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Chapter 8: Problem 1

As in Example 1, use the ratio test to find the radius of convergence \(R\) for the given power series. $$ \sum_{n=0}^{\infty} \frac{t^{n}}{2^{n}} $$

Short Answer

Expert verified
Answer: The radius of convergence for the given power series is R = 2.

Step by step solution

01

Write down the general term of the series and the ratio test formula

The general term of the series is \(a_n = \frac{t^n}{2^n}\). The ratio test states that the series converges if \(\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} < 1\) and diverges if the limit is greater than 1.
02

Substitute the next term and the current term in the formula

To apply the ratio test, we need to find the ratio of absolute values of consecutive terms: $$\frac{ |a_{n+1}| }{ |a_n| } = \frac{|\frac{t^{n+1}}{2^{n+1}}|}{|\frac{t^n}{2^n}|}$$
03

Simplify the expression

We can simplify the expression as shown below: $$ \frac{|\frac{t^{n+1}}{2^{n+1}}|}{|\frac{t^n}{2^n}|} = \frac{t^{n+1}2^n}{t^n 2^{n+1}} = \frac{t}{2} |t|$$
04

Find the limit as n approaches infinity

Note that the simplified expression does not actually depend on the value of \(n\). Therefore, it holds for any \(n\): $$ \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \frac{1}{2} |t| $$
05

Determine the range of values for t using the ratio test

According to the ratio test, the series converges if the limit is less than 1. We can set up an inequality: $$\frac{1}{2}|t| < 1$$ To solve this inequality, we just need to divide both sides by 1/2: $$ |t| < 2$$
06

Find the radius of convergence

Since the series converges for \(|t| < 2\), the radius of convergence is \(R = 2\).

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Key Concepts

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

Ratio Test
Understanding the ratio test is crucial for determining the convergence of series with terms that can be complex functions of the variable involved. In essence, the ratio test provides a criterion for the convergence by analyzing the limit of the absolute ratio of successive terms in the series.

Here's how the ratio test works: given a sequence \(a_n\), compute the limit \(L\) of \(\frac{|a_{n+1}|}{|a_n|}\) as \(n\) approaches infinity. If \(L < 1\), the series converges absolutely; if \(L > 1\), it diverges; and if \(L = 1\), the test is inconclusive, meaning we cannot determine convergence or divergence from this test alone.
Power Series
A power series is an infinite series of the form \(\sum_{n=0}^{\infty} a_n (x - c)^n\), where \(a_n\) are the coefficients of the series, \(x\) is a variable, and \(c\) is the center of the series. Power series are incredibly versatile in mathematics, particularly for representing functions as infinite polynomials. They can be used to approximate functions and solve differential equations.

One of the significant problems when working with power series is determining the interval or radius within which the series converges, known as the radius of convergence. This is a measure of the largest interval around the center \(c\) within which the series converges to a finite number.
Limit
In calculus, a limit is a fundamental concept that describes the value that a function or sequence 'approaches' as the input or index approaches some value. Limits help us define and understand behavior in points of discontinuity, infinity, and near inflection points.

The concept of limits is not restricted to functions alone but also applies to sequences, as seen in the ratio test for series convergence. Limits can sometimes be straightforward, but often require deeper analysis or application of specialized rules (such as L'Hôpital's rule) to compute.
Convergence of Series
The convergence of series is a topic at the heart of analysis, which deals with the conditions under which an infinite series will sum to a finite value. A convergent series, unlike a divergent one, does not 'explode' to infinity or oscillate without settling down to a particular value.

To check for series convergence, various tests can be applied, such as the comparison test, the integral test, and the root test. Each test has specific scenarios where it is most effective. The convergence or divergence of a series has profound implications in mathematical analysis and related fields, as it often dictates the behavior of functions and the success of numerical methods for solving equations.

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Most popular questions from this chapter

In each exercise, determine the polynomial \(P(t)\) of smallest degree that causes the given differential equation to have the stated properties. \(\begin{array}{ll}y^{\prime \prime}+\frac{1}{t} y^{\prime}+\frac{1}{P(t)} y=0 & \begin{array}{l}\text { There is a regular singular point at } t=0 . \text { All other } \\ \text { points are ordinary points. }\end{array}\end{array}\)

Exercises 31 and 32 outline the proof of parts (a) and (b) of Theorem 8.2, respectively. In each exercise, consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), where \(p\) and \(q\) are continuous on the domain \((-b,-a) \cup(a, b), a \geq 0\). Now let \(p\) and \(q\) be analytic at \(t=0\) with a common radius of convergence \(R>0\), where \(p\) is an odd function and \(q\) is an even function. (a) Let \(f_{1}(t)\) and \(f_{2}(t)\) be solutions of the given differential equation, satisfying initial conditions \(f_{1}(0)=1, f_{1}^{\prime}(0)=0, f_{2}(0)=0, f_{2}^{\prime}(0)=1\). What does Theorem \(8.1\) say about the solutions \(f_{1}(t)\) and \(f_{2}(t)\) ? (b) Use the results of Exercise 31 to show that \(f_{1}(-t)\) and \(f_{2}(-t)\) are also solutions on the interval \(-R

In each exercise, an initial value problem is given. Assume that the initial value problem has a solution of the form \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\), where the series has a positive radius of convergence. Determine the first six coefficients, \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\). Note that \(y(0)=a_{0}\) and that \(y^{\prime}(0)=a_{1}\). Thus, the initial conditions determine the arbitrary constants. In Exercises 40 and 41 , the exact solution is given in terms of exponential functions. Check your answer by comparing it with the Maclaurin series expansion of the exact solution. $$ y^{\prime \prime}+(1+t) y^{\prime}+y=0, \quad y(0)=-1, \quad y^{\prime}(0)=1 $$

As in Example 1, use the ratio test to find the radius of convergence \(R\) for the given power series. $$ \sum_{n=0}^{\infty} \frac{\sqrt{n}}{2^{n}}(t-4)^{n} $$

Consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\). In some cases, we may be able to find a power series solution of the form \(y(t)=\sum_{n=0}^{\infty} a_{n}\left(t-t_{0}\right)^{n}\) even when \(t_{0}\) is not an ordinary point. In other cases, there is no power series solution. (a) The point \(t=0\) is a singular point of \(t y^{\prime \prime}+y^{\prime}-y=0\). Nevertheless, find a nontrivial power series solution, \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\), of this equation. (b) The point \(t=0\) is a singular point of \(t^{2} y^{\prime \prime}+y=0\). Show that the only solution of this equation having the form \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\) is the trivial solution.
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