Question

∫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<=t<=2pie)

Answer 1

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
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