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Chapter 10: Problem 101

Assume that each sequence converges and find its limit. $$ a_{1}=2, \quad a_{n+1}=\frac{72}{1+a_{n}} $$

Short Answer

Expert verified
The limit of the sequence is 8.

Step by step solution

01

Understand the Recursive Formula

The sequence is defined by the recursive formula: \( a_{n+1} = \frac{72}{1 + a_n} \). This means each term depends on the previous term in the sequence.
02

Assume the Sequence Converges

We are given that the sequence converges to some limit, say \( L \). Therefore, as \( n \to \infty \), \( a_n \to L \). In this case, both \( a_n \) and \( a_{n+1} \) approach the same limit \( L \).
03

Set Up the Limit Equation

From the recursive formula, as \( n \to \infty \) and \( a_n \to L \), we substitute into the recursive formula to get \( L = \frac{72}{1 + L} \). This gives us an equation in terms of \( L \).
04

Solve the Limit Equation

To solve \( L = \frac{72}{1 + L} \), first multiply both sides by \( 1 + L \), which gives \[ L(1 + L) = 72 \]. Then, expand and rearrange the equation to get \( L^2 + L - 72 = 0 \), which is a quadratic equation.
05

Solve the Quadratic Equation

Solve the quadratic equation \( L^2 + L - 72 = 0 \) using the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -72 \). This gives: \[ L = \frac{-1 \pm \sqrt{1 + 288}}{2} = \frac{-1 \pm \sqrt{289}}{2} = \frac{-1 \pm 17}{2} \].
06

Select the Valid Limit

Calculate the solutions from the quadratic formula: \( L = \frac{16}{2} = 8 \) and \( L = \frac{-18}{2} = -9 \). Since a sequence that converges to a negative limit (in this context) does not make practical sense, the valid limit is \( L = 8 \).

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Key Concepts

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

Recursive Formula
A recursive formula is a type of mathematical equation that defines each term in a sequence based on the previous term(s). In the given sequence, the formula is defined as \( a_{n+1} = \frac{72}{1+a_n} \). This recursive approach means that to find any term in the sequence, you rely on the term before it.

Think of a recursive formula as a set of instructions. You start with an initial term鈥攈ere, \( a_1 = 2 \)鈥攁nd use the equation to find the following terms:
  • First, substitute the known term \( a_n \) into the formula.
  • Calculate to find the next term \( a_{n+1} \).
This method is essential for sequences that extend indefinitely. It allows patterns to emerge as you repeatedly apply the formula.
Limit of a Sequence
When we say a sequence has a limit, we're talking about the value that the terms of the sequence get closer to as the sequence progresses. Mathematically, if \( a_n \rightarrow L \) as \( n ightarrow \infty \), then \( L \) is the limit of the sequence. In simple words, no matter how long the sequence goes on, the terms will keep getting closer and closer to the limit value, \( L \).

In our scenario, we assume the sequence converges, meaning there exists some \( L \) such that as \( n \) grows infinitely large, \( a_n \) tends towards \( L \). The main idea here is understanding that converging sequences don't just bumble forever; they have a particular destination value.
Quadratic Equation
A quadratic equation is a second-degree polynomial generally written in the form of \( ax^2 + bx + c = 0 \). The exercise brought us to a quadratic equation after using the limit equation derived from the recursive formula: \( L^2 + L - 72 = 0 \).

To find the roots (solutions) of this equation, we use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -72 \). Solving this gives us potential limits, but only the positive solution \( x = 8 \) is practical for this sequence, as a negative limit doesn't fit the context.

Recognizing when a quadratic equation is necessary for finding sequence limits can be a powerful tool in mathematical reasoning.
Convergence Assumption
The convergence assumption is the presumption that a given sequence will settle to a single value, its limit, as it progresses towards infinity. This assumption simplifies the process of finding the limit by making it feasible to set \( a_n = a_{n+1} = L \) in the limit equation.

This assumption is crucial when solving certain sequence problems as it sets the stage for approaching the problem systematically:
  • Assume \( a_n \rightarrow L \), so we equate the recursive formula at infinity to find such \( L \).
  • It allows us to use the sequence's format to deduce the behavior as it approaches infinity.
Without this assumption, understanding how the sequence evolves long-term would be difficult. It provides a clear goal post and anchors our calculations.

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Most popular questions from this chapter

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) Find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=1 / \sqrt{1-x^{2}}\)

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\) . Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\), \begin{equation} \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right). \end{equation}

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 \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=(123456)^{1 / n} $$

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}.$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

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