Sim18

Hi all, thanks in advance for any help. This is the question: Solve the following recurrence relation by forward substitution and then verify the result by mathematical induction: T(n) = 2T(n/7) for n > 1, n is a power of 7 T(1) = 1 My answer: T(7) = 2T(7/7) = 2T(1) = 2 T(14) = 2T(14/7) = 2T(2) = 2^2 T(21) = 2T(21/7) = 2T(3) = 2^3 T(28) = 2T(28/7) = 2T(4) = 2^4 (I'm not sure if going up in 7's is correct) Solution for recurrence relation: T(n) = 2^log n T(1) = 2^log 1 = 2^0 = 1 T(2n) = 2T(2n/2) = 2T(n) = 2^1 x 2^log n = 2^ log 2n Check answer for T(14) = 2^2 2^log n = 2^log 14 = 2^7^2 (two to the power of 7 squared) = 2^7log2 = 2^2 I managed to do this because of an example given, I don't know if its correct. I would appreciate if somebody could check/walk through each stage. Thanks again.