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Chapter 11: Problem 4

Find the radius of convergence of each of the series in Exercises 1-12. $$ \sum_{n=0}^{\infty} \sqrt{n} x^{n+1} $$

Short Answer

Expert verified
The radius of convergence is 1.

Step by step solution

01

Identify the Series Form

The given series is \( \sum_{n=0}^{\infty} \sqrt{n} x^{n+1} \). This can be rewritten as \( \sum_{n=0}^{\infty} \sqrt{n} x^{n} \cdot x \), implying each term is \( a_n = \sqrt{n} x^{n+1} \).
02

Apply the Ratio Test

To find the radius of convergence, we use the ratio test. For the sequence \( a_n \), compute the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \):\[a_{n+1} = \sqrt{n+1} x^{n+2} \]\(a_n = \sqrt{n} x^{n+1}\)\[\text{Ratio} = \left| \frac{\sqrt{n+1} x^{n+2}}{\sqrt{n} x^{n+1}} \right| = \left| x \right| \frac{\sqrt{n+1}}{\sqrt{n}}\]
03

Simplify the Ratio

The ratio becomes \( \left| x \right| \frac{\sqrt{n+1}}{\sqrt{n}} = \left| x \right| \sqrt{1 + \frac{1}{n}} \). As \( n \to \infty \), \( \sqrt{1 + \frac{1}{n}} \to 1 \), hence this simplifies to \( |x| \times 1 \).
04

Set Up the Convergence Inequality

By the Ratio Test, the series converges if \( |x| < 1 \). Thus, the inequality is \( |x| < 1 \).
05

Determine the Radius of Convergence

The radius of convergence \( R \) is determined as the value such that \( |x| < R \) for convergence. Hence, \( R = 1 \).

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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 powerful tool for determining the convergence of infinite series. It involves comparing the absolute value of the ratio of successive terms. In general, for a series \( \sum a_n \), the Ratio Test states:
  • Calculate \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In the context of our exercise, the Ratio Test helps us determine the radius of convergence of the series \( \sum_{n=0}^{\infty} \sqrt{n} x^{n+1} \). By applying the test to our series, we found that the limit \( \lim_{n \to \infty} \left| x \right| \sqrt{1 + \frac{1}{n}} = |x| \). Thus, for convergence, \( |x| < 1 \). This provides us with vital information about the behavior of the series depending on \( x \).
Series Convergence
Understanding series convergence involves determining whether the sum of an infinite series approaches a finite number. Not all series converge.
  • A series is convergent if its terms approach zero as \( n \) approaches infinity and the partial sums approach a specific finite value.
  • If these conditions aren't met, the series is said to be divergent.
In the given problem, the condition \( |x| < 1 \) implies convergence due to the Ratio Test. Within this interval, the partial sums of the series \( \sum_{n=0}^{\infty} \sqrt{n} x^{n+1} \) will approach a finite limit as opposed to diverging. To determine convergence, tools like the Ratio Test or the Root Test often come in handy. However, always remember that each series might exhibit unique behaviors based on its composition.
Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are coefficients and \( a \) is the center of the series.
  • Power series convergence depends heavily on the value of \( x \) relative to the center \( a \).
  • The interval within which a power series converges is determined by its radius of convergence, \( R \).
  • The series will converge for all \( x \) such that \( |x-a| < R \).
In the context of our exercise, the power series is \( \sum_{n=0}^{\infty} \sqrt{n} x^{n+1} \) with center 0. After using the Ratio Test, we discovered that \( R = 1 \). Hence, the series converges for \( |x| < 1 \). Understanding power series is crucial because it helps us approximate complicated functions and solve differential equations.
Differential Equations
Differential equations are mathematical equations that describe how a function changes over time or space, often related to physical phenomena. Here's how they interact with series:
  • Many solutions to differential equations can be expressed as power series.
  • Power series provide an effective means to approximate solutions where standard techniques fail, especially for non-linear differential equations.
  • Finding the radius of convergence for a power series linked to a differential equation helps ensure the solution is valid over a desired interval.
In many cases, solving a differential equation might involve finding a power series solution, then determining its convergence radius. Even in our exercise, recognizing the radius of convergence has implicit connections to differential equations. It assures that within the determined interval, the approximations or solutions remain valid and meaningful.

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Most popular questions from this chapter

Show that \(y(x)=(1+x)^{p}\), where \(p\) is an arbitrary constant, is the unique solution of the initial value problem $$ (1+x) y^{\prime}=p y, \quad y(0)=1 . $$ Find a series solution of the initial value problem, showing that $$ \begin{aligned} (1+x)^{p}=& 1+p x+\frac{p(p-1)}{2 !} x^{2}+\cdots \\ &+\frac{p(p-1) \cdots(p-n+1)}{n !} x^{n}+\cdots \end{aligned} $$ Notice that if \(p\) is a positive integer, the series truncates at \(n=p\), and the result is the binomial formula. The series for general \(p\) is called the binomial series. What is the radius of convergence of the binoimal series?

In Exercises 19-24 find the power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) for the function \(f(x)\). $$ f(x)=e^{x}+\sum_{n=2}^{\infty} \frac{x^{n-2}}{n(n+1)} $$

In Exercises 11-25, find two Frobenius series solutions. $$ 2\left(x+x^{2}\right) y^{\prime \prime}+y^{\prime}-y=0 $$

In Exercises 15-20, verify that \(x_{0}=0\) is an ordinary point of the given differential equation. Then find two linearly independent solutions to the differential equation valid near \(x_{0}=0\). Estimate the radius of convergence of the solutions. $$ y^{\prime \prime}+x y^{\prime}-y=0 $$

Many differential equations which arise in applications can be transformed into Bessel's equation. In Exercises 8-12 we will explore some of the possibilities. Show that the substitution \(y(x)=u(\mu x)\) transforms the equation \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(\mu^{2} x^{2}-p^{2}\right) y=0\) into Bessel's equation of order \(p\) and the general solution is \(y(x)=\) \(a J_{p}(\mu x)+b Y_{p}(\mu x)\)
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