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Chapter 8: Problem 11

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

Short Answer

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

Step by step solution

01

Write down the given power series

The given power series is: $$ \sum_{n=0}^{\infty} \frac{\sqrt{n}}{2^{n}}(t-4)^{n}$$
02

Apply the ratio test

The ratio test involves taking the limit as n approaches infinity of the ratio of the (n+1)-th term to the nth term of the series. Let \(a_n = \frac{\sqrt{n}}{2^{n}}(t-4)^{n}\) be the nth term of the series. Then the ratio test gives us: $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{\frac{\sqrt{n+1}}{2^{n+1}}(t-4)^{n+1}}{\frac{\sqrt{n}}{2^{n}}(t-4)^{n}}$$
03

Simplify the expression and find the limit

Simplify the expression by canceling out the common terms and grouping the remaining terms together: $$\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}} \cdot \frac{(t-4)^{n+1}}{(t-4)^{n}} \cdot \frac{2^n}{2^{n+1}}$$ Further simplify the expression: $$\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}} \cdot (t-4) \cdot \frac{1}{2}$$ For the series to converge, this limit must be less than 1: $$\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}} \cdot (t-4) \cdot \frac{1}{2} < 1$$
04

Calculate the limit and solve for t

The limit of the fraction is 1, so our inequality becomes: $$(t-4) \cdot \frac{1}{2} < 1$$ Multiply both sides by 2 to isolate (t-4): $$(t-4) < 2$$ Add 4 to both sides to solve for t: $$t < 6$$
05

Determine the radius of convergence

The interval of convergence is centered at 4, and the inequality we derived states that t is less than 6. The radius of convergence (R) can be found by calculating the distance between the center (4) and the endpoint (6): $$R = 6 - 4 = 2$$ The radius of convergence R is 2.

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Key Concepts

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

Ratio Test
The Ratio Test is a method used in mathematics, particularly in calculus, to determine the convergence or divergence of infinite series. When you're given a power series, such as \( \sum_{n=0}^\infty a_n(t-a)^n \), where \( a_n \) represents the terms in the series and \(a\) is the center point of the series, the ratio test can be a vital tool in analyzing its behavior.

Applying the Ratio Test begins by examining the limit of the ratio of the successive terms in the series. In other words, you need to find \( \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} \). If this limit is less than 1, the series converges; if the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive, and you may need to use other tests to determine the convergence of the series.

In the exercise, the Ratio Test is applied to the power series \( \sum_{n=0}^\infty \frac{\sqrt{n}}{2^n}(t-4)^n \) to find the radius of convergence. Simplifying the expression involves canceling common factors and applying limit properties to capture the series' behavior as \(n\) approaches infinity. It's important to remember that the Ratio Test doesn't just conclude convergence but also helps find the radius of the convergence zone around the center point of the series.
Power Series
Power series are expressions formed by an infinite sum of terms, each containing a variable raised to a successive power and multiplied by a coefficient. They look like \( \sum_{n=0}^\infty a_n(t-a)^n \), where the base \(t-a\) signifies the series' expansion around the point \(a\), and \(a_n\) represents the coefficient of the \(n\)-th term.

One of the key properties of power series is that they may not converge for all values of the variable; instead, they often have a specific interval or radius within which they are convergent, known as the radius of convergence. This concept is crucial for understanding functions and their representations as power series, as it dictates the series' scope and utility.

Power series can represent a wide array of functions, and finding the radius of convergence tells us about the function's stability and reliability over that range. The purpose of computing this radius is to identify the values of the variable for which the sum of the series is well-defined and finite. When you're working on developing an understanding of power series, consider them as building blocks that can replicate functions within a certain interval around a central point.
Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus and analysis that deals with the behavior of a sequence of numbers as the index increases indefinitely. In the formal definition, a sequence \( \{a_n\} \) is said to have a limit \( L \) as \( n \) approaches infinity, denoted as \( \lim_{n\to\infty} a_n = L \), if the terms in the sequence get arbitrarily close to \( L \) as \( n \) becomes very large.

To find the limit of a sequence, you often need to simplify the expression and apply limit laws. For example, the limit of the sequence \( \{\sqrt{n+1}/\sqrt{n}\} \) as \( n \) approaches infinity can be found by recognizing that the terms inside the square roots grow without bound and that \( \sqrt{n+1}/\sqrt{n} \) approaches 1.

Grasping the concept of the limit of a sequence is crucial because it helps not only in finding the radius of convergence for power series but also in understanding the behavior of functions and the output of mathematical models for large values of input. It is a notion that not only helps in theoretical mathematics but also has practical applications in engineering, physics, and other sciences where predicting the behavior of systems at large scales is key.

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Most popular questions from this chapter

The Laguerre \({ }^{10}\) differential equation \(t y^{\prime \prime}+(1-t) y^{\prime}+\alpha y=0\) has a regular singular point at \(t=0\). (a) Determine the indicial equation and show that the roots are \(\lambda_{1}=\lambda_{2}=0\). (b) Find the recurrence relation. Show that if \(\alpha=N\), where \(N\) is a nonnegative integer, then the series solution reduces to a polynomial. Obtain the polynomial solution when \(N=5\). The polynomial solutions of this differential equation, when properly normalized, are called Laguerre polynomials. (c) Is the polynomial obtained in part (b) for \(\alpha=N=5\) an even function, an odd function, or neither? Would you expect even and odd solutions of the differential

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. $$ y^{\prime \prime}-2 y^{\prime}+y=0, \quad y(0)=0, \quad y^{\prime}(0)=2, \quad y(t)=2 t e^{t} $$

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.

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

In each exercise, find the singular points (if any) and classify them as regular or irregular. $$ |t| y^{\prime \prime}+y^{\prime}+y=0 $$
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