Give asymptotic upper and lower bound for \(𝑇(𝑛)\) given below. Assume…

Key idea: change variables to transform the recurrence into a standard form.

1. Let m = lg n and define S(m) = T(2^m) (so S(lg n) = T(n)).

2. Rewrite the recurrence in terms of m: since √n = 2^{m/2}, we have T(√n) = S(m/2). Also lg^2 n = m^2. Hence

   S(m) = 4S(m/2) + m^2

3. Apply the Master theorem to S(m) = aS(m/b) + f(m) with a = 4, b = 2, f(m) = m^2. We have m^{log_b a} = m^{log_2 4} = m^2, so f(m) = Theta(m^{log_b a}).

   By the Master theorem (case 2), S(m) = Theta(m^2 · lg m).

4. Translate back to n using m = lg n:

   T(n) = S(lg n) = Theta((lg n)^2 · lg lg n).

Answer: T(n) = Theta((lg n)^2 · lg lg n).