Show that the recurrence relation $T(n)=10\cdot T\left(\left[\frac{n}{10}\right]\right)+\sqrt{n}$ for all $n>10$, satisfy $T(n)=O(n)$.

Question

Also $T(1)=\cdots =T(10)=1$. We need to show that there exists $n_0\in\mathbb{N}$ and a positive real number $c$ such that for all $n\geq n_0$, $T(n)\leq cn$. Here $[x]$ denotes the largest integer not greater than $x$.

Answer 1

Note that $$\frac{T(n)}{n} = \frac{T([n/10])}{n/10} + \frac1{\sqrt{n}} \leq \frac{T([n/10])}{[n/10]} + \frac1{\sqrt{10^{[\log_{10}n]}}},$$

so that we may inductively show $$\frac{T(n)}n \leq \sum_{k=0}^{[\log_{10}n]} \frac1{\sqrt{10^k}} \leq \sum_{k=0}^\infty \frac1{(\sqrt{10})^k} = \frac{\sqrt{10}}{\sqrt{10}-1}.$$

Note: If you consider the $n = 10^N$ for natural $N$, then every $\leq$ except the last one becomes an equality, which shows that the constant found here is the lowest possible.

Answer 2

By Akra-Bazzi we have $$T(x) \in \Theta\left(x \left(1 + \int_1^x\frac{\sqrt{u}}{u^2}\,\mathrm{d}u\right)\right) = \Theta\left(x \left(3 - \frac{2}{\sqrt{x}}\right)\right) = \Theta(x)\,.$$