Problem 21 - Thomas Calculus in SI Units

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 \(x^{1 / 1.000 .000}>\ln x .\) You might start by observing that when \(x>1\) the equation \(\ln x=x^{1 / 1,000,000}\) 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

a. \(\ln x\) grows slower than \(x^{1/n}\) as \(x \to \infty\). b. \(x \approx e^{1,000,000}\). c. \(x \approx 10^7\). d. Graphing may fail beyond software limits.

Step 1: Analyze Growth Rate for Part a

To demonstrate that \( \ln x \) grows slower than \( x^{1/n} \), compare their limits as \( x \to \infty \). Compute the limit \( \lim_{x \to \infty} \frac{\ln x}{x^{1/n}} \). Applying L'Hôpital's Rule, since both numerator and denominator approach infinity:\[ \lim_{x \to \infty} \frac{\ln x}{x^{1/n}} = \lim_{x \to \infty} \frac{1/x}{(1/n)x^{1/n - 1}} = \lim_{x \to \infty} \frac{n}{x^{(n-1)/n}} = 0 \].The limit is 0, which shows that \( \ln x \) grows slower than \( x^{1/n} \) as \( x \to \infty \).

Step 2: Solve Part b - Find x Overtaking Point for Small n

We need \( x^{1/1,000,000} > \ln x \). This requires solving \( \ln x = x^{1/1,000,000} \) or equivalently \( \ln (\ln x) = (\ln x) / 1,000,000 \).To do this, one can set \( y = \ln x \) and solve \( \ln y = y / 1,000,000 \) using numerical methods or iterative checking.A calculator reveals that the crossover happens around \( x \approx e^{1,000,000} \) as the solution \( y \approx 1,000,000 \) gives \( x \approx e^{0.000001^{1,000,000}} \).

Step 3: Solve Part c - Crossing Point with Calculator

We seek when \( x^{1/10} = \ln x \) or equivalently \( 10 \ln(\ln x) = \ln x \). Use a calculator to find \( x \) such that the above equality holds.Begin by estimating between powers of 10:- Evaluate at \( x = 10^6, 10^7 \) etc.- Refine by halving intervals, e.g., testing upper/lower bounds, such as \( x = 3.16^6 \).Determination shows crossover around \( x \approx 10^7 \) using iterative halving.

Step 4: Verify Equipment Limits for Part d

Attempt to find \( x \) using tools and equipment. You may notice that for large \( x \), some graphers or calculators may not handle \( \ln x = 10 \ln (\ln x) \) due to representation limits.Try software like Python or advanced calculators. If hardware limitations prevent exact resolution, validate by manual calculations or superior software accuracy for \( x \approx 10^7.5 \) with computational tests.

Key Concepts

Logarithmic Functions

Logarithmic functions have a pivotal role in mathematics, serving as the inverse operations of exponential functions. The natural logarithm, denoted by \( \ln(x) \), specifically refers to the logarithm with base \( e \), where \( e \approx 2.718281828 \). It helps in solving equations where the variable is an exponent, transforming the equation into a linear form. Unlike exponential growth, logarithmic growth is slower and eventually levels off, which is crucial when comparing functions like \( \ln(x) \) and \( x^{1/n} \).
When working with the natural logarithm, remember:

• \( \ln(1) = 0 \) because \( e^0 = 1 \)
• \( \ln(e) = 1 \) since \( e^1 = e \)
• The function is undefined for \( x \leq 0 \) as you cannot take the logarithm of a non-positive number in the real domain

Understanding these properties allows us to effectively compare \( \ln(x) \) with other functions and solve complex problems in calculus.

Limits

In calculus, limits help us understand the behavior of a function as it approaches a certain point. Specifically, limits are crucial when evaluating growth rates of functions like \( \ln(x) \) as \( x \to \infty \). To compare these functions to \( x^{1/n} \), we look at the limit:

• \( \lim_{x \to \infty} \frac{\ln x}{x^{1/n}} \)

If this limit approaches 0, it indicates that \( \ln(x) \) grows slower than \( x^{1/n} \).
This concept of limits allows us to formally prove the asymptotic behavior of functions, making it fundamental in calculus. Remember, the notation \( \lim_{x \to c} f(x) \) signifies the value that \( f(x) \) approaches as \( x \) gets arbitrarily close to \( c \).

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits of fractions, where both the numerator and denominator approach 0 or infinity. When faced with \( \lim_{x \to \infty} \frac{\ln x}{x^{1/n}} \), this rule allows us to differentiate both the numerator and denominator separately:

• \( \frac{d}{dx}(\ln x) = \frac{1}{x} \)
• \( \frac{d}{dx}(x^{1/n}) = \frac{1}{n}x^{1/n - 1} \)

Applying L'Hôpital's Rule turns our expression into:

• \[ \lim_{x \to \infty} \frac{1/x}{(1/n)x^{1/n - 1}} = \lim_{x \to \infty} \frac{n}{x^{(n-1)/n}} = 0 \]

This method is ideal for resolving indeterminate forms, making it highly valuable when comparing the growth rates of different functions.

Numerical Methods

Numerical methods provide a way to approximate solutions to mathematical problems, especially when analytical solutions are difficult to reach or do not exist. In the context of the exercise, numerical methods are crucial for finding values of \( x \) where \( x^{1/1,000,000} > \ln x \) or determining when \( x^{1/10} = \ln x \).
For instance, when the equation \( \ln(x) = 10 \ln(\ln x) \) does not allow easy factorization, numerical methods like successive approximations or using calculators allow us to experiment with different large powers of ten. These approaches might involve:

• Estimating values within a range by halving intervals
• Increasing the precision of guesses by evaluating smaller increments
• Utilizing software tools to handle computations beyond calculator precision

This practice of bracketing and then narrowing down estimates allows students to visualize solutions and understand the dynamics between complex functions.