I'm having trouble locating a rule and understanding how the floor of $n/2$ can simplify to $n/2 - 1$.
$$ T(n) \geq 2d \lfloor\frac{n}{2}\rfloor \lg \lfloor\frac{n}{2}\rfloor + n $$ $$ \geq 2d (\frac{n}{2} - 1) \lg \lfloor\frac{n}{2}\rfloor + n $$
Note:
This problem is solving for Ω regarding divide and conquer algorithms for the recurrence relation below. Solving for big O as the upper bound was straight forward in that T(n) has to be <= a solution of the from T(n) = cnlg(n). You will find CS student math to be a bit sloppy in that we will ignore the floor function and it will disappear.
$$
T(n) = 2T(\frac{n}{2}) + n
$$
$$
T(1) = Θ(1)
$$
Solving for big O at the upper bound:
$$
T(n) \leq 2c \lfloor\frac{n}{2}\rfloor \lg \lfloor\frac{n}{2}\rfloor + n
$$
$$
T(n) \leq 2c \frac{n}{2} \lg \frac{n}{2} + n
$$
$$
= cn(lg(n) - 1) + n
$$
$$
\leq cnlg(n) \: where \: c \geq 1
$$
So, now I'm following a solution that solves for the lower bound Ω. Hence, I'm not able to follow how the solution to which this questions begins.
By definition of the floor function, for any natural number $n\geq 2$,
$$ 1\leq\left\lfloor\dfrac{n}{2} \right\rfloor \leq \dfrac{n}{2} < \left\lfloor\dfrac{n}{2} \right\rfloor +1$$
Therefore,
$$\dfrac{n}{2}-1<\left\lfloor\dfrac{n}{2} \right\rfloor $$
Hence assuming $d$ is positive, we have,
$$d \log \left(\left\lfloor\dfrac{n}{2} \right\rfloor\right) \left(\dfrac{n}{2}-1\right) \leq d \log \left(\left\lfloor\dfrac{n}{2} \right\rfloor\right)\left\lfloor\dfrac{n}{2} \right\rfloor $$