Problem 21 - Thomas Calculus

Question

a. Show that \(\ln x\) grows slower as \(x \rightarrow \infty\) than \(x^{1 / n}\) for any positive integer \(n,\) even \(x^{1 / 1,000,000}\) . b. Although the values of \(x^{1 / 1,000,000}\) eventually overtake the values of \(\ln x,\) you have to go way out on the \(x\) -axis before this happens. Find a value of \(x\) greater than 1 for which \(x > 1 / 1,00,000 > \ln x .\) You might start by observing that when \(x > 1\) the equation \(\ln x=x^{1 / 1,00000}\) is equivalent to the equation \(\ln (\ln x)=(\ln x) / 1,000,000\) c. Even \(x^{1 / 10}\) takes a long time to overtake \(\ln x\) . Experiment with a calculator to find the value of \(x\) at which the graphs of \(x^{1 / 10}\) and \(\ln x\) cross, or, equivalently, at which \(\ln x=10 \ln (\ln x)\) Bracket the crossing point between powers of 10 and then close in by successive halving. .d. (Continuation of part \((c) . )\) . The value of \(x\) at which \(\ln x=10 \ln (\ln x)\) is too far out for some graphers and root finders to identify. Try it on the equipment available to you and see what happens.

Step-by-Step Solution

Answer

The solution involves comparing growth rates and using numerical methods for precise evaluation as \(x \to \infty\).

Step 1: Establish the growth rates

We need to compare the growth rates of \(\ln x\) and \(x^{1/n}\). Consider the limits: \(\lim_{{x \to \infty}} \frac{\ln x}{x^{1/n}}\). This is equivalent to considering \(\lim_{{x \to \infty}} \frac{1/x}{(1/n)x^{1/n - 1}}\) using L'Hôpital's Rule. Simplifying, we keep applying L'Hôpital's Rule until the exponent makes the denominator dominate. Make sure \(n\) is large enough so that as \(x \to \infty\), the result tends to \(0\), proving \(\ln x\) grows slower.

Step 2: Solve \(\ln x = x^{1/1,000,000}\)

Substitute \(u = \ln x\) into the equation \(\ln x = x^{1/1,000,000}\) to get \(u = e^{u/1,000,000}\). This becomes \(u \ln u = u/1,000,000\) after taking logarithms. It is challenging to solve analytically, so numerical methods or graphing tools are required to find the solution where \(u\) gets large.

Step 3: Use trial and error or graphical methods

To find when \(\ln x\) and \(x^{1/10}\) meet, use a calculator or plotting software. The goal is to solve \(\ln x = 10 \ln(\ln x)\). Start by checking \(x\) at powers of 10, narrowing down via halving intervals. Eventually find \(x\) where this equality holds.

Step 4: Continue trial and improvement for accuracy

Once you find an approximate \(x\) using a graphing calculator, further refine by testing smaller intervals around the point where \(\ln x\) and \(x^{1/10}\) converge. This requires iteration and possibly software that can handle large numbers for precision.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used for finding the limits of indeterminate forms. Indeterminate forms such as 0/0 or ∞/∞ can often appear when evaluating the limits of functions. L'Hôpital's Rule helps us in such situations by allowing us to differentiate the numerator and the denominator separately. This rule states that if the limit of the functions as they approach a point gives an indeterminate form, then:

• \[\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}\]

provided the limit on the right side exists.

For example, in our exercise, we compare the growth rates of \(\ln x\) and \(x^{1/n}\). Initially, the form is \(\frac{\ln x}{x^{1/n}}\) as \(x \to \infty\), which is an indeterminate form \(\infty/\infty\). By applying L'Hôpital's Rule repeatedly, we can evaluate it to zero, confirming that \(\ln x\) grows slower than \(x^{1/n}\) for any positive integer \(n\). This is particularly important when proving which function dominates as \(x\) increases towards infinity.

Logarithmic Functions

Logarithmic functions, denoted as \(\ln x\), where \(\ln\) stands for the natural logarithm, are inverse to exponentiation. In simpler terms, if you have \(e^{y} = x\), then \(\ln x = y\). This function is widely used due to its unique properties:

• It grows very slowly compared to polynomial and exponential functions.
• The logarithm of a product is the sum of logarithms: \(\ln(ab) = \ln a + \ln b\).
• The logarithm of a power is a multiple: \(\ln(a^b) = b\ln a\).

These properties make logarithms crucial for simplifying complex equations, especially those involving exponentiation.

In our exercise, the logarithmic function \(\ln x\) is compared with growth functions like \(x^{1/1000000}\) for very large \(x\). Despite \(\ln x\) being overtaken by these functions as \(x\) becomes extremely large, \(\ln x\) starts growing notably slow, which is why you need to go significantly far on the \(x\)-axis for these powers of \(x\) to surpass \(\ln x\). The behaviour of \(\ln x\) helps us understand different growth rates effectively.

Growth Rates

In mathematics, growth rates refer to how quickly or slowly a function increases or decreases. Understanding growth rates is particularly important when analyzing functions approaching large values of \(x\). For example, in our exercise, we compare growth rates of \(\ln x\) and other functions like \(x^{1/10}\) or \(x^{1/1000000}\).

• \(\ln x\) grows slower than any power of \(x\).
• A function like \(x^{1/n}\), even with a large \(n\), will eventually outgrow \(\ln x\).

To compare these, calculating the limit using L'Hôpital’s Rule allows us to express their relative growth behavior. This limit calculation derives the conclusion that though \(\ln x\) grows slowly, it is bounded to be overtaken by polynomial forms of \(x\). Calculating such limits is useful in understanding the dominant function as \(x\) tends towards infinity.
Moreover, experimenting with numerical methods or graphical tools provides insight into specific crossover points where the change in dominance occurs. Growth rates are essential in various fields, from computer science algorithms to economics, where predicting the behavior of models over large scales is crucial.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. It represents repeated multiplication of the base. For instance, expressing \(b\) raised to the power \(n\) is given as \(b^n = b \cdot b \cdot \ldots \cdot b\) (with \(n\) times the base \(b\)). Some important characteristics of exponentiation include:

• It grows much faster than logarithmic functions for any positive base greater than one.
• Exponentiation underlies the concept of powers, roots, and the famous exponential function \(e^x\).

In the context of our problem, exponentiation and its related functions like \(x^{1/10}\) and \(x^{1/1000000}\) are key. The exercise shows the stark contrast in growth rates between \(\ln x\) and these exponential functions, where evidencing \(x^{1/1000000}\) will eventually surpass \(\ln x\). Despite the fact that these functions start overtaking logarithms much further along the \(x\)-axis, exponentiation inherently has a rapid growth pattern that transcends most calculated limits.