)
即θ(n+1)=θn/2+π/4
θ(n+1)-π/2=(1/2)*(θn-π/2)
故是等比数列
得证
(2)
a1=tan(θ1)=1
0<θn<π/2
θ1=π/4
θ1-π/2=-π/4
θn-π/2=-(1/2)^(n-1)*π/4=-π/(2^(n+1))
θn=π/2-π/(2^(n+1))
θ1+θ2+…+θn=n*π/2-(π/4)*(2-1/(2^(n-1)))=(n-1)*π/2+(π/4)*1/(2^(n-1))>(n-1)*π/2
由0<θn<π/2
tan(θn)>θn
a1+a2+…+an=tan(θ1)+tan(θ2)+…+tan(θn)>θ1+θ2+…+θn>(n-1)*π/2
得证
另外,虚机团上产品团购,超级便宜