Question

The iterated logarithms are defined by \(L_{0}(x)=x\) and $$ L_{n}(x)=\log \left(L_{n-1}(x)\right), \quad x>a_{n}, \quad n \geq 1 $$ where \(a_{1}=0\) and \(a_{n}=e^{a_{n-1}}, n \geq 1 .\) Show that (a) \(L_{n}(x)=L_{n-1}(\log x), \quad x>a_{n}, \quad n \geq 1\) (b) \(L_{n-1}\left(a_{n}+\right)=0\) and \(L_{n}\left(a_{n}+\right)=-\infty\). (c) \(\lim _{x \rightarrow a_{n}+}\left(L_{n-1}(x)\right)^{\alpha} L_{n}(x)=0\) if \(\alpha>0\) and \(n \geq 1 .\) (d) \(\lim _{x \rightarrow \infty}\left(L_{n}(x)\right)^{\alpha} / L_{n-1}(x)=0\) if \(\alpha\) is arbitrary and \(n \geq 1\).

Short answer

In this exercise, we have worked with iterated logarithms and proved properties related to them. We were given the definition of the iterated logarithm \(L_n(x)\) and a sequence of constants \(a_n\). We proved the following four properties: (a) We showed that \(L_n(x) = L_{n-1}(\log x)\) for any \(x > a_n\) and \(n \geq 1\). (b) We showed that \(L_{n-1}(a_n+) = 0\) and \(L_n(a_n+)= -\infty\). (c) We showed that \(\lim_{x \rightarrow a_n+} (L_{n-1}(x))^{\alpha} L_{n}(x) = 0\) if \(\alpha > 0\) and \(n \geq 1\). (d) We showed that \(\lim_{x \rightarrow \infty} (L_{n}(x))^{\alpha} / L_{n-1}(x) = 0\) if \(\alpha\) is arbitrary and \(n \geq 1\). By proving these properties, we have gained a deeper understanding of iterated logarithms and their behavior as they approach certain points.

Step-by-step solution

Prove property (a)

Since the iterated logarithm is defined recursively, we will use the definition to prove the first property. Given that \(L_{n}(x) = \log(L_{n-1}(x))\) for \(x > a_n\) and \(n \geq 1\), we have: $$ L_{n}(x) = \log(L_{n-1}(x)) = \log(L_{n-1}(\log e^{x})), $$ since \(x = \log e^{x}\). From the definition of the iterated logarithm, this is equal to \(L_{n-1}(\log x)\). Thus, we have shown the property, that \(L_{n}(x) = L_{n-1}(\log x)\) for \(x > a_n\) and \(n \geq 1.\)

Prove property (b)

We are given that \(a_n = e^{a_{n-1}}\), and we have to prove two properties: \(L_{n-1}(a_n+) = 0\) and \(L_n(a_n+) = -\infty\). Since \(L_{n}(x) = \log(L_{n-1}(x))\), we can write \(L_{n-1}(a_n) = \log(L_{n}(a_{n}))\). Now, let's consider the limit as \(x \rightarrow a_n+\): 1. \(\lim_{x \rightarrow a_n+} L_{n-1}(x) = \lim_{x \rightarrow a_n+} \log(L_{n}(x)) = \log(\lim_{x \rightarrow a_n+} L_{n}(x)) = 0\), by definition. Thus, we have shown that \(L_{n-1}(a_n+) = 0\). 2. \(L_{n}(a_n) = \log(L_{n-1}(a_{n})) \rightarrow -\infty\) as \(x \rightarrow a_n+\). Since \(\lim_{x \rightarrow a_n+} L_{n-1}(x) = 0\), we have that \(\lim_{x \rightarrow a_n+} L_{n}(x) = -\infty\). Therefore, we have shown that \(L_n(a_n+)=-\infty\).

Prove property (c)

Given \(\alpha > 0\) and \(n \geq 1\), we have to show that \(\lim_{x \rightarrow a_n+}(L_{n-1}(x))^{\alpha}L_{n}(x) = 0\). Now, as \(x \rightarrow a_n+\), we have that \(L_{n-1}(x) \rightarrow 0\) and \(L_{n}(x) \rightarrow -\infty\). Therefore, we have: $$ \lim_{x \rightarrow a_n+}(L_{n-1}(x))^{\alpha}L_{n}(x) = (\lim_{x \rightarrow a_n+} (L_{n-1}(x))^{\alpha}) (\lim_{x \rightarrow a_n+} L_{n}(x)). $$ Since \(0^{\alpha} = 0\) and \((\lim_{x \rightarrow a_n+} L_{n}(x)) = -\infty\), we have: $$ \lim_{x \rightarrow a_n+}(L_{n-1}(x))^{\alpha}L_{n}(x) = 0 \cdot (-\infty) = 0. $$ Thus, we have shown the desired property (c).

Prove property (d)

We are given that \(\alpha\) is arbitrary, and \(n \geq 1\). We have to show that \(\lim_{x \rightarrow \infty}(L_{n}(x))^{\alpha} / L_{n-1}(x) = 0\). Now, using property (a) that we previously proved, we have: $$ \lim_{x \rightarrow \infty}(L_{n}(x))^{\alpha} / L_{n-1}(x) = \lim_{x \rightarrow \infty} (L_{n-1}(\log x))^{\alpha} / L_{n-1}(x). $$ As \(x \rightarrow \infty\), we have that \(\log x \rightarrow \infty\) and hence, we can set \(y = \log x\). Then, we have: $$ \lim_{y \rightarrow \infty}(L_{n-1}(y))^{\alpha} / L_{n-1}(e^y). $$ Now, since \(L_{n-1}(x)\) is an increasing function as \(x\) increases, as \(y \rightarrow \infty\), we have that \((L_{n-1}(y))^\alpha \rightarrow \infty\) and \(L_{n-1}(e^y) \rightarrow \infty\). Therefore, the limit becomes an indeterminate form of type \(\frac{\infty}{\infty}\). Using L'Hôpital's Rule, we get: $$ \lim_{y \rightarrow \infty} \frac{(L_{n-1}(y))^{\alpha} / \alpha (L_{n-1}(y))^{\alpha - 1} L_{n-1}'(y)}{L_{n-1}'(e^y) e^y}. $$ Since \((L_{n-1}(y))^{\alpha - 1}\) is finite and nonzero for all \(y > 0\), we can cancel out the term \((L_{n-1}(y))^{\alpha - 1}\) from the expression. We are left with: $$ \lim_{y \rightarrow \infty} \frac{(L_{n-1}(y))^{\alpha} L_{n-1}'(y)}{L_{n-1}'(e^y) e^y} = 0, $$ since the factors in the numerator and denominator both approach infinity but at different rates. Thus, we have shown the desired property (d).

Key concepts

Real Analysis

Real analysis is a branch of mathematics dealing with real numbers and real-valued functions. It's critical for understanding the behavior of functions, their properties, and the types of operations you can perform on them.

A fundamental concept in real analysis is the notion of a limit, which is a way of expressing the behavior of a function as its input approaches a certain value. The iterated logarithm problem mentioned in the exercise is deeply rooted in these concepts, particularly because it deals with recursive functions that have limits at infinity or at certain constants, showcased by the values of \(a_n\).

Understanding the behavior of such recursive functions near their limits is crucial because it allows us to predict and describe the function's properties. In cases such as iterated logarithms, we delve deeper into the subject by studying how quickly, or slowly, the function values descend to zero or diverge to infinity, informing us about the nature of convergence or divergence of sequences and series derived from these functions.

Limit of a Function

The limit of a function is a fundamental concept in calculus and real analysis. It describes the value that a function approaches as the input approaches a particular point from either side of that point.

In our example of iterated logarithms, limits help us understand the behavior of functions as they iterate to deeper levels. Specifically, when analyzing the limit as \(x \rightarrow a_n+\), we are essentially scoping into the behavior of \(L_{n-1}(x)\) and \(L_n(x)\) as they approach the critical point \(a_n\) from the right. The notion of approaching \(a_n+\) symbolizes getting infinitesimally close to \(a_n\) but never actually reaching it, which is crucial for evaluating the outcome of iterated calculations.

Proving properties like \(L_{n-1}(a_n+) = 0\) and \(L_n(a_n+) = -\text{\infty}\) requires a careful understanding of the behavior of logarithmic functions as their arguments tend towards zero or infinity. The alpha power in property (c) introduces another layer where limits can further be manipulated, creating compound expressions to study these tendencies.

Recursive Functions

Recursive functions are those that are defined in terms of themselves. They use their previous output as the input for the next step in their process. They are ubiquitous in mathematics and computer science because they can describe complex phenomena and calculate values in a compact form.

In the case of the iterated logarithm problem, the definition of \(L_n(x)\) is inherently recursive. Each \(L_n(x)\) is built off the previous \(L_{n-1}(x)\), creating a chain of logarithmic transformations. This recursive nature is precisely what makes iterated logarithms fascinating and somewhat tricky to work with when evaluating their limits.

Furthermore, recursive functions such as these can have particularly interesting properties at different points of iteration, as shown by property (d). Evaluating their limits—when \(x \rightarrow \text{\infty}\) or \(x \rightarrow a_n+\)—highlights the recursive depth and the fact that even though the functions get infinitely large or small, their ratios or products can still tend towards a finite limit, a concept that may seem counterintuitive without a firm understanding of recursion and limits.

Recursive functions, therefore, are potent tools for decomposing complex problems into simpler, more manageable parts.