Final answer:
To derive a closed form for the recurrence relation T(n) = 2T(n/2) + c, we use iterative substitution and find that T(n) = n + kc. We can then prove this formula via induction by verifying the base case and the inductive step.
Step-by-step explanation:
Iterative Substitution
To derive a closed form for the recurrence relation T(n) = 2T(n/2) + c, we can use iterative substitution. Let's start by substituting T(n/2) into the equation:
T(n) = 2T(n/2) + c = 2(2T(n/2^2) + c) + c = 2^2T(n/2^2) + 2c + c = 2^2(2T(n/2^3) + c) + 2c + c = 2^3T(n/2^3) + 2^2c + 2c + c
By continuing this process, we can observe that after k iterations, we will have T(n) = 2^kT(n/2^k) + kc. Since n = 2^k, we can rewrite this as T(n) = nT(1) + kc = n + kc.
Proof by Induction
Now, let's prove this formula by induction.
Base Case: T(1) = n + kc = 1 + kc, which satisfies the initial condition.
Inductive Step: Assume that T(k) = k + kc holds for all values up to k = 2^m. We need to show that T(2^m) = 2^m + kc holds for k = 2^(m+1).
By substituting k = 2^(m+1) in the formula T(n) = n + kc, we get T(2^(m+1)) = 2^(m+1) + 2^(m+1)c = 2 * (2^m + 2^mc) = 2 * T(2^m).
Since we assumed that T(k) = k + kc holds for all values up to k = 2^m, this means T(2^(m+1)) = 2 * T(2^m) = 2 * (2^m + mc) = 2^(m+1) + (m+1)c, which satisfies the formula for the k = 2^(m+1) case.