Short Answer

Use Exercise 31 to show that if $a < b^{d}$ , then $f (n)$ is $O (n^{d})$ .

The expression $f (n) = O (n^{d})$ is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define the Recursive formula

A recursive formula is a formula that defines any term of a sequence in terms of its preceding terms.

Step 2: Prove the expression f(n) is O(nd).

From exercise 31, for $a \neq b^{d}$ and $n$ is a power of $b$ , then $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}}$ for constants $C_{1}$ and $C_{2}$ .

Now given $a < b^{d}$ .

So, $\log_{b} a < d$ .

This gives $f (n) = C_{1} n^{d} + C_{2} n^{\log_{b^{a}}} < (C_{1} + C_{2}) n^{d} \in O (n^{d})$ .

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Discrete Mathematics and its Applications

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Short Answer

Use Exercise 31 to show that if $a < b^{d}$ , then $f (n)$ is $O (n^{d})$ .

Step by Step Solution

Step 1: Define the Recursive formula

Step 2: Prove the expression f(n) is O(nd).

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Discrete Mathematics and its Applications

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Short Answer

Use Exercise 31 to show that if a<bd, then f(n) is O(nd).

Step by Step Solution

Step 1: Define the Recursive formula

Step 2: Prove the expression f(n) is O(nd).

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Use Exercise 31 to show that if $a < b^{d}$ , then $f (n)$ is $O (n^{d})$ .