/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Any real number in the interval ... [FREE SOLUTION] | 91Ó°ÊÓ

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Any real number in the interval [0,1] can be represented by a decimal (not necessarily unique) as \(0 . d_{1} d_{2} d_{3} d_{4} \ldots=\frac{d_{1}}{10}+\frac{d_{2}}{10^{2}}+\frac{d_{3}}{10^{3}}+\frac{d_{4}}{10^{4}}+\cdots\), where \(d_{i}\) is one of the integers \(0,1,2,3, \ldots, 9 .\) Prove that the series on the right-hand side always converges.

Short Answer

Expert verified
The series converges by the comparison test with a convergent geometric series.

Step by step solution

01

Understand the Series Structure

The given series is the sum of an infinite number of terms of the form \( \frac{d_i}{10^i} \). Each \( d_i \) represents a digit from 0 to 9, and each term represents a fraction of decreasing value, contributing less as \( i \) increases.
02

Analyze the Convergence of the Infinite Series

Consider the general term \( a_i = \frac{d_i}{10^i} \), where \( d_i \) is between 0 and 9. Since \( 0 \le d_i \le 9 \), it follows that \( 0 \le \frac{d_i}{10^i} \le \frac{9}{10^i} \). Each term gets smaller as \( i \) increases because they are divided by increasing powers of 10.
03

Use Comparison Test for Convergence

To evaluate convergence, compare \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \) with \( \sum_{i=1}^{\infty} \frac{9}{10^i} \). The series \( \sum_{i=1}^{\infty} \frac{9}{10^i} \) is a geometric series where each term is 1/10th of the previous, starting at 0.9. The sum of this geometric series is \( \frac{0.9}{1-0.1} = 1 \), which converges.
04

Conclude the Original Series Convergence

Since \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \le \sum_{i=1}^{\infty} \frac{9}{10^i} \) and the latter converges to 1, it follows from the comparison test that \( \sum_{i=1}^{\infty} \frac{d_i}{10^i} \) also converges. Therefore, the series representation of any real number in [0,1] converges.

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Key Concepts

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

Decimal Representation
Decimal representation is a way to express real numbers using the base-10 numeral system. In the context of the exercise, every real number between 0 and 1 can be expressed in the form \(0.d_1d_2d_3...\). The value of this number is determined by the infinite series \[ \frac{d_1}{10} + \frac{d_2}{10^2} + \frac{d_3}{10^3} + \cdots \] where each \(d_i\) is a digit between 0 and 9.
Each digit represents a fractional part of a whole number, contributing smaller and smaller values as you move to the right. This setup is fundamentally important because it reflects the decimal system's way of dividing a unit into ten equal parts.
  • The first digit, \(d_1\), represents tenths.
  • The second digit, \(d_2\), represents hundredths, and so on.
These fractional parts sum up to create a unique decimal representation, allowing every rational and some irrational numbers in the interval [0, 1] to be expressed easily.
Geometric Series
A geometric series is a series where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. For a geometric series to converge (i.e., sum to a specific number), this common ratio must be between -1 and 1, excluding endpoints.
In this exercise, our focus is the geometric series \[ \sum_{i=1}^{\infty} \frac{9}{10^i} \] which plays a key role in proving convergence. This series has a first term of 0.9 and a common ratio of \(\frac{1}{10}\), which is well within the necessary range for convergence.
The sum of a convergent geometric series can be quickly calculated using the formula:\[ S = \frac{a}{1-r} \]where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
  • For our example: \(a = 0.9\) and \(r = 0.1\), resulting in a sum of \(1\).
This demonstrates how certain infinite series can simplify into finite answers, an essential takeaway from geometric series.
Comparison Test
The comparison test is a powerful tool in determining the convergence of an infinite series by comparing its terms to those of a known series.
The comparison typically involves showing that the given series' terms are smaller than the terms of a known convergent series.
  • If \(a_i \le b_i\) and \(\sum b_i\) converges, then \(\sum a_i\) converges as well.
For this task, we demonstrate the convergence of the series \[ \sum_{i=1}^{\infty} \frac{d_i}{10^i} \] by comparing it to the convergent geometric series \[ \sum_{i=1}^{\infty} \frac{9}{10^i} \]shown to sum to 1. Since \[ 0 \le \frac{d_i}{10^i} \le \frac{9}{10^i} \] for \(0 \leq d_i \leq 9\), the original series is always less than the geometric series, ensuring its convergence.
Infinite Series
An infinite series is the sum of infinitely many terms, which can often converge to a finite value. Understanding when and why these series converge is crucial for many areas of mathematics. In this exercise, we explore the concept using the series representation of decimal fractions.
Infinite series have some fascinating properties:
  • Not all infinite series converge. Some diverge, meaning they don't settle on a specific sum.
  • Convergence depends on the behavior of the terms as they progress towards infinity.
  • Understanding convergence helps in approximating complex calculations with simpler numbers.
In our specific case, we show that a decimal representation series converges, demonstrating that every real number between 0 and 1 can be precisely defined. This example of decimal fractions highlights how infinite series can represent numbers accurately, showcasing their practical application in mathematical analysis.

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(2,2+\frac{1}{2}, 2+\frac{1}{2+\frac{1}{2}}, 2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}, \dots\)

Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

a. Use the binomial series and the fact that $$ \frac{d}{d x} \sin ^{-1} x=\left(1-x^{2}\right)^{-1 / 2} $$ to generate the first four nonzero terms of the Taylor series for \(\sin ^{-1} x\). What is the radius of convergence? b. Series for \(\cos ^{-1} x \quad\) Use your result in part (a) to find the first five nonzero terms of the Taylor series for \(\cos ^{-1} x\)

Show that if \(M_{1}\) and \(M_{2}\) are least upper bounds for the sequence \(\left\\{a_{n}\right\\},\) then \(M_{1}=M_{2}\) That is, a sequence cannot have two different least upper bounds.

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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