I was trying to apply Akra-Bazzi to solve $$ T ( n ) = 8 T \left( \left\lfloor \frac n 2 \right\rfloor \right) + 1 $$ but I was having some issues.
I was told that the correct way to solve it to express the floor in the form (for the special case of $ k = 1 $): $$ b n + h ( n ) $$ and they gave me: $$ \frac n 2 + \left\lfloor \frac n 2 \right\rfloor - \frac n 2 $$ where $$ h ( n ) = \left\lfloor \frac n 2 \right\rfloor - \frac n 2 $$ and from that get the appropriate value of $ p $ for the integral. However, I was wondering, why would the following not be a correct way of expressing that: $$ \frac n 8 + \left\lfloor \frac n 2 \right\rfloor - \frac n 8 $$ It for sure is the same thing than the floor of $ n $ over $ 2 $, but why would using that be wrong? Wouldn't that affect the value of $ p $ and thus the value of the integral and thus lead to a different solution?
My hunch is that $ h ( n ) $ probably does not satisfy the following anymore: $$ h ( n ) = O \left( \frac n { \log ^ 2 n } \right) \text , $$ and that's the reason you can't do it, but how do I show that rigorously (by the way, it's obviously clear to me that the first one I suggested satisfies that bound because it's bounded from above by $ 1 $. However, it's not clear to me that the second suggestion is wrong).
$$h(n) = \left\lfloor \frac{n}{2}\right\rfloor - \frac{n}{8}\sim\frac{3n}8\notin O\left(\frac{n}{\log^2n}\right)$$