Problem 4 - Thomas Calculus in SI Units

Question

Which of the following functions grow faster than \(x^{2}\) as \(x \rightarrow \infty ?\) Which grow at the same rate as \(x^{2} ?\) Which grow slower? a. \(x^{2}+\sqrt{x}\) b. \(10 x^{2}\) c. \(x^{2} e^{-x}\) d. \(\log _{10}\left(x^{2}\right)\) e. \(x^{3}-x^{2}\) f. \((1 / 10)^{x}\) g. \((1.1)^{x}\) h. \(x^{2}+100 x\)

Step-by-Step Solution

Answer

Functions a, b, and h grow at the same rate; c, d, and f grow slower; e and g grow faster.

Step 1: Identify the Function Dominance

Begin by analyzing each function provided in the options to determine the dominant term as \(x\) approaches infinity. This will help in understanding their growth rates relative to \(x^2\).

Step 2: Growth Analysis of x^{2}+\sqrt{x}

Consider the function \(x^2 + \sqrt{x}\). The term \(x^2\) dominates over \(\sqrt{x}\) as \(x\rightarrow\infty\), because \(x^2\) increases much faster than \(\sqrt{x}\). Therefore, \(x^2 + \sqrt{x}\) grows at the same rate as \(x^2\).

Step 3: Growth Analysis of 10x^{2}

For \(10x^2\), the function is a constant multiple of \(x^2\). Multiplying by a constant does not change the growth rate significantly. Thus, it grows at the same rate as \(x^2\).

Step 4: Growth Analysis of x^{2}e^{-x}

Consider the function \(x^2e^{-x}\). As \(x\rightarrow\infty\), \(e^{-x}\) approaches zero much faster than \(x^2\) increases. Thus, \(x^2e^{-x}\) diminishes to zero, suggesting it grows slower than \(x^2\).

Step 5: Growth Analysis of \log _{10}(x^{2})

\(\log _{10}(x^2) = 2\log_{10}(x)\). Logarithmic functions \(\log_{10}(x)\) grow much slower than polynomials. Hence, \(\log _{10}(x^2)\) grows slower than \(x^2\).

Step 6: Growth Analysis of x^{3}-x^{2}

For \(x^3 - x^2\), as \(x\rightarrow \infty\), \(x^3\) dominates because \(x^3\) grows faster than \(x^2\). Therefore, \(x^3-x^2\) grows faster than \(x^2\).

Step 7: Growth Analysis of (1/10)^{x}

The function \((1/10)^{x}\) is an exponential decay function, which approaches zero as \(x\rightarrow\infty\). This means it grows slower than \(x^2\).

Step 8: Growth Analysis of (1.1)^{x}

The function \((1.1)^{x}\) is an exponential growth function, growing significantly faster than any polynomial. Thus, it grows faster than \(x^2\).

Step 9: Growth Analysis of x^{2}+100x

For \(x^2 + 100x\), the term \(x^2\) again dominates as \(x\rightarrow\infty\). So, \(x^2 + 100x\) grows at the same rate as \(x^2\).

Key Concepts

Polynomial Dominance

In the world of asymptotic growth, polynomial functions have a special place in determining the pace at which functions increase. When comparing two polynomial functions, the term with the highest degree controls the growth. For example, in the given functions like \(x^{2} + \sqrt{x}\) and \(x^{2} + 100x\), the dominant term is \(x^2\). This means these functions grow at the same rate as \(x^2\) when \(x\) becomes very large. It's crucial to identify the dominant term, as it simplifies comparison. Remember, constant factors don't affect polynomial dominance significantly, which is why \(10x^2\) still grows at the same rate as \(x^2\). The same principle applies to \(x^3 - x^2\), where \(x^3\) dominates, showing faster growth compared to \(x^2\).

Exponential Growth

Exponential functions like \((1.1)^{x}\) showcase rapid growth rates, outpacing polynomials as \(x\) increases. Unlike polynomial functions, exponentials grow based on a constant raised to the power of \(x\). This results in a function that not only increases quickly but accelerates as \(x\) expands further. In the given exercise, \((1.1)^{x}\) exemplifies this explosive growth, outpacing the more tempered pace of \(x^2\). On the flip side, \((1/10)^{x}\) illustrates exponential decay, shrinking rapidly towards zero compared to any polynomial, including \(x^2\). Understanding exponential functions means recognizing that even small base increases, as in moving from 1.0 to 1.1, result in significant growth over large values of \(x\).

Logarithmic Growth

Logarithmic functions grow much more slowly than both polynomial and exponential functions. For instance, \(\log_{10}(x^2) = 2\log_{10}(x)\) reveals that the logarithmic term increases gradually as \(x\) enlarges. While polynomials could still be increasing dramatically, the growth of logarithmic functions remains controlled, appearing almost flat by comparison. This means that \(\log_{10}(x^2)\) grows slower than \(x^2\). Logs are useful in mapping multiplicative processes into additive ones, and their slow pace makes them suitable for depicting processes that require more manageable scaling. A key takeaway is that logarithms, no matter their basis, will always exhibit measured growth, emphasizing their slower rate compared to other functions.

Comparative Growth Rates

When analyzing growth rates of functions, the key lies in comparing the dominant terms, especially as \(x\) heads towards infinity. This comparison helps answer questions about which functions grow faster, slower, or at the same rate as given benchmarks like \(x^2\).
For instance, \(x^3-x^2\) grows faster due to the higher degree of \(x^3\). Meanwhile, \(x^2 + \sqrt{x}\) grows at the same rate as \(x^2\), since the dominant term is still \(x^2\). Functions such as \(x^2e^{-x}\) and \(\log_{10}(x^2)\) show slower growth compared to \(x^2\).
Understanding comparative growth rates helps one prioritize efforts in deriving long-term behaviors of complex functions. It provides a clearer view of how they might scale compared to others as limits push towards infinity, helping in fields as varied as algorithm analysis and economic forecasting.