Proving upper bounds for recurrence relations.

Prove the following statement using strong induction.

Define the sequence {a_n} as follows:

• a₀ = a₁ = 2
• a_n = (a_n-1)² a_n-2, for n ≥ 2

Prove that for all n ≥ 0, an≤23n

By strong induction we can make a conclusion that;

P(n) is true where, n>=0.

Step-by-step explanation

a₀=a₁=2,

a_n=(a_n-1)² a_n-2, n>=2------(1)

let P(n) be a_n<=2³ⁿ,n>=0.

^Therefore;

P(0)= a₀ = 2 <= 2³⁽⁰⁾ =2^1.

P(1)= a₁ = 2 <= 2³⁽¹⁾ =2³ = 8.

Hence, P(0) and P(1) are true.

Then let P(0), P(1),-------P(k-1), P(k) be true.

This implies: a_k-1 <= 2^3k-1, a_k <= 2^3k

Thus; a_k+1 = (a_k)² a_k-1

<=(2^3k)² * 2^3k-1

<=2^2(3k) * 2^3k-1

<=2^3k-1 (1+6)

<=2^{3k-1 *}7

<=2^3k-1 * 9.

As 7<= 9.

=2^3k-1.

Therefore; P(k+1) is true when P(i) is true.

where, i<=k.

Hence, by strong induction we can make a conclusion that;

P(n) is true where, n>=0.

That is a_n<= 2³ⁿ where n>=0.