91Ó°ÊÓ

Deal with iterations of the logarithm function. Let log \(n\) denote the logarithm of \(n\) to the base \(2,\) as usual. The function \(\log ^{(k)} n\) is defined recursively by $$\log ^{(k)} n=\left\\{\begin{array}{cl}{n} & {\text { if } k=0} \\ {\log \left(\log ^{(k-1)} n\right)} & {\text { if } \log ^{(k-1)} n \text { is defined }} \\ {} & {\text { and positive }} \\ {\text { undefined }} & {\text { otherwise. }}\end{array}\right.$$ The iterated logarithm is the function log" \(n\) whose value at \(n\) is the smallest nonnegative integer \(k\) such that \(\log ^{(k)} n \leq 1\) Find the value of \(\log ^{*} n\) for these values of \(n\) a) 2 b) 4 c) 8 d) 16 e) 256 f) 65536 g) \(2^{2048}\)

Step by step solution

01

Understanding the Problem

Given the function \(\log ^{(k)} n\) defined recursively, the goal is to find the smallest nonnegative integer \(\) such that \(\log ^{(k)} n \leq 1\). Let's apply this to each provided value of \(\): 2, 4, 8, 16, 256, 65536, and \(<2^{2048}>\).

02

Calculate \(\log ^{(k)} 2\)

Start with \(\log^{(0)} 2 = 2\). Since \(\log_2 2 = 1\), we have \(\log^{(1)} 2 = 1\). \( = 1\) satisfies the condition.

03

Calculate \(\log ^{(k)} 4\)

Start with \(\log^{(0)} 4 = 4\). \(\log_2 4 = 2\), so \(\log^{(1)} 4 = 2\). Then \(\log_2 2 = 1\), so \(\log^{(2)} 4 = 1\). \( = 2\) satisfies the condition.

04

Calculate \(\log ^{(k)} 8\)

Start with \(\log^{(0)} 8 = 8\). \(\log_2 8 = 3\), so \(\log^{(1)} 8 = 3\). \(\log_2 3 \approx 1.58496\), so \(\log^{(2)} 8 \approx 1.58496\). Finally, \(\log_2 1.58496 \le 1\), so \(\log^{(3)} 8 = 1\). \( = 3\) satisfies the condition.

05

Calculate \(\log ^{(k)} 16\)

Start with \(\log^{(0)} 16 = 16\). \(\log_2 16 = 4\), so \(\log^{(1)} 16 = 4\). \(\log_2 4 = 2\), so \(\log^{(2)} 16 = 2\). Finally, \(\log_2 2 = 1\), so \(\log^{(3)} 16 = 1\). \( = 3\) satisfies the condition.

06

Calculate \(\log ^{(k)} 256\)

Start with \(\log^{(0)} 256 = 256\). \(\log_2 256 = 8\), so \(\log^{(1)} 256 = 8\). \(\log_2 8 = 3\), so \(\log^{(2)} 256 = 3\). Then \(\log_2 3 \approx 1.58496\), so \(\log^{(3)} 256 \approx 1.58496\). Finally, \(\log_2 1.58496 \le 1\), so \(\log^{(4)} 256 = 1\). \( = 4\) satisfies the condition.

07

Calculate \(\log ^{(k)} 65536\)

Start with \(\log^{(0)} 65536 = 65536\). \(\log_2 65536 = 16\), so \(\log^{(1)} 65536 = 16\). \(\log_2 16 = 4\), so \(\log^{(2)} 65536 = 4\). \(\log_2 4 = 2\), so \(\log^{(3)} 65536 = 2\). Finally, \(\log_2 2 = 1\), so \(\log^{(4)} 65536 = 1\). \( = 4\) satisfies the condition.

08

Calculate \(\log ^{(k)} 2^{2048}\)

Start with \(\log^{(0)} 2^{2048} = 2^{2048}\). \(\log_2 2^{2048} = 2048\), so \(\log^{(1)} 2^{2048} = 2048\). \(\log_2 2048 = 11\), so \(\log^{(2)} 2^{2048} = 11\). \(\log_2 11 \approx 3.45943\), so \(\log^{(3)} 2^{2048} \approx 3.45943\). Finally, \(\log_2 3.45943 \approx 1.78588\), so \(\log^{(4)} 2^{2048} \approx 1.78588\). Lastly, \(\log_2 1.78588 \le 1\), so \(\log^{(5)} 2^{2048} = 1\). \( = 5\) satisfies the condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Functions

Logarithmic functions are a crucial mathematical concept, particularly in the study of growth and scaling behaviors. A logarithm answers the question: 'To what exponent must we raise a base number to obtain a given number?' For example, \(\text{log}_2 8 = 3\) because \2^3 = 8\. In simpler terms, the logarithm tells how many times we need to multiply the base (here, 2) to reach the target number (8 in this case). In the context of the iterated logarithm, we're not just calculating a single logarithm but repeatedly applying the logarithm function, reducing the size of the number step by step.

Recursion

Recursion is a programming and mathematical concept where a function calls itself as a subroutine. This is useful for solving complex problems by breaking them down into more manageable sub-problems. In the given exercise, the iterated logarithm function \(\text{log } ^{(k)} n\) is defined recursively. The function is broken into base cases: if \k = 0\, then \text{log } ^{(0)} n = n\; otherwise, \text{log } ^{(k)} n = \text{log } (\text{log } ^{(k-1)} n)\. Recursion allows us to repeatedly apply the logarithm function until the number becomes less than or equal to 1.

Discrete Mathematics

Discrete mathematics deals with distinct and separate values, often involving integers. It's the foundation for computer science and algorithms. In this exercise, the iterative logarithm function involves integer steps rather than continuous progression. The importance of discrete mathematics lies in its ability to help us analyze algorithms, data structures, and understand iterative and recursive processes clearly. These concepts are critical in optimizing computations and making computer programs efficient.

Integer Computation

Integer computation involves arithmetic operations on whole numbers. Since logarithms and their iterations are often applied in computational contexts, understanding integer computation is essential. For the iterated logarithm problem, we perform logarithmic calculations and keep track of the number of steps (iterations) required to reduce the original number to a value less than or equal to 1. This application showcases how integer computation and logarithmic functions intertwine to solve practical mathematical problems.