/*! 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 78 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

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Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$a_{n}=n\left(1-\cos \frac{1}{n}\right)$$

Short Answer

Expert verified
The sequence converges to 0.

Step by step solution

01

Recognize the Sequence

We are provided with the sequence \(a_n = n(1 - \cos \frac{1}{n})\). Our task is to determine if this sequence converges or diverges as \(n \to \infty\).
02

Simplify Using Taylor Expansion for Cosine

As \(n\to\infty\), \(\frac{1}{n}\to 0\). We can use the Taylor expansion \(\cos x \approx 1 - \frac{x^2}{2}\) when \(x\to 0\). Applying this to \(\cos\frac{1}{n}\), we have \(1 - \cos \frac{1}{n} \approx \frac{1}{2n^2}\).
03

Substitute Approximation and Simplify

Substitute the approximation from Step 2 into the sequence, we get:\[a_n \approx n \cdot \frac{1}{2n^2} = \frac{1}{2n}.\]
04

Determine the Limit of the Simplified Sequence

The simplified sequence \(b_n = \frac{1}{2n}\) is a basic sequence that converges to zero as \(n\to\infty\).
05

Conclude the Convergence Result

Since the simplified form \(b_n = \frac{1}{2n}\) converges to 0 as \(n \to \infty\), this shows that the original sequence \(a_n = n(1 - \cos\frac{1}{n})\) also converges to 0.

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Key Concepts

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

Taylor Expansion
The Taylor expansion is a powerful mathematical tool used to approximate complex functions with simpler polynomials. When we're dealing with small values of a variable, Taylor expansion provides an effective way to estimate the value of functions like sine, cosine, and other analytic functions.
In simple terms, it breaks down complex functions into an infinite sum of terms calculated from the derivatives of the function at a single point. For practical calculations, often just the first few terms are enough to give a good approximation.
For the cosine function, particularly as an angle approaches zero, we use the expansion:
  • \(\cos x \approx 1 - \frac{x^2}{2}\)
This approximation is helpful in problems where we encounter values close to zero, such as \(\cos \frac{1}{n}\) as in our sequence context.
Using this approximation makes it easier to simplify and analyze sequences and functions that would otherwise be cumbersome to handle.
Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus that describes the value a sequence approaches as its index (often denoted as n) goes to infinity. Knowing how to find the limit helps us determine whether a sequence converges or diverges.
In simpler terms, if the terms of a sequence keep getting closer to a certain number as the sequence progresses, the sequence has a limit. If the terms don't get closer to any specific number and either grow indefinitely or oscillate, the sequence diverges.
In our sequence example, after simplifying with the Taylor expansion, the sequence \(a_n = \frac{1}{2n}\) makes it clear that as \(n \to \infty\), \(a_n \to 0\).
  • Here each term becomes smaller and closer to 0, illustrating that the sequence converges.
  • Understanding this concept is key for solving many calculus problems involving sequences.
Cosine Approximation
Approximating the cosine function is invaluable in simplifying complex problems that include trigonometric expressions. Cosine approximation often uses Taylor series to estimate its value around x = 0, as we did in the original exercise.
For very small angles (or close to zero), \(\cos x \approx 1 - \frac{x^2}{2}\) serves as an excellent approximation. This is particularly useful when dealing with sequences or series that involve cosine of small arguments, as it allows us to replace the trigonometric function with a polynomial, which is much simpler to handle.
In the sequence example given, transforming \(\cos \frac{1}{n}\) made it possible to simplify \(a_n\) to \(\frac{1}{2n}\), showcasing how approximation transforms the problem solving process. This simplification helps identify the behavior of the sequence as \(n\) tends to infinity, ultimately confirming its convergence to zero.

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Most popular questions from this chapter

Show by example that \(\Sigma\left(a_{n} / b_{n}\right)\) may converge to something other than \(A / B\) even when \(A=\Sigma a_{m}, B=\Sigma b_{n} \neq 0,\) and no \(b_{n}\) equals 0

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=(123456)^{1 / n}$$

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

Sequences generated by Newton's method \(\quad\) Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\). b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(1+x)^{3 / 2}, \quad-\frac{1}{2} \leq x \leq 2$$

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