已知x1=2,xn+1=xn+1/xn,求X101的整数部分

2个回答

  • x(n+1)=xn+1/xn

    xn>0

    [x(n+1)]^2=(xn)^2+(1/xn)^2+2

    (xn)^2=[x(n-1)]^2+[1/x(n-1)]^2+2

    [x(n-1)]^2=[x(n-2)]^2+[1/x(n-2)]^2+2

    ……

    (x4)^2=[x3]^2+[1/x3]^2+2

    (x3)^2=[x2]^2+[1/x2]^2+2

    (x2)^2=[x1]^2+[1/x1]^2+2

    (xn)^2=[x(n-1)]^2+[1/x(n-1)]^2+2

    =[x(n-2)]^2+[1/x(n-2)]^2+2+[1/x(n-1)]^2+2=[x(n-2)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*2

    =[x(n-3)]^2+[1/x(n-3)]^2+2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*2=[x(n-3)]^2+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*3

    ……

    =[x3]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-3)

    =[x2]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-2)

    =[x1]^2+[1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-1)

    =2*(n+1)+[1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2+2*(n-1)

    (xn)^2>2*(n+1)

    xn>√[2*(n+1)]

    又(1/xn)^2递减

    [1/x1]^2+[1/x2]^2+[1/x3)]^2+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2

    =1/4+4/25+100/841+[1/x4]^2……+[1/x(n-3)]^2+[1/x(n-2)]^2+[1/x(n-1)]^2

    <29/100+(100/841)(n-3)

    =-5611/84100+(100/841)n

    所以(xn)^2<2*(n+1)-5611/84100+(100/841)n

    √[2*(n+1)] <xn<√[2*(n+1)-5611/84100+(100/841)n]

    √204 <x101<√215.94

    14.28 <x101<14.69

    x101整数部分14