Problem 7 Suppose that $s_{0}>0$ and ... [FREE SOLUTION]

91影视

Introduction to Real Analysis

William F. Trench

$Math Studyset 91影视 Explanations$ Math

2013 Edition

Chapter 1: Problem 7

Suppose that $s_{0}>0$ and $s_{n}=1-e^{-s_{n-1}}, n \geq 1 .$ Show that \(0

Short Answer

Expert verified

Question: Prove that for the recursive sequence $s_n = 1 - e^{-s_{n-1}}$ with the initial condition $s_0 > 0$, the inequality $0 < s_n < 1$ holds for all $n \geq 1$. Answer: Using induction and the properties of exponential functions, we showed that the inequality holds true for $n=1$, and if it holds true for some $n$, it also holds for $n+1$. Therefore, the inequality $0 < s_n < 1$ holds for all $n \geq 1$.

Step by step solution

Verify the inequality for $n=1$

First, we need to show that the inequality holds true for $n=1$. We know that $s_0 > 0$. Using the given formula, we can find $s_1$: $s_1 = 1 - e^{-s_0}$ Since $s_0 > 0$, $-s_0 < 0$ and $e^{-s_0} > 0$. Moreover, $e^0 = 1$, thus, $0 < e^{-s_0} < 1$. So, we have: $0 < 1 - e^{-s_0} < 1$ Hence, $0 < s_1 < 1$.

Inductive hypothesis and step

Now, assume that the inequality $0 < s_n < 1$ holds true for some $n \geq 1$. We want to show that the inequality also holds for $n+1$. Using the given formula, we have: $s_{n+1} = 1 - e^{-s_n}$ Since $0 < s_n < 1$ (by our inductive hypothesis), we have that $-s_n < 0$ and $e^{-s_n} > 0$. Moreover, $e^0 = 1$, thus $0 < e^{-s_n} < 1$. So, we have: $0 < 1 - e^{-s_n} < 1$ Hence, $0 < s_{n+1} < 1$.

Conclusion

By induction, we have shown that the inequality $0 < s_n < 1$ holds true for all $n \geq 1$. This completes the proof.

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Key Concepts

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

Inequality Proof

An inequality proof shows that a statement involving inequalities holds true. In this exercise, we demonstrate that for a given sequence $s_n$, all terms lie strictly between 0 and 1. To achieve this, we use **mathematical induction**. This powerful technique involves proving a statement in two main steps:

**Base Case**: Verify the inequality for the initial term (for $n=1$, we check if $0 < s_1 < 1$).
**Inductive Step**: Assume the inequality holds for an arbitrary $n$, and then prove it for $n+1$.

Mathematical induction is like a domino effect. When the first domino falls (base case), if each knocks over the next (inductive step), all dominoes fall if you start the chain.By verifying the inequality for the base case $n=1$, and using our hypothesis to show it holds for $n+1$, we conclude that the sequence's terms always satisfy the inequality.

Real Analysis

Real analysis is a foundational branch of mathematics dealing with real numbers and real-valued functions. It involves concepts such as:

**Limits and Continuity**: Understanding the behavior of sequences and functions as they approach certain values.
**Differentiation and Integration**: Dealing with rates of change and accumulation of quantities.
**Sequences and Series**: Studying ordered collections of numbers and their summation.

This specific exercise involves sequences, an integral part of real analysis. The problem explores a sequence $s_n$ defined recursively, meaning each term builds on the previous one. A solid grasp of limits is needed, as it ensures the sequence is bounded and behaves predictably. In this way, real analysis provides the rigorous framework needed to understand, verify, and analyze sequences like $s_n$, ensuring they remain within certain limits, such as between 0 and 1.

Sequences and Series

A sequence is an ordered list of numbers following a specific rule. In this exercise, the sequence $s_n$ is defined recursively: each term depends on the previous one through the formula $s_n = 1 - e^{-s_{n-1}}$. Key concepts of sequences include:

**Recursive Definition**: Each term is defined based on previous terms, offering a step-by-step construction.
**Convergence**: A sequence converges if its terms approach a certain number as they progress.
**Boundedness**: A sequence is bounded if it stays within a fixed range for all of its terms.

In this exercise, the recursive nature and boundedness of $s_n$ are essential. A sequence that is bounded, like $0 < s_n < 1$, is constrained within specific limits, thus ensuring that the terms do not diverge wildly or become undefined. This stability is critical in real analysis, helping mathematicians understand the behavior of sequences and ensure they conform to expectations.

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91影视

Suppose that \(s_{0}>0\) and \(s_{n}=1-e^{-s_{n-1}}, n \geq 1 .\) Show that \(0

Short Answer

Step by step solution

Verify the inequality for \(n=1\)

Inductive hypothesis and step

Conclusion

Key Concepts

Inequality Proof

Real Analysis

Sequences and Series

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