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Approximations that Get Worse and Worse Apply Newton's method to \(f(x)=x^{1 / 3}\) with \(x_{1}=1,\) and calculate \(x_{2}\) , \(x_{3}, x_{4},\) and \(x_{5} .\) Find a formula for \(\left|x_{n}\right| .\) What happens to \(\left|x_{n}\right|\) as \(n \rightarrow \infty ?\) Draw a picture that shows what is going on.

Short Answer

Expert verified
The sequence \(x_n\) generated by Newton's method for \(f(x) = x^{1/3}\) with initial guess \(x_1=1\) oscillates between -2 and 1. We observe that the absolute value of \(x_n\), \(|x_n|\), alternates between 2 and 1. Therefore, \(|x_n|\) does not have a limit as \(n\) approaches infinity.

Step by step solution

01

Apply Newton's Method for the First Iteration

Start by using the formula for Newton's method, which is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). The derivative of \(f(x) = x^{1/3}\) is \(f'(x) = \frac{1}{3x^{2/3}}\). Now apply Newton's method for \(x_1=1\) to get \(x_2=1 - \frac{1^{1/3}}{\frac{1}{3(1)^{2/3}}} = 1 -3 = -2\).
02

Continue with Newton's Method for Further Iterations

Now apply Newton's method to get \(x_3,x_4,x_5\). Doing so gives \(x_3=-2 - \frac{(-2)^{1/3}}{\frac{1}{3(-2)^{2/3}}}= -2 - (-3)=1\), \(x_4=1 - \frac{1^{1/3}}{\frac{1}{3(1)^{2/3}}} = 1 -3 = -2\) and \(x_5=-2 - \frac{(-2)^{1/3}}{\frac{1}{3(-2)^{2/3}}}=-2 - (-3)=1\). Here we observe that the sequence repeated from \(x_3\). So the cycle is -2, 1.
03

Find a Formula for \(|x_n|\)

Considering the absolute values, \(|x_n|\) alternates between 1 and 2, that is, \(|x_n| = 1\) if \(n\) is odd and \(|x_n| = 2\) if \(n\) is even.
04

Analyze Behavior as \(n \rightarrow \infty\)

As \(n \rightarrow \infty\), there is no limit for \(|x_n|\) since it oscillates between 1 and 2.
05

Draw a Picture

The figure would show an oscillation between -2 and 1 for the values of \(x_n\) as \(n\) increases. This could be represented graphically by a series of vertical jumps between -2 and 1. Unfortunately, the output doesn't support images.

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Key Concepts

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

Convergence and Divergence
In calculus, convergence and divergence are key concepts when considering the behavior of sequences and series. The convergence of a sequence or function expresses the idea that as we progress through the terms (i.e., as the index n increases), the terms get closer and closer to a specific value, known as the limit. Conversely, divergence implies that as we move through the terms, they do not settle down to any single value. Instead, they may oscillate, increase, or decrease without bounds.

In the case of the exercise with Newton's Method, we see that the sequence does not converge. Interestingly, the method, which is a powerful tool for finding roots, does not always guarantee convergence. The provided sequence \<|x_n|\> oscillates between 1 and 2, meaning it neither approaches a single value nor goes to infinity; it simply keeps jumping back and forth indefinitely. This behavior is an example of divergence in that the sequence lacks a limit as \
Iterative Methods
Newton's Method is one of the many iterative methods used in calculus and numerical analysis to find approximations to solutions. Iterative methods involve the repetition of a specific process where each subsequent iteration is built upon the previous one. These methods are particularly useful because they can provide successively better approximations to solutions of equations, especially when an exact solution is difficult to obtain.

With each step of Newton's Method, a new guess \(x_{n+1}\) is calculated based on the current guess \(x_n\) and the function values and derivatives at that point. This process leverages the power of derivatives to inform us how to adjust our guess to approximate the root of a function. While Newton's Method can be highly efficient when conditions are favorable, it does rely on the function and initial guess being well-suited to the method. The divergence in the given exercise exhibits a pitfall of iterative methods—without careful consideration, they can lead to sequences that do not converge to a solution.
Limits and Infinity
A fundamental concept in calculus is the notion of limits and infinity. When we talk about \(n \rightarrow \infty\), we're interested in the behavior of sequences as \(n\) becomes very large. In some cases, sequences display a pattern as \(n\) grows without bound; they might approach a fixed value, becoming arbitrarily close to that value as \(n\) increases—that would be the limit of the sequence.

In other scenarios, as demonstrated in our exercise, the terms do not close in on a fixed value, and thus, we say the limit does not exist. Divergence to infinity is another possible behavior where the terms grow larger and larger without bound; however, in our case, the terms don't grow; they just oscillate between two fixed values. This oscillatory nature informs us that the sequence will never settle at a single point, highlighting the importance of understanding limits and infinite behavior in sequence analysis and calculus.
Derivative Application
The derivative is a tool in calculus that measures how a function changes as its input changes. Its essential role in Newton's Method showcases a fascinating application of derivatives beyond just finding slopes of tangents. In the iterative formula of Newton's Method, \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), the derivative \(f'(x_n)\) tells us about the behavior of the function near \(x_n\) and how we should adjust our approximation to get closer to the function's root.

In the alternating sequence revealed in the textbook problem, the derivatives play a crucial part not only in the iteration steps but also in creating the pattern of oscillation that we observe. This example underscores the derivative's importance in estimating and adjusting during the iterative process, and it provides a critical lesson: the application of derivatives must be coupled with a thorough understanding of both the function's behavior and the methodology for the results to be reliable and meaningful.

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Most popular questions from this chapter

Industrial Production (a) Economists often use the expression expression "rate of growth" in relative rather than absolute terms. For example, let \(u=f(t)\) be the number of people in the labor force at time \(t\) in a given industry. (We treat this function as though it were differentiable even though it is an integer-valued step function.) Let \(v=g(t)\) be the average production per person in the labor force at time \(t .\) The total production is then \(y=u v\) . If the labor force is growing at the rate of 4\(\%\) per year year \((d u / d t=\) 0.04\(u\) ) and the production per worker is growing at the rate of 5\(\%\) per year \((d v / d t=0.05 v),\) find the rate of growth of the total production, y. (b) Suppose that the labor force in part (a) is decreasing at the rate of 2\(\%\) per year while the production per person is increasing at the rate of 3\(\%\) per year. Is the total production increasing, or is it decreasing, and at what rate?

Stiffness of a Beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. (a) Find the dimensions of the stiffest beam that can be cut from a 12-in. diameter cylindrical log. (b) Writing to Learn Graph \(S\) as a function of the beam's width \(w,\) assuming the proportionality constant to be \(k=1 .\) Reconcile what you see with your answer in part (a). (c) Writing to Learn On the same screen, graph \(S\) as a function of the beam's depth \(d,\) again taking \(k=1 .\) Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of \(k ?\) Try it.

Monotonic Functions Show that monotonic increasing and decreasing functions are one-to-one.

Parallel Tangents Assume that \(f\) and \(g\) are differentiable on \([a, b]\) and that \(f(a)=g(a)\) and \(f(b)=g(b) .\) Show that there is at least one point between \(a\) and \(b\) where the tangents to the graphs of \(f\) and \(g\) are parallel or the same line. Illustrate with a sketch.

Let \(y=f(x)=x^{3}-4 x\) If \(d x / d t=-2 \mathrm{cm} / \mathrm{sec},\) find \(d y / d t\) at the point where (a) \(x=-3 . \quad\) (b) \(x=1 . \quad\) (c) \(x=4\)

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