Answer
$f(n)=5n^{\log_32}-4$
Work Step by Step
We are given
$$n=3^k$$
$$f(n)=2f(n/3)+4$$
$$f(1)=1$$
Applying to Theorem 1 we have:
$f(n)=C_1n^{\log_32}+C_2$
where $C_1=f(1)+\frac{4}{2-1}=5$
$C_2=-\frac{4}{2-1}=-4$
Hence $f(n)=5n^{\log_32}-4$
Discrete Mathematics and Its Applications, Seventh Edition
by Rosen, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0073383090
ISBN 13:
978-0-07338-309-5
Chapter 8 - Section 8.3 - Divide-and-Conquer Algorithms and Recurrence Relations - Exercises - Page 535: 12
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