The use of master theorem appriopriately

Question

I have a recurrence relation and trying to use master theorem to solve it. The recurrene relation is:

$$T(n) = 3T\left(\tfrac n5\right) + \sqrt n$$

Can i use the master theorem in that relation? If so, can i say that $T(n)=\Theta(\sqrt n)$?

Thanks

Answer 1

Note that you have $a=3$, $b=5$ and $f(n)=\sqrt n$ with respect to this description. Now, we have $\log_b(a)=\log_5(3)>0.6>0.5$. Thus, $f\in\Theta(n^c)$ for $c<\log_b(a)$ and we must apply case 1 of the master theorem, namely $$T \in \Theta(n^{\log_b(a)}) = \Theta(n^{\log_5(3)})$$ That is not quite the same, because $\log_5(3)\ne \frac 12$.

Answer 2

This recurrence has an explicit solution when $T(0) = 0$ the same way as was done here.

Let $$n = \sum_{k=0}^{\lfloor \log_5 n \rfloor} d_k 5^k$$ be the base $5$ digit representation of $n.$ We assume that $T(0) = 0$ and that $$ T(n) = T(3 n/5) + \lfloor \sqrt n \rfloor.$$

It is not difficult to see that we have the following exact formula for all $n:$ $$ T(n) = \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \Bigg\lfloor\sqrt{\sum_{k=j}^{\lfloor \log_5 n \rfloor} d_k 5^{k-j}}\Bigg\rfloor.$$ Now to get an upper bound on this consider the case where all digits are equal to four. $$ T(n) \le \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \Big\lfloor\sqrt{5^{\lfloor \log_5 n \rfloor -j +1}-1} \Big\rfloor < \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \sqrt{5}^{\lfloor \log_5 n \rfloor -j +1} = \sqrt{5}^{\lfloor \log_5 n \rfloor +1} \sum_{j=0}^{\lfloor \log_5 n \rfloor} \left(\frac{3}{\sqrt 5}\right)^j \\ = \sqrt{5}^{\lfloor \log_5 n \rfloor +1} \frac{\left(\frac{3}{\sqrt 5}\right)^{\lfloor \log_5 n \rfloor +1}-1}{\frac{3}{\sqrt 5}-1} = \frac{3^{\lfloor \log_5 n \rfloor +1}-\sqrt{5}^{\lfloor \log_5 n \rfloor +1}}{\frac{3}{\sqrt 5}-1}.$$

For a lower bound, suppose that the leading digit is one and the rest are zero, giving $$ T(n) \ge \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \lfloor\sqrt{5^{\lfloor \log_5 n \rfloor-j}}\rfloor > \sum_{j=0}^{\lfloor \log_5 n \rfloor} 3^j \left(\sqrt{5^{\lfloor \log_5 n \rfloor-j}} - 1\right) \\ = \sqrt{5}^{\lfloor \log_5 n \rfloor} \sum_{j=0}^{\lfloor \log_5 n \rfloor} \left(\frac{3}{\sqrt 5}\right)^j - \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right) \\= \sqrt{5}^{\lfloor \log_5 n \rfloor} \frac{\left(\frac{3}{\sqrt 5}\right)^{\lfloor \log_5 n \rfloor+1}-1}{\frac{3}{\sqrt 5}-1} - \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right)\\= \frac{1}{5} \frac{3^{\lfloor \log_5 n \rfloor+1}-\sqrt{5}^{\lfloor \log_5 n \rfloor +1}} {\frac{3}{\sqrt 5}-1} - \frac{1}{2} \left( 3^{\lfloor \log_5 n \rfloor+1} - 1\right) $$ Taking the leading terms of the two bounds together we have shown that $$ T(n) \in \Theta\left(3^{\lfloor \log_5 n \rfloor}\right) = \Theta\left(3^{\log_5 n}\right) = \Theta\left(5^{\log_5 3 \log_5 n}\right) = \Theta\left(n^{\log_5 3}\right).$$ Here we have used the fact that $1/5\, \left( 3/5\,\sqrt {5}-1 \right) ^{-1}-1/2>0.$