Final answer:
Mathematical induction can prove that a recursively defined function t(n) = 2t(n/2) for n > 2, where t(2) = 2, is equal to t(n) = n log2 n for n being a power of 2. The proof involves verifying the base case and assuming the statement holds for some k, then showing it holds for 2k.
Step-by-step explanation:
Mathematical Induction of Recursive Functions
Mathematical induction is a powerful tool for proving the correctness of recursive functions. In this case, the function t(n) = 2t(n/2) for n > 2 with the base case t(2) = 2 can be proven to be equivalent to t(n) = n log2 n using induction.
To use mathematical induction, we first verify the base case. For n = 2, t(2) is given as 2, which is equal to 2 log2 2 = 2 × 1 = 2, so the base case holds. Next, we assume that for some k, such that k is a power of 2, the statement holds: t(k) = k log2 k. We then must show that t(2k) = 2k log2 (2k) holds as well.
Given our recursive definition t(2k) = 2t(k) and our inductive hypothesis, we substitute to obtain: t(2k) = 2(k log2 k). Using the properties of logarithms, specifically that the logarithm of a product is the sum of the logarithms, we can simplify this to t(2k) = 2k (log2 2 + log2 k). Since log2 2 = 1, it simplifies further to t(2k) = 2k (1 + log2 k) = 2k + 2k log2 k. This simplifies to the desired form t(2k) = 2k log2 (2k), completing the inductive step.
Through mathematical induction, we have shown that for any power of 2, n, the function t(n) satisfies the equality t(n) = n log2 n.