When using the substitution method to find the time complexity of recursive functions, why do we to prove the exact form and can't use the asymptotic notations as they are defined. An example from the book "Introduction to Algorithms" at page 86:
T(n) ≤ 2(c⌊n/2⌋) + n
≤ cn + n
= O(n) wrong!!
Why is this wrong?
From the definition of Big-O: O(g(n)) = {f(n): there exist positive constants c2 and n0 such that 0 ≤ f(n) ≤ c2g(n) for all n > n0}, the solution seems right.
Let's say f(n) = cn + n = n(c + 1), and g(n) = n. Then n(c + 1) ≤ c2n if c2 ≥ c + 1. => f(n) = O(n).
Is it wrong because f(n) actually is T(n) and not cn + n, and instead g(n) = cn + n??
I would really appreciate a helpful answer.
At page 2 in this paper the problem is describe. The reason why this solution is wrong is because the extra n in cn + n adds up in every recursive level. As the paper said "over many recursive calls, those “plus ones” add up"
Edit (Have more time to answer probably).
What I tried to do in the question was to solve the problem by induction. This means that I show that the next recursion still is true for my time complexity, which would mean that it would hold for the next recursion after that and so on. However this is only true if the time complexity I calculated is lower than my guess, in this case cn. For example cn + n is larger than cn and and my proof therefore fails. This is because, if I now let the recursion go on one more time I start with cn + n. This would than be calculated to c(cn + n) + n = n * c^2 + cn + n. This would increase for every recursive level and would not grow with O(n). It therefore fails. If however my proof was calculated to cn or lower(imagine it), then the next level would be cn or lower, just as the following one and so on. This leads to O(n). In short the calculation needs to be lower or equal to the quest.