/*! 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 29 a. Use the Taylor series for \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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a. Use the Taylor series for \(\sin x\) and the Alternating Series Estimation Theorem to show that $$ 1-\frac{x^{2}}{6}<\frac{\sin x}{x}<1, \quad x \neq 0 $$ b. Graph \(f(x)=(\sin x) / x\) together with the functions \(y=1-\left(x^{2} / 6\right)\) and \(y=1\) for \(-5 \leq x \leq 5 .\) Comment on the relationships among the graphs.

Short Answer

Expert verified
Taylor series and Alternating Series Theorem confirm the inequalities; graphing them shows how \((\sin x)/x\) is bounded by 1 and \(1 - x^2/6\).

Step by step solution

01

Taylor Series for \(\sin x\)

The Taylor series expansion for \( \sin x \) around 0 is given by:\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]This series alternates between positive and negative terms. The terms we are concerned with here are the first few odd powers of \( x \).
02

Divide \(\sin x\) by \(x\)

Consider the function \( \frac{\sin x}{x} \). Using the Taylor series for \( \sin x \), we have:\[ \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \]This series also alternates, and is only defined for \( x eq 0 \).
03

Apply the Alternating Series Estimation Theorem

According to the Alternating Series Estimation Theorem, for a convergent alternating series \[ S = a_1 - a_2 + a_3 - a_4 + \cdots \] the remainder \( R_n \) of using the first \( n \) terms to approximate the sum \( S \) is bounded by the absolute value of the first neglected term.In our case, the next term after \( 1 - \frac{x^2}{6} \) is:\[ \frac{x^4}{120} \]Thus:\[ \left| R \right| < \frac{x^4}{120} \]
04

Establish the Inequalities

Given the remainder's estimation, we have:\[ 1 - \frac{x^2}{6} < \frac{\sin x}{x} < 1 \]This matches our requirement in the problem statement.
05

Graph the Functions

Graph \( f(x) = \frac{\sin x}{x} \), \( y = 1 \), and \( y = 1 - \frac{x^2}{6} \) over the interval \(-5 \le x \le 5\). Notice how \( \frac{\sin x}{x} \) stays between \( y = 1 \) and \( y = 1 - \frac{x^2}{6} \) for values of \( x \) other than zero. This visualizes the alternating series bound.

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Key Concepts

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

Alternating Series Estimation Theorem
The Alternating Series Estimation Theorem is a powerful tool for analyzing convergent series where the signs alternate between positive and negative. This theorem provides a way to determine how close a partial sum of an alternating series is to the actual sum.

For an alternating series of the form:
  • \( S = a_1 - a_2 + a_3 - a_4 + \cdots \)
the theorem states that if the series is decreasing in magnitude and approaches zero, the error or remainder \( R_n \) when estimating \( S \) using the first \( n \) terms is less than or equal to the first omitted term \( a_{n+1} \).
  • In simpler terms, this means if you stop the series at the \( n \)th term, the difference between your partial sum and the true sum \( S \) will be less than the next term following \( n \).
In the context of the Taylor series for \( \sin x \), the alternating series estimation helps to prove that \( 1 - \frac{x^2}{6} < \frac{\sin x}{x} < 1 \). By neglecting the term \( \frac{x^4}{120} \) and noting that its magnitude is small, we can see that it's the key to bounding the error of the partial sum, thus ensuring the inequality holds true for \( x eq 0 \).
Graphical Analysis
Graphical analysis provides a visual way to understand mathematical concepts like inequalities and bounds. When you graph the function \( f(x) = \frac{\sin x}{x} \) along with the functions \( y = 1 \) and \( y = 1 - \frac{x^2}{6} \), you get a clear view of how \( f(x) \) behaves relative to these bounds.

Within the interval
  • \( -5 \leq x \leq 5 \), we observe the graph of \( \frac{\sin x}{x} \) weaving between the other two lines.
  • The function \( \frac{\sin x}{x} \) approaches 1 as \( x \) approaches zero.
  • Outside of x = 0, it remains constrained by the red (or theoretical) boundary \( 1 - \frac{x^2}{6} \).
Graphical analysis doesn't just tell us where functions lie but also emphasizes their behavior across an interval visually. Through this perspective, it's clear how \( \frac{\sin x}{x} \) interacts with its surrounding bounds, affirming the inequality produced by the series estimation.
Inequalities involving Functions
Inequalities involving functions are essential in helping us comprehend limits and approximations in calculus. These inequalities often depict the relationships between different mathematical expressions or functions.

When proving inequalities such as
  • \( 1 - \frac{x^2}{6} < \frac{\sin x}{x} < 1 \),
we seek to understand how closely a function approximates another or its bound.
  • This task involves mathematical tools such as Taylor series expansions and error estimations.
  • These tools allow us to provide valid ranges or intervals for which certain properties hold true.
In this particular exercise, the Taylor series and the Alternating Series Estimation Theorem effectively illustrate how \( \frac{\sin x}{x} \) fits neatly between 1 and \( 1 - \frac{x^2}{6} \). This is achieved by taking the Taylor series for \( \sin x \), simplifying it, and observing where the smaller terms fall in relation to these bounds.

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Most popular questions from this chapter

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1\)

In Exercises \(45-48\) , estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. $$ \frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^{n} t^{n}, \quad 0 < t < 1 $$

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)

Prove that \(\lim _{n \rightarrow \infty} x^{1 / n}=1,(x>0)\).

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