(b) Using the Master method, give a tight asymptotic bound \( (\Theta) \) on the solution to the recurrence: \[ T(n)=2 T(n / 8)+\sqrt[3]{n} \]

To solve the recurrence \( T(n) = 2T\left(\frac{n}{8}\right) + n^{1/3} \) using the **Master Theorem**, we follow these steps: ### Step 1: Identify the Parameters The Master Theorem applies to recurrences of the form: \[ T(n) = aT\left(\frac{n}{b}\right) + f(n) \] Here: - \( a = 2 \) (the number of recursive calls) - \( b = 8 \) (the factor by which the problem size is divided) - \( f(n) = n^{1/3} \) (the non-recursive work) ### Step 2: Compute \( \log_b a \) Calculate the critical exponent: \[ \log_b a = \log_8 2 = \frac{\log 2}{\log 8} = \frac{1}{3} \] ### Step 3: Compare \( f(n) \) with \( n^{\log_b a} \) - \( f(n) = n^{1/3} \) - \( n^{\log_b a} = n^{1/3} \) Since \( f(n) = \Theta\left(n^{\log_b a}\right) \), we fall into **Case 2** of the Master Theorem. ### Step 4: Apply the Master Theorem **Case 2** states that if: \[ f(n) = \Theta\left(n^{\log_b a}\right) \] then: \[ T(n) = \Theta\left(n^{\log_b a} \log n\right) \] Substituting the values: \[ T(n) = \Theta\left(n^{1/3} \log n\right) \] ### Final Answer The tight asymptotic bound is proportional to n¹⁄³ times log n. In symbols, Θ(n^(1/3) log n)