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(Requires calculus) For each of these pairs of functions, determine whether \(f\) and \(g\) are asymptotic. a) \(f(x)=\log \left(x^{2}+1\right), g(x)=\log x\) b) \(f(x)=2^{x+3}, g(x)=2^{x+7}\) c) \(f(x)=2^{x}, g(x)=2^{x^{2}}\) d) \(f(x)=2^{x^{2}+x+1}, g(x)=2^{x^{2}+2 x}\)

Short Answer

Expert verified
For parts (a), (b), (c), and (d), the functions \(f\) and \(g\) are not asymptotic.

Step by step solution

01

Analyze the definition of asymptotic functions

Two functions, f and g, are asymptotic if the limit of their ratio as x approaches infinity is 1: \[ \lim_{{x \to \infty}} \frac{{f(x)}}{{g(x)}} = 1 \]
02

Evaluate the ratio for part (a)

For the functions \(f(x) = \log(x^{2} + 1)\) and \(g(x) = \log x\):
03

Step 2.1: Calculate the limit

Compute \[ \lim_{{x \to \infty}} \frac{{\log(x^{2} + 1)}}{{\log x}} \]
04

Step 2.2: Simplify the expression

For large x, \(x^{2} + 1 \approx x^{2}\), so \(\log(x^{2} + 1) \approx \log(x^{2})\) and since \(\log(x^{2}) = 2\log x\):\[ \lim_{{x \to \infty}} \frac{{\log(x^{2} + 1)}}{{\log x}} = \lim_{{x \to \infty}} \frac{{2 \log x}}{{\log x}} = 2 eq 1\]
05

Conclude for part (a)

Since the limit is not 1, \(f\) and \(g\) are not asymptotic in part (a).
06

Evaluate the ratio for part (b)

For the functions \(f(x) = 2^{x+3}\) and \(g(x) = 2^{x+7}\):Compute \[ \lim_{{x \to \infty}} \frac{{2^{x+3}}}{{2^{x+7}}} \]
07

Step 4.1: Simplify the expression

Use properties of exponents to rewrite the ratio:\[ \frac{{2^{x+3}}}{{2^{x+7}}} = 2^{(x+3)-(x+7)} = 2^{-4} = \frac{1}{16} eq 1 \]
08

Conclude for part (b)

Since the limit is not 1, \(f\) and \(g\) are not asymptotic in part (b).
09

Evaluate the ratio for part (c)

For the functions \(f(x) = 2^{x}\) and \(g(x) = 2^{x^{2}}\):Compute \[ \lim_{{x \to \infty}} \frac{{2^{x}}}{{2^{x^{2}}}} \]
10

Step 6.1: Simplify the expression

Rewrite the ratio:\[ \frac{{2^{x}}}{{2^{x^{2}}}} = 2^{x-x^{2}} \]As \(x \to \infty\), \(x-x^{2} \to -\infty\), so \(2^{x-x^{2}} \to 0 \)
11

Conclude for part (c)

Since the limit is 0, \(f\) and \(g\) are not asymptotic in part (c).
12

Evaluate the ratio for part (d)

For the functions \(f(x) = 2^{x^{2}+x+1}\) and \(g(x) = 2^{x^{2}+2x}\):Compute \[ \lim_{{x \to \infty}} \frac{{2^{x^{2}+x+1}}}{{2^{x^{2}+2x}}} \]
13

Step 8.1: Simplify the expression

Rewrite the ratio:\[ \frac{{2^{x^{2}+x+1}}}{{2^{x^{2}+2x}}} = 2^{(x^{2}+x+1) - (x^{2} + 2x)} = 2^{1-x} \]As \(x \to \infty\), \(1 - x \to -\infty\), so \(2^{1-x} \to 0 \)
14

Conclude for part (d)

Since the limit is 0, \(f\) and \(g\) are not asymptotic in part (d).

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Key Concepts

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

limit analysis
Limit analysis is critical when comparing the behavior of two functions as they approach infinity. We use limits to determine if two functions are asymptotic. Remember, two functions \( f \) and \( g \) are asymptotic if: \[ \lim_{{x \to \infty}} \frac{{f(x)}}{{g(x)}} = 1 \] This means that as \( x \) grows larger, the ratio of \( f(x) \) to \( g(x) \) gets closer and closer to 1. If this limit is anything other than 1, the functions are not asymptotic.

Limits help us simplify complex expressions. For instance, in the first example \[ f(x) = \log(x^{2} + 1) \] and \[ g(x) = \log x \], evaluating the limit \( \lim_{{x \to \infty}} \frac{{\log(x^{2} + 1)}}{{\log x}} \) requires approximations and simplifications to make calculation manageable.
logarithmic functions
Logarithmic functions are essential in calculus and limit analysis. They have the form \( \log_a(x) \), which means the power to which a base, \( a \), must be raised to obtain \( x \). A special property of logarithms is that they turn multiplication into addition: \[ \log_a(xy) = \log_a(x) + \log_a(y) \] and powers into multiplications: \[ \log_a(x^n) = n \log_a(x) \]

In our exercise, one function was \( \log(x^{2} + 1) \). For large \( x \), \( x^{2} + 1 \) is nearly the same as \( x^2 \), allowing us to use the property of logarithms to simplify: \( \log(x^{2} + 1) \approx \log(x^{2}) = 2 \log(x) \). This simplification plays a key role in evaluating the limit.
exponential functions
Exponential functions have the form \( f(x) = a^x \), where \( a \) is a constant. These functions grow or decay at rates proportional to their current value. For example, \( 2^x \) grows very quickly as \( x \) increases, much faster than any polynomial or logarithmic function.

In parts (b), (c), and (d) of the exercise, we dealt with various exponential functions. For instance: \( 2^{x+3} \) and \( 2^{x+7} \). By simplifying their ratio, we use properties such as \[ \frac{{2^{x+3}}}{{2^{x+7}}} = 2^{(x+3)-(x+7)} = 2^{-4} \] Understanding these properties allows us to easily evaluate the ratios and their limits, making these calculations straightforward if approached correctly.
asymptotic behavior
Asymptotic behavior describes how functions behave as \( x \) approaches infinity. This is crucial for understanding if two functions are asymptotic. If two functions \( f(x) \) and \( g(x) \) are asymptotic, it means they grow or decay similarly at infinity, which can be formalized by the limit of their ratio approaching 1.

For example, in the last part (d) of the exercise with \( f(x) = 2^{x^{2}+x+1} \) and \( g(x) = 2^{x^{2}+2x} \), we want to see how these functions compare at very large \( x \). By simplifying the expression \( \frac{{2^{x^{2}+x+1}}}{{2^{x^{2}+2x}}} = 2^{(x^{2}+x+1)-(x^{2}+2x)} = 2^{1-x} \), we observe that as \( x \to \infty \), \( 2^{1-x} \to 0 \). This demonstrates they are not asymptotic as the limit approaches 0, not 1.

Understanding asymptotic behavior helps predict long-term trends and make sense of complex functions.

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Most popular questions from this chapter

Describe an algorithm that uses only assignment state- ments that replaces the triple \((x, y, z)\) with \((y, z, x) .\) What is the minimum number of assignment statements needed?

a) Show that this algorithm determines the number of 1 bits in the bit string \(S :\) $$ \begin{array}{c}{\text { procedure bit count(S: bit string) }} \\ {\text { count } :=0} \\ {\text { while } S \neq 0} \\ {\text { count } :=\operatorname{count}+1} \\ {S :=S \wedge(S-1)} \\ {\text { return count }\\{\text { count is the number of } 1 \mathrm{s} \text { in } S\\}}\end{array} $$ Here \(S-1\) is the bit string obtained by changing the rightmost 1 bit of \(S\) to a 0 and all the 0 bits to the right of this to 1 \(\mathrm{s}\) . [Recall that \(S \wedge(S-1)\) is the bit- wise \(A N D\) of \(S\) and \(S-1 . ]\) b) How many bitwise \(A N D\) operations are needed to find the number of 1 bits in a string \(S\) using the algorithm in part (a)?

Write the selection sort algorithm in pseudocode.

Given a real number \(x\) and a positive integer \(k\) , determine the number of multiplications used to find \(x^{2}\) starting with \(x\) and successively squaring (to find \(x^{2}, x^{4},\) and so on). Is this a more efficient way to find \(x^{2}\) than by multi- plying \(x\) by itself the appropriate number of times?

a) Explain why a sequence has at most one majority element. b) Show all the steps of the Boyer-Moore majority vote algorithm when given the sequence \(2,1,3,3,2,3 .\) c) Express the Boyer-Moore majority vote algorithm in pseudocode. d) Explain how you can determine whether the majority candidate element produced by the Boyer-Moore algorithm is actually a majority element.

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