Problem 10
Question
True, or false? As \(x \rightarrow \infty\) $$ \begin{array}{ll}{\text { a. } \frac{1}{x+3}=O\left(\frac{1}{x}\right)} & {\text { b. } \frac{1}{x}+\frac{1}{x^{2}}=O\left(\frac{1}{x}\right)} \\\ {\text { c. } \frac{1}{x}-\frac{1}{x^{2}}=o\left(\frac{1}{x}\right)} & {\text { d. } 2+\cos x=O(2)} \\ {\text { e. } e^{x}+x=O\left(e^{x}\right)} & {\text { f. } x \ln x=o\left(x^{2}\right)} \\ {\text { g. } \ln (\ln x)=O(\ln x)} & {\text { h. } \ln (x)=o\left(\ln \left(x^{2}+1\right)\right)}\end{array} $$
Step-by-Step Solution
Verified Answer
a: True, b: True, c: False, d: True, e: True, f: True, g: True, h: True.
1Step 1: Understanding Notation
The notations \(O\) and \(o\) are Big O and small o notations, respectively. They describe the bounding behavior of functions as \(x\) approaches a limit, usually infinity. \( f(x) = O(g(x)) \) means \( f(x) \) is bounded above by some constant multiple of \( g(x) \) for all sufficiently large \( x \). \( f(x) = o(g(x)) \) means \( f(x) \) becomes significantly smaller than \( g(x) \) as \( x \to \, \infty \).
2Step 2: Analyzing Option a
Compare \( \frac{1}{x+3} \) and \( \frac{1}{x} \). As \( x \to \infty \), the term \(3\) becomes negligible compared to \(x\). Hence, \( \frac{1}{x+3} \approx \frac{1}{x} \), satisfying \(O(\frac{1}{x})\). Thus, it's true.
3Step 3: Analyzing Option b
\( \frac{1}{x} + \frac{1}{x^2} \) can be approximated by \( \frac{1}{x} \) for large \( x \) because \( \frac{1}{x^2} \) is much smaller than \( \frac{1}{x} \). Thus, \( \frac{1}{x} + \frac{1}{x^2} = O(\frac{1}{x}) \). This statement is true.
4Step 4: Analyzing Option c
For \( \frac{1}{x} - \frac{1}{x^2} \), \( \frac{1}{x^2} \) is negligible as compared to \( \frac{1}{x} \) for large \( x \), meaning \( \frac{1}{x} - \frac{1}{x^2} \approx \frac{1}{x} \). As \( x \to \infty \), this doesn't equal \(o(\frac{1}{x})\); hence, false.
5Step 5: Analyzing Option d
\( 2 + \cos(x) \) is bounded between \(1\) and \(3\) for all \( x \). Thus, \(2 + \cos(x) = O(2)\), since it remains bounded by a constant multiple of \(2\). This statement is true.
6Step 6: Analyzing Option e
Since the exponential function \(e^x\) grows much faster than the linear function \(x\), we have \(e^x + x \approx e^x\). This implies that \(e^x + x = O(e^x)\). Therefore, it is true.
7Step 7: Analyzing Option f
\(x \ln x\) grows faster than \(x\) but much slower than \(x^2\) as \(x \to \infty\). Thus, \(x \ln x = o(x^2)\). This is true.
8Step 8: Analyzing Option g
Since \(\ln(\ln x)\) is a double logarithm, it grows slower than \(\ln x\). Hence, \(\ln(\ln x) = O(\ln x)\). This is true.
9Step 9: Analyzing Option h
\(\ln(x) \) grows slower than \( \ln(x^2+1) \), which behaves like \(\ln(x^2) = 2\ln(x)\). Therefore, \(\ln(x) = o(\ln(x^2+1))\) is true.
Key Concepts
Asymptotic NotationLimits in CalculusBounding Functions
Asymptotic Notation
Asymptotic notation is a mathematical concept used to describe the limiting behavior of functions. It is especially useful in analyzing the efficiency of algorithms. The two most common forms of asymptotic notation include Big O notation, represented as \( O \), and small o notation, represented as \( o \).
- **Big O Notation (\( O \)):** This notation is used to express an upper bound on the growth rate of a function. It means that a function \( f(x) \) is bounded above by a constant multiple of another function \( g(x) \) for all sufficiently large \( x \). Mathematically, \( f(x) = O(g(x)) \) implies that there exist positive constants \( c \) and \( n_0 \) such that for all \( x > n_0 \), \( |f(x)| \leq c|g(x)| \).
- **Small o Notation (\( o \)):** This notation describes a function that grows significantly slower than the comparison function. For \( f(x) = o(g(x)) \), \( f(x) \) becomes negligible compared to \( g(x) \) as \( x \) approaches infinity. Formally, for any positive constant \( c \), there exists an \( n_0 \) such that for all \( x > n_0 \), \( |f(x)| < c|g(x)| \).
Limits in Calculus
Limits in calculus are a core concept used to define derivatives, integrals, and the asymptotic behavior of functions. When calculating the limit, we're interested in the behavior of a function as the input approaches a certain value, which can often be infinity.
- The notation \( \lim\limits_{x \to a} f(x) = L \) is read as "the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \)."
- When discussing asymptotic notation, we often look at limits as \( x \) approaches infinity. This is vital for understanding the growth rates of functions.
Bounding Functions
Bounding functions is a concept central to asymptotic notation and algorithm analysis. It involves finding functions that are consistently larger or smaller in magnitude, which help analyze the efficiency or growth behavior of algorithms.
- **Upper Bounds:** These functions cap the maximum growth rate of another function. In Big O notation, upper bounds are expressed mathematically as \( f(x) = O(g(x)) \). This means for sufficiently large \( x \), \( f(x) \) does not grow faster than a constant multiple of \( g(x) \).
- **Lower Bounds:** Correspondingly, lower bounds express how slowly a function can grow. This is captured using Omega notation, represented as \( \Omega \), which was not used in our original problem, but it's important to understand that it denotes lower bounds in terms of function growth.
- **Tight Bounds:** These represent both upper and lower bounds, meaning \( f(x) \) grows at the same rate as \( g(x) \). It's often denoted using Theta notation, \( \Theta \).
Other exercises in this chapter
Problem 10
In Exercises \(5-10,\) solve for \(y\) in terms of \(t\) or \(x,\) as appropriate. $$ \ln \left(y^{2}-1\right)-\ln (y+1)=\ln (\sin x) $$
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The U.S. population The Museum of Science in Boston displays a running total of the U.S. population. On May \(11,1993\) , the total was increasing at the rate o
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In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \frac{10}{x} $$
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Solve the equations. \(\ln e+4^{-2 \log _{4}(x)}=\frac{1}{x} \log _{10}(100)\)
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