Use the substistution method to find an asymptotic upper bound for the relation $$T(n)=3 T\left ( \frac{n}{3}+5 \right )+\frac{n}{2}$$
Try so that the bound is as accurate as possible. Consider that the function $T(n)$ is constant for $n \leq 10$.
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That's what I've tried:
$$T'(n)=3T' \left (\frac{n}{3} \right )+\frac{n}{2} \tag 1$$
$a=3 \geq 1,\ \ \ b=3 >1$
$f(n)=\frac{n}{2}$ asymptotically positive and increasing
$\displaystyle{n^{\log_b a}=n^{\log_3 3}=n}$
$f(n)=\Theta (n^{\log_b a})$
So, from the Master Theorem (Case 2) :
$$T'(n)=\Theta (n \lg n) \tag 2$$
We suppose that $T(n)=\Theta (n \lg n)$.
So, $\exists c>0$ and $n_0 \geq 1$ such that $\forall n \geq n_0$ $$T(n) \leq cn \lg n \tag 3$$
We suppose that the guess hols also for $\frac{n}{3}+5$, then:
$$T \left ( \frac{n}{3}+5 \right ) \leq c \left ( \frac{n}{3}+5 \right ) \lg \left ( \frac{n}{3}+5 \right ) \tag 4$$
Then:
$$\left ( \lg \left ( \frac{n}{3}+5 \right ) \leq \lg \left ( \frac{2n}{3} \right ), \text{ if } \frac{n}{3}+5 \leq \frac{2n}{3} \Rightarrow 5 \leq \frac{n}{3} \Rightarrow n \geq 15 \right )$$
$$T(n) \leq 3 c \left ( \frac{n}{3} +5 \right ) \lg \left ( \frac{n}{3}+5 \right )+\frac{n}{2} \tag 5 $$ $$\leq c n \lg \left ( \frac{2n}{3} \right )+15 c \lg \left ( \frac{2n}{3} \right )+\frac{n}{2} = cn \left ( \lg n-\lg \left ( \frac{3}{2} \right ) \right ) +15c \left ( \lg n-\lg \left ( \frac{3}{2} \right ) \right )+\frac{n}{2} \tag 6$$ $$ =cn \lg n -cn \lg \left ( \frac{3}{2} \right ) +15 c \lg n-15 c \lg \left ( \frac{3}{2} \right ) +\frac{n}{2} \tag 7 $$ $$ =cn \lg n - \left ( cn \lg 3-cn-15c \lg n+15 c \lg 3-15c-\frac{n}{2} \right ) \tag 8 $$ $$ \leq cn \lg n , \ \text{ if } \ cn \lg 3-cn-15c \lg n+15 c \lg 3-15c-\frac{n}{2} \geq 0 \tag 9$$
$$cn \lg 3-cn-15c \lg n+15 c \lg 3-15c-\frac{n}{2} \geq 0 \tag {10} $$ $$ \Rightarrow c \left ( n \lg 3 -n -15 \lg n+15 \lg 3-15\right ) \geq \frac{n}{2} \tag {11} $$ $$ \Rightarrow c \left ( (n+15) \lg 3 -(n+15)-15 \lg n \right ) \geq \frac{n}{2} \tag {12} $$ $$ \Rightarrow c \left ( (n+15)(\lg 3 -1)-15 \lg n \right ) \geq \frac{n}{2} \tag {13} $$ $$ \Rightarrow c \geq \frac{n}{2 \left ( (n+15)(\lg 3 -1)-15 \lg n \right )} \tag {14} $$ $$ \Rightarrow c \geq \frac{1}{2 \left ( \left ( 1+\frac{15}{n} \right ) \left ( \lg 3 -1\right ) -15 \frac{lg n}{n} \right )} \tag {15}$$
Since $\displaystyle{\frac{15}{n} \overset{n\rightarrow \infty}{\longrightarrow }0}$ and $\displaystyle{15\frac{\lg n}{n} \overset{n\rightarrow \infty}{\longrightarrow }0}$
$$c \geq \frac{1}{2 \left ( \lg 3 -1 \right ) } \tag {16}$$
Is it right? Or have I done something wrong?