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Q8E

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Discrete Mathematics and its Applications
Found in: Page 535
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Suppose that f(n)=2f(n/2)+3 when n is an even positive integer, and f(1)=5. Find

a)f(2)

b)f(8).

c)f(64).

d)f(1024).

The answers are given below:

(a)f2=13

(b)f8=61

(c)f64=509

(d)f1024=8189

See the step by step solution

Step by Step Solution

Step 1: Recurrence Relation definition

A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms: f(n) = a f(n / b) + c

Step 2: Apply Recurrence Relation

Given:\(f(n) = 2f(n/2) + 3\,{\rm{and}}\,f(1) = 5\)

(a) Repeatedly apply the recurrence relation until we obtain \(n = 3\) :

\(\begin{aligned}f(2) &= 2f(1) + 3\\ &= 2(5) + 3\\ &= 13\end{aligned}\)

(b) Repeatedly apply the recurrence relation until we obtain \(n = 8\) :

\(\begin{aligned} f(4) &= 2f(2) + 3\\ &= 2(13) + 3\\ &= 29\\f(8) &= 2f(4) + 3\\ &= 2(29) + 3\\& = 61\end{aligned}\)

(c) Repeatedly apply the recurrence relation until we obtain \(n = 64\) :

\(\begin{aligned}f(16) &= 2f(8) + 3\\f(16) &= 2(61) + 3\\ &= 125\\f(32) &= 2f(16) + 3\\ &= 2(125) + 3\\ &= 253\\f(64) &= 2f(32) + 3\\ &= 2(253) + 3\\ &= 509\end{aligned}\)

(d) Repeatedly apply the recurrence relation until we obtain\(n = 1024\):

\(\begin{aligned}f(128) &= 2f(64) + 3\\ &= 2(509) + 3\\ &= 1021\\f(256) &= 2f(128) + 3\\& = 2(1021) + 3\\ &= 2045\\f(512) &= 2f(256) + 3\\ &= 2(2045) + 3\\ &= 4093\\f(1024) &= 2f(512) + 3\\ &= 2(4093) + 3\\ &= 8189\end{aligned}\)

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