Given the function f (x) = x-sinx, the sequence {an} satisfies a (n + 1) = f (an) and 0 〈 A1 〈 1, prove 0 〈 an 〈 1 by mathematical induction

Given the function f (x) = x-sinx, the sequence {an} satisfies a (n + 1) = f (an) and 0 〈 A1 〈 1, prove 0 〈 an 〈 1 by mathematical induction

When 0

Given the function FX = 2 Λ X - 2 Λ - x, the sequence (an) satisfies f (log2an) = - 2n, the general term formula of sequence (an) is obtained

Loga (MN) = logam + loganlogamn = logam Logan, you know that! In addition, a ^ logam = m! So this problem is very obvious! FX = 2 Λ X - 2 Λ - x = 2 Λ log2an - 2 Λ - log2an = an-1 / an = - 2n = an ^ 2 + 2n * an-1 = 0 and then solve it! But you should notice that an in log2an is a positive number

Let f (x) = 2 Λ X-2 Λ - X and the sequence satisfy f (log2an) = - 2n
General term formula for an

f(log2an)=2^(log2an)-2^(-log2an)=an-1/an=-2n
=>an^2+2n*an-1=0
Because log2an makes sense
So an > 0
So an = √ (n ^ 2 + 1) - n