/*! 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 7 In this exercise, we examine one... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 7

In this exercise, we examine one of the conditions of the Alternating Series Test. Consider the alternating series $$ 1-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{9}+\frac{1}{4}-\frac{1}{16}+\cdots $$ where the terms are selected alternately from the sequences \(\left\\{\frac{1}{n}\right\\}\) and \(\left\\{-\frac{1}{n^{2}}\right\\}\). a. Explain why the \(n\) th term of the given series converges to 0 as \(n\) goes to infinity. b. Rewrite the given series by grouping terms in the following manner: $$ (1-1)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{9}\right)+\left(\frac{1}{4}-\frac{1}{16}\right)+\cdots $$ Use this regrouping to determine if the series converges or diverges. c. Explain why the condition that the sequence \(\left\\{a_{n}\right\\}\) decreases to a limit of 0 is included in the Alternating Series Test.

Short Answer

Expert verified
The nth term converges to 0; regrouped series suggests potential divergence; the decrease condition ensures proper term behavior leading to convergence.

Step by step solution

01

Show that the nth term converges to 0

Consider the sequences \(\frac{1}{n}\) and \(-\frac{1}{n^{2}}\). As \(n\) approaches infinity, \(\frac{1}{n}\) approaches 0 and \(-\frac{1}{n^{2}}\) also approaches 0. Therefore, the general nth term of the series, which is an alternating term from these two sequences, will also approach 0 as \(n\) goes to infinity.
02

Group the terms of the series

Rewrite the series by grouping: \((1-1) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{9}\right) + \left(\frac{1}{4} - \frac{1}{16}\right) + \cdots\).
03

Simplify each grouped term

Simplify each group in the series: \((1-1) = 0\) \(\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{4}\) \(\left(\frac{1}{3} - \frac{1}{9}\right) = \frac{2}{9}\) \(\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16}\).
04

Determine convergence or divergence

Consider the simplified terms: \(0, \frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \cdots\). This sequence of positive terms tends to become smaller but does not necessarily sum to a finite value. As the denominator grows, the positive terms tend to be small but non-zero, implying potential divergence since the series forms a harmonic-like series.
05

Explain the condition of the Alternating Series Test

The Alternating Series Test requires that the terms of the series decrease to 0. The provided series has parts that satisfy this as \(-\frac{1}{n^{2}}\) decreases, but \(\frac{1}{n}\) does not. This is why the test ensures that terms approaching zero indicate a potential convergence based on the smaller negative contributions dominating.

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Key Concepts

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

Series Convergence
When we talk about **Series Convergence**, we're looking at whether adding up all the terms in a series will give us a finite number. This is important in mathematics because it helps us understand long-term behavior of sequences and series, rather than just individual terms or small subsets.

For a series \(\textstyle \sum a_n \) to converge, the sum of its terms as it progresses towards infinity must approach a specific number. This doesn't mean the individual terms approach a set value, but rather their cumulative addition trends towards convergence.

Take the given series, for example. To determine if it converges, we regroup the terms and simplify them: \(1 - 1 = 0 \), \( \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \), \( \frac{1}{3} - \frac{1}{9} = \frac{2}{9} \). Each term is positive and decreasing but continues to grow smaller without reaching zero. This tells us the series might not sum to a finite number, hence hints at **divergence**.
Mathematical Sequences
To understand **Mathematical Sequences**, begin with the idea of *ordered lists* of numbers that follow a specific pattern. For instance, the sequences given in the exercise \( \frac{1}{n} \) and \( -\frac{1}{n^2} \) both consist of terms that decrease as \( n \) increases.

Here's a closer look:
  • **Sequence \( \frac{1}{n} \)**: This is a *harmonic sequence*, where each term is the reciprocal of an integer. As \( n \) increases, \( \frac{1}{n} \) gets smaller and approaches **0**.
  • **Sequence \( -\frac{1}{n^2} \)**: These are negative fractions where the numerator is always -1 and the denominator is the square of an integer. As \( n \) increases, \( -\frac{1}{n^2} \) also tends towards **0**.
Loving sequences means appreciating their predictable behaviors as they progress. Their convergence behavior often gives insight into the larger series they contribute to.
Harmonic Series Analysis
The **Harmonic Series** is a specific type of series known for its interesting properties related to convergence. Mathematically, it's expressed as:
\(\textstyle \sum \frac{1}{n} \) where \( n = 1, 2, 3, ...\).

Here's the catch: even though each term \( \frac{1}{n} \) gets smaller and smaller, the **total sum** never settles on a finite number. This is a classic case of **divergence**.

In the provided exercise, each component sequence - harmonic \( \frac{1}{n} \) and the quadratic \( \frac{1}{n^2} \) contribute to an overall behavior. Harmonic terms \( \frac{1}{n} \) decrease slower compared to negative quadratic terms \( -\frac{1}{n^2} \). Therefore, when we combine them in an alternating way, we have to be cautious. The positive harmonic terms might dominate leading to divergence unless controlled by stricter conditions.

Understanding the harmonic series is crucial because it shows how sequences can behave differently. Even as individual terms appear negligible, their accumulative sum can tell an opposite story. This insight is key to mastering concepts of series and sequence convergence.

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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 \(\sum_{k=0}^{\infty} \frac{2^{k}+3^{k}}{5^{k}}\).

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.

In this exercise we will begin with a strange power series and then find its sum. The Fibonacci sequence \(\left\\{f_{n}\right\\}\) is a famous sequence whose first few terms are$$ f_{0}=0, f_{1}=1, f_{2}=1, f_{3}=2, f_{4}=3, f_{5}=5, f_{6}=8, f_{7}=13, \cdots, $$ where each term in the sequence after the first two is the sum of the preceding two terms. That is, \(f_{0}=0, f_{1}=1\) and for \(n \geq 2\) we have $$ f_{n}=f_{n-1}+f_{n-2} $$ Now consider the power series $$ F(x)=\sum_{k=0}^{\infty} f_{k} x^{k} $$ We will determine the sum of this power series in this exercise. a. Explain why each of the following is true. $$ \text { i. } x F(x)=\sum_{k=1}^{\infty} f_{k-1} x^{k} $$ ii. \(x^{2} F(x)=\sum_{k=2}^{\infty} f_{k-2} x^{k}\) b. Show that $$ F(x)-x F(x)-x^{2} F(x)=x $$ c. Now use the equation $$ F(x)-x F(x)-x^{2} F(x)=x $$ to find a simple form for \(F(x)\) that doesn't involve a sum. d. Use a computer algebra system or some other method to calculate the first 8 derivatives of \(\frac{x}{1-x-x^{2}}\) evaluated at \(0 .\) Why shouldn't the results surprise you?

Suppose you drop a golf ball onto a hard surface from a height \(h\). The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height \(\frac{3}{4}\) of the height from which it dropped. Let \(h_{n}\) be the height of the ball on the \(n\) th bounce, with \(h_{0}=h .\) In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance. a. Determine a formula for \(h_{1}\) in terms of \(h\). b. Determine a formula for \(h_{2}\) in terms of \(h\). c. Determine a formula for \(h_{3}\) in terms of \(h\). d. Determine a formula for \(h_{n}\) in terms of \(h\). e. Write an infinite series that represents the total distance traveled by the ball. Then determine the sum of this series. f. Next, let's determine the total amount of time the ball is in the air. i) When the ball is dropped from a height \(H,\) if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time \(t\) is given by $$ H-\frac{1}{2} g t^{2} $$ Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height \(H\). ii) Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air.

In this exercise we investigation the Taylor series of polynomial functions. a. Find the 3 rd order Taylor polynomial centered at \(a=0\) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Does your answer surprise you? Explain. b. Without doing any additional computation, find the 4 th, 12 th, and 100 th order Taylor polynomials (centered at \(a=0\) ) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Why should you expect this? c. Now suppose \(f(x)\) is a degree \(m\) polynomial. Completely describe the \(n\) th order Taylor polynomial (centered at \(a=0\) ) for each \(n\).
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