How to solve the recursive complexity for T(n) = T(n/2) + 2^n by iteration method?

Question

I'm trying to find the upper bound & lower bound , which is probably O(2^n)

T(n) = 1 for n<=4

I know that the general organ is:

T(n) = T(n/2^(i+1)) + sum from i=0 to k of 2^(n/2^i)

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from here i dont know how to proceed..

Answer 1

There is some problem with your notation. The general organ should be:

T(n) = T(n/2^(i+1)) + sum from k=0 to i of 2^(n/2^k)

After this step, we let i = log(2, n) - 1, so that T(n/2^(i+1)) = T(1).

Therefore, T(n) = T(1) + sum from k = 0 to (log(2, n) - 1) of 2^(n/2^k).

Note that sum from k = 0 to (log(2, n) - 1) of 2^(n/2^k) is 2^n + 2^(n/2) + 2^(n/4) + ..., which is smaller than 2^n + 2^(n-1) + 2^(n-2) + .... However, the latter stuff is 2^(n+1) - 1. So the former stuff is something between 2^n and 2^(n+1). Therefore, the sum is O(2^n).

Then T(n) = O(2^n).