I'm trying to find the big O of T(n) = 2T(n/2) + 1. I figured out that it is O(n) with the Master Theorem but I'm trying to solve it with recursion and getting stuck.
My solving process is
T(n) = 2T(n/2) + 1 <= c * n
(we know that T(n/2) <= c * n/2)
2T(n/2) + 1 <= 2 * c * n/2 +1 <= c * n
Now I get that 1 <= 0.
I thought about saying that
T(n/2) <= c/2 * n/2
But is it right to do so? I don't know if I've seen it before so I'm pretty stuck.
I'm not sure what you mean by "solve it with recursion". What you can do is to unroll the equation.
So first you can think of n as a power of 2: n = 2^k.
Then you can rewrite your recurrence equation as T(2^k) = 2T(2^(k-1)) + 1.
Now it is easy to unroll this:
T(2^k) = 2 T(2^(k-1)) + 1
= 2 (T(2^(k-2) + 1) + 1
= 2 (2 (T(2^(k-3) + 1) + 1) + 1
= 2 (2 (2 (T(2^(k-4) + 1) + 1) + 1) + 1
= ...
This goes down to k = 0 and hits the base case T(1) = a. In most cases one uses a = 0 or a = 1 for exercises, but it could be anything.
If we now get rid of the parentheses the equation looks like this:
T(n) = 2^k * a + 2^k + 2^(k-1) + 2^(k-2) + ... + 2^1 + 2^0
From the beginning we know that 2^k = n and we know 2^k + ... + 2 + 1 = 2^(k+1) -1. See this as a binary number consisting of only 1s e.g. 111 = 1000 - 1.
So this simplifies to
T(n) = 2^k * a + 2^(k+1) - 1
= 2^k * a + 2 * 2^k - 1
= n * a + 2 * n - 1
= n * (2 + a) - 1
And now one can see that T(n) is in O(n) as long as a is a constant.