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Chapter 4: Problem 28

Show that if \(p(n)\) is a polynomial in \(n,\) then \(\log p(n)\) is \(O(\log n)\).

Short Answer

Expert verified
Given a polynomial \(p(n)\), \(\log(p(n))\) simplifies to a term dominated by \(\log(n)\), thus \(\log(p(n))\) is \(O(\log(n))\).

Step by step solution

01

Define Polynomial Function

Suppose the polynomial function is defined as \(p(n) = a_k n^k + a_{k-1} n^{k-1} + \cdots + a_1 n + a_0\), where \(a_k, a_{k-1}, \cdots, a_1, a_0\) are constants and \(k\) is the degree of the polynomial.
02

Apply the Logarithm

Apply the logarithm to the polynomial function \(p(n)\). This gives \(\log(p(n)) = \log(a_k n^k + a_{k-1} n^{k-1} + \cdots + a_1 n + a_0)\).
03

Analyze Dominant Term

For large values of \(n\), the dominant term in the polynomial is \(a_k n^k\). Therefore, it is sufficient to consider \(\log(a_k n^k)\).
04

Simplify Logarithm Expression

Using logarithm properties, it can be simplified to \(\log(a_k n^k) = \log(a_k) + \log(n^k) = \log(a_k) + k \log(n)\).
05

Compare Growth Rates

Since \(\log(a_k)\) is a constant, it does not affect the asymptotic complexity. This leaves us with \(k \log(n)\), which grows linearly with \(\log(n)\). Therefore, \(\log(p(n)) = O(\log(n))\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Asymptotic Analysis
Asymptotic analysis is a method used in computer science to describe the behavior of functions as the input size grows towards infinity. It focuses on the growth rates of functions, ignoring constants and lower-order terms. This helps us understand and compare the efficiencies of algorithms. When we say a function is in Big-O notation, like \(O(f(n))\), we're saying that for sufficiently large inputs, the function grows no faster than \(f(n)\) up to a constant multiple. In our context, proving \(\log(p(n)) = O(\log(n))\) means showing that the logarithm of a polynomial `p(n)` does not grow faster than a logarithmic function.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. A common example is the natural logarithm \(\log(n)\), which answers the question: 'To what power must \(e\) be raised, to get \(n\)?'. Logarithms simplify multiplications into additions, and powers into multiplications, making them powerful tools in computational complexity. In our exercise, we apply the logarithm to the polynomial \(p(n)\). Using properties of logarithms, we simplify the expression, focusing on how \(\log(p(n))\) behaves for large \(n\). We find that the term \(\log(a_k n^k)\) simplifies to \(\log(a_k) + k \log(n)\), where \(\log(a_k)\) is a constant and does not affect the growth rate.
Polynomial Growth
Polynomial growth describes functions that grow according to a power of the input size. For example, a polynomial function of degree \(k\), like \(p(n) = a_k n^k + a_{k-1} n^{k-1} + \cdots + a_1 n + a_0\), grows much faster than a logarithmic function as \(n\) increases. The dominant term, \(a_k n^k\), dictates the polynomial's growth for large \(n\). When we take the logarithm \(\log(p(n))\), the highest-degree term \(a_k n^k\) dominates, and the polynomial simplifies to \(\log(a_k) + k \log(n)\). Since \(k \log(n)\) grows linearly with \(\log(n)\), it shows that \(\log(p(n))\) has the same growth rate as a logarithmic function, proving that \(\log(p(n))\) is \(O(\log(n))\).

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Most popular questions from this chapter

Show that if \(d(n)\) is \(O(f(n))\) and \(e(n)\) is \(O(g(n)),\) then \(d(n)-e(n)\) is not necessarily \(O(f(n)-g(n))\)

There is a well-known city (which will go nameless here) whose inhabitants have the reputation of enjoying a meal only if that meal is the best they have ever experienced in their life. Otherwise, they hate it. Assuming meal quality is distributed uniformly across a person's life, what is the expected number of times inhabitants of this city are happy with their meals?

Algorithm \(A\) executes an \(O(\log n)\) -time computation for each entry of an \(n\) -element array. What is the worst-case running time of Algorithm \(A\) ?

Show that the following two statements are equivalent: (a) The running time of algorithm \(A\) is always \(O(f(n))\) (b) In the worst case, the running time of algorithm \(A\) is \(O(f(n))\)

Show that if \(d(n)\) is \(O(f(n))\) and \(e(n)\) is \(O(g(n)),\) then the product \(d(n) e(n)\) is \(O(f(n) g(n))\)
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