/*! 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 6 We have shown that if \(\sum(-1)... [FREE SOLUTION] | 91Ó°ÊÓ

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We have shown that if \(\sum(-1)^{k+1} a_{k}\) is a convergent alternating series, then the sum \(S\) of the series lies between any two consecutive partial sums \(S_{n}\). This suggests that the average \(\frac{S_{n}+S_{n+1}}{2}\) is a better approximation to \(S\) than is \(S_{n}\). a. Show that \(\frac{S_{n}+S_{n+1}}{2}=S_{n}+\frac{1}{2}(-1)^{n+2} a_{n+1}\). b. Use this revised approximation in (a) with \(n=20\) to approximate \(\ln (2)\) given that $$ \ln (2)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} . $$ Compare this to the approximation using just \(S_{20} .\) For your convenience, \(S_{20}=\frac{155685007}{232792560}\).

Short Answer

Expert verified
Using the revised approximation gives a closer value \( \frac{161227687}{232792560} \) compared to \( S_{20} = \frac{155685007}{232792560} \).

Step by step solution

01

- Establish the Series Definition and Partial Sums

Define the alternating series sum and its partial sums: \[ S_n = \sum_{k=1}^{n} (-1)^{k+1} a_k \] and \[ S_{n+1} = \sum_{k=1}^{n+1} (-1)^{k+1} a_k = S_n + (-1)^{n+2} a_{n+1} \]
02

- Average of Consecutive Partial Sums

Write the average of two consecutive partial sums: \[ \frac{S_n + S_{n+1}}{2} \] Substitute the expression for \(S_{n+1}\): \[ \frac{S_n + (S_n + (-1)^{n+2} a_{n+1})}{2} \]
03

- Simplify the Expression

Simplify the expression for the average: \[ = \frac{2S_n + (-1)^{n+2} a_{n+1}}{2} = S_n + \frac{1}{2}(-1)^{n+2} a_{n+1} \] This confirms the required equation: \[ \frac{S_n + S_{n+1}}{2} = S_n + \frac{1}{2}(-1)^{n+2} a_{n+1} \]
04

- Special Case: Logarithm Series

Given the series \[ \ln(2) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} \] with \( a_k = \frac{1}{k} \), calculate the approximation for \(n=20\).
05

- Applying the Result

Using \( n = 20 \), \[ S_{20} = \frac{155685007}{232792560} \]. Compute \[ \frac{S_{20} + S_{21}}{2} = S_{20} + \frac{1}{2}(-1)^{22} a_{21} \] Since \( (-1)^{22} = 1 \) and \( a_{21} = \frac{1}{21} \), the result is: \[ \frac{155685007}{232792560} + \frac{1}{2 \cdot 21} = \frac{155685007}{232792560} + \frac{1}{42} \]
06

- Evaluate and Compare

Evaluate \( \frac{1}{42} = \frac{232792560}{42} = 5542680 \), so \[ S_{20} + \frac{1}{42} = \frac{155685007 + 5542680}{232792560} = \frac{161227687}{232792560} \] Compare this to \( S_{20} = \frac{155685007}{232792560} \). The approximation \( \frac{161227687}{232792560} \) is closer to \( \ln(2) \) than \( \frac{155685007}{232792560} \).

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Key Concepts

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

Convergent Series
A convergent series is a series where the sum of its terms approaches a specific value as more terms are added. In mathematical terms, a series \(\[ \sum_{k=1}^{\text{∞}} a_k \]\) is said to be convergent if its sequence of partial sums converges to a limit. This means that as you add up the terms one by one, the total gets closer and closer to this limit.

Alternating series, like the one in our problem, often show this behavior. For an alternating series to be convergent, the absolute value of the terms must decrease over time and approach zero. This is often confirmed using the Alternating Series Test.

Identifying whether a series is convergent is crucial. If a series is convergent, its sum exists and can be approximated using various methods, as we will see.
Partial Sum Approximation
When dealing with endless series, calculating the exact sum is often impossible. Instead, we rely on partial sums to approximate the total sum. A partial sum is simply the sum of the first n terms of a series. We denote it as \( S_n = \sum_{k=1}^{n} a_k \) where n is the number of terms included.

The interesting property of alternating series is that the sum of the series \( S \) lies between any two consecutive partial sums, \( S_n \) and \( S_{n+1} \). By averaging these two partial sums, \( \frac{S_n + S_{n+1}}{2} \), we obtain a better approximation of the true series sum. This is because the oscillation in the values is smoothed out, providing a closer estimate to the actual limit.
Natural Logarithm Approximation
One fascinating application of series summation is approximating functions like the natural logarithm. In the given problem, we use an alternating series to approximate \( \ln(2) \). The series representing this is: \( \ln(2) = \sum_{k=1}^{\text{∞}} (-1)^{k+1} \frac{1}{k} \).

This alternating series converges because its terms decrease in absolute value and approach zero. By calculating partial sums up to a specific term and applying our averaging method, we get an approximation of \( \ln(2) \) that becomes more accurate the further we go. For instance, in the problem, we use up to the 20th term (\( S_{20} \)) and notice that the revised approximation provides a value closer to \( \ln(2) \) than using just the 20th partial sum.
Series Summation
Summation of series—especially infinite series—can be daunting. The key lies in understanding the properties of the series and applying appropriate techniques to approximate sums.

In practice, we often work with partial sums. For example, given an infinite series \( \sum_{k=1}^{\text{∞}} a_k \), the nth partial sum \( S_n \) is the sum of the first n terms: \( S_n = \sum_{k=1}^{n} a_k \). To approximate the entire series, we examine these partial sums as n increases.

Alternating series pose unique characteristics where the exact sum lies between consecutive partial sums. This enables the use of averaging to improve estimations. By computing \( \frac{S_n + S_{n+1}}{2} \), we smooth out the oscillations between sums, achieving a superior approximation to the sum of an alternating series like the approximation of \( \ln(2) \) shown in the worked example.

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Most popular questions from this chapter

The associative and distributive laws of addition allow us to add finite sums in any order we want. That is, if \(\sum_{k=0}^{n} a_{k}\) and \(\sum_{k=0}^{n} b_{k}\) are finite sums of real numbers, then $$ \sum_{k=0}^{n} a_{k}+\sum_{k=0}^{n} b_{k}=\sum_{k=0}^{n}\left(a_{k}+b_{k}\right) $$ However, we do need to be careful extending rules like this to infinite series. a. Let \(a_{n}=1+\frac{1}{2^{n}}\) and \(b_{n}=-1\) for each nonnegative integer \(n\). \- Explain why the series \(\sum_{k=0}^{\infty} a_{k}\) and \(\sum_{k=0}^{\infty} b_{k}\) both diverge. \- Explain why the series \(\sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right)\) converges. \- Explain why $$ \sum_{k=0}^{\infty} a_{k}+\sum_{k=0}^{\infty} b_{k} \neq \sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right) $$ This shows that it is possible to have to two divergent series \(\sum_{k=0}^{\infty} a_{k}\) and \(\sum_{k=0}^{\infty} b_{k}\) but yet have the series \(\sum_{k=0}^{\infty}\left(a_{k}+b_{k}\right)\) converge. b. While part (a) shows that we cannot add series term by term in general, we can under reasonable conditions. The problem in part (a) is that we tried to add divergent series. In this exercise we will show that if \(\sum a_{k}\) and \(\sum b_{k}\) are convergent series, then \(\sum\left(a_{k}+b_{k}\right)\) is a convergent series and $$ \sum\left(a_{k}+b_{k}\right)=\sum a_{k}+\sum b_{k} $$ - Let \(A_{n}\) and \(B_{n}\) be the \(n\) th partial sums of the series \(\sum_{k=1}^{\infty} a_{k}\) and \(\sum_{k=1}^{\infty} b_{k}\), respectively. Explain why $$ A_{n}+B_{n}=\sum_{k=1}^{n}\left(a_{k}+b_{k}\right) $$ \- Use the previous result and properties of limits to show that $$ \sum_{k=1}^{\infty}\left(a_{k}+b_{k}\right)=\sum_{k=1}^{\infty} a_{k}+\sum_{k=1}^{\infty} b_{k} . $$ (Note that the starting point of the sum is irrelevant in this problem, so it doesn't matter where we begin the sum.) c. Use the prior result to find the sum of the series \(\sum_{k=0}^{\infty} \frac{2^{k}+3^{k}}{5^{k}}\).

The examples we have considered in this section have all been for Taylor polynomials and series centered at 0 , but Taylor polynomials and series can be centered at any value of \(a\). We look at examples of such Taylor polynomials in this exercise. a. Let \(f(x)=\sin (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=\frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(x=\frac{\pi}{2}\). Why should you have expected the result? b. Let \(f(x)=\ln (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=1\). Then find the Taylor series for \(f(x)\) centered at \(x=1\).

Based on the examples we have seen, we might expect that the Taylor series for a function \(f\) always converges to the values \(f(x)\) on its interval of convergence. We explore that idea in more detail in this exercise. Let \(f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}} & \text { if } x \neq 0, \\ 0 & \text { if } x=0 .\end{array}\right.\) a. Show, using the definition of the derivative, that \(f^{\prime}(0)=0\). b. It can be shown that \(f^{(n)}(0)=0\) for all \(n \geq 2\). Assuming that this is true, find the Taylor series for \(f\) centered at 0 . c. What is the interval of convergence of the Taylor series centered at 0 for \(f ?\) Explain. For which values of \(x\) the interval of convergence of the Taylor series does the Taylor series converge to \(f(x) ?\)

We can use power series to approximate definite integrals to which known techniques of integration do not apply. We will illustrate this in this exercise with the definite integral \(\int_{0}^{1} \sin \left(x^{2}\right) d s\) a. Use the Taylor series for \(\sin (x)\) to find the Taylor series for \(\sin \left(x^{2}\right) .\) What is the interval of convergence for the Taylor series for \(\sin \left(x^{2}\right) ?\) Explain. b. Integrate the Taylor series for \(\sin \left(x^{2}\right)\) term by term to obtain a power series expansion for \(\int \sin \left(x^{2}\right) d x\) c. Use the result from part (b) to explain how to evaluate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\). Determine the number of terms you will need to approximate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\) to 3 decimal places.

Match the formulas with the descriptions of the behavior of the sequence as \(n \rightarrow \infty\). 1\. \(s_{n}=n(n+1)-1\) 2\. \(s_{n}=1 /(n+1)\) 3\. \(s_{n}=3-1 / n\) 4\. \(s_{n}=n \sin (n) /(n+1)\) 5\. \(s_{n}=(n+1) / n\) A. does not converge, but doesn'\operatorname{tg} o ~ t o ~ \(\pm \infty\) B. converges to three from below C. diverges to \(\infty\) D. converges to one from above E. converges to zero through positive numbers

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