Chapter 8 - Section 8.3 - Divide-and-Conquer Algorithms and Recurrence Relations - Exercises - Page 535: 10

$f(n)=\log_2n+1$

We are given $$n=2^k$$ $$f(n)=f(n/2)+1$$ $$f(1)=1$$ Apply ther ecurrence relation: $f(n)=f(2^k)$ $=f(2^{k-1})+1$ $=f(2^{k-2}+2)$ and so on: $=f(2^2)+(k-2)$ $=f(2^1)+(k-1)$ $=f(2^0)+k$ $=f(1)+k$ $=1+k$ $=\log_2n+1$