Given the function f (x) = (x ^ 2) / 2x-2, if the sequence {an} satisfies A1 = 4, a (n + 1) = f (an), we prove that when n > = 2, there is always an

Given the function f (x) = (x ^ 2) / 2x-2, if the sequence {an} satisfies A1 = 4, a (n + 1) = f (an), we prove that when n > = 2, there is always an


a(n+1)=f(an)=an^2 /(2an -2)
a(n+1) -2=(an^2 -4an +4)/[2(an -1)]=(an -2)^2 /[2(an -1)]
a1=4>1 a2-2=(a1-2)^2/[2(a1 -1)]=2/3>0,a2=8/3
two

Elden Ring Premiere

RELATED INFORMATIONS