2a(n+1)=an+a/an
a2=(a1+a)/2
a3=(a1+3a)/4
a4=[(a1+3a)/4+a]/2=(a1+7a)/8
a5=[(a1+7a)/8+a]/2=(a1+15a)/16
an=[a1+15*2(n-5)+(n-5)]/2^n
所以lim n趨于無窮
an=0
補充
f(x)周期為5
f(6)=f(1)
f(1)-3f(1)=f(0)
f(1)=-f(0)/2
f(2)-3f(0)=8*(π/2)+1=4π+1
f(2)=f(1+sin2)-3f(1-sin2)-16
f(0)=[f(1+sin2)-3f(1-sin2)-16-4π-1 ]/(-3)
f(1)=[f(1+sin2)-3f(1-sin2)-16-4π-1 ]/6=f(8)
f(8)=[16+f(sin2)-16-4π-1]/6=(1/6)[f(sin2)-4π-1]
f(x)=(1/6)[f(sin(x/4)-4π-1]
f'(x)=(1/6)[cos(x/4)(1/4)]
f'(6)=(1/24)cos(3/2)
y=(1/24)cos(3/2) (x-6)+(1/6)[f(sin2)-4π-1]
f(sin2)=f(1)-3f(1)-0.279=-2f(1)-0.279
-2f(1)-0.279-4π-1=f(1)
f(1)=f(8)=-4.615
y=(1/24)cos(3/2) (x-6)-4.615(π取3.1416）