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Determine the radius of convergence of the given power series. $$ \sum_{n=1}^{\infty} \frac{\left(x-x_{0}\right)^{n}}{n} $$

Short Answer

Expert verified
Answer: The radius of convergence for the given power series is 1.

Step by step solution

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01

Identify the terms of the series

The given power series is $$ \sum_{n=1}^{\infty} \frac{\left(x-x_{0}\right)^{n}}{n}. $$
02

Apply the Ratio Test

To apply the Ratio Test, we need to find the limit of the ratio of consecutive terms as \(n\) goes to infinity. We compute the absolute value of the ratio of the \((n+1)\)th term to the \(n\)th term: $$ \lim_{n\to\infty} \left|\frac{\frac{(x-x_0)^{n+1}}{n+1}}{\frac{(x-x_0)^n}{n}}\right|. $$
03

Simplify the limit

Simplify the limit expression by dividing the numerator by the denominator inside the absolute value: $$ \lim_{n\to\infty} \left|\frac{(x-x_0)^{n+1}n}{(x-x_0)^n (n+1)}\right|. $$ Then, simplify the powers of \((x-x_0)\): $$ \lim_{n\to\infty} \left|\frac{(x-x_0)n}{n+1}\right|. $$
04

Compute the limit

As \(n\) goes to infinity, the limit becomes: $$ \lim_{n\to\infty} \left|\frac{(x-x_0)n}{n+1}\right|=\left|x-x_0\right|. $$
05

Determine the radius of convergence

For the series to converge, the limit obtained in Step 4 must be less than 1. So, we have the inequality: $$ \left|x-x_0\right| < 1. $$ This inequality defines an interval of convergence centered at \(x_0\) with a radius of convergence equal to 1. Therefore, the radius of convergence for the given power series is \(1\).

Key Concepts

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

Power Series
A power series is an infinite series of the form \[ \sum_{n=0}^{\infty} a_n (x - x_0)^n \.\] Here, \(a_n\) represents the coefficient of the nth term, \(x\) is the variable, and \(x_0\) is the center of the series. As a sum of infinitely many terms, power series can be thought of as polynomials with an infinite number of terms. These series are crucial because they can represent functions in areas of the complex plane where the function is differentiable.

Understanding the power series revolves around knowing where it converges, which means where the sum of its terms forms a finite value. The convergence is determined within a certain radius from the center \(x_0\), known as the radius of convergence. The interval within this radius is where the power series behaves nicely, meaning it converges to a function. Outside this interval, the series may diverge, meaning it does not settle to a finite sum.

In practice, to find the radius of convergence, tests like the Ratio Test are commonly used, which brings us to the next essential concept.
Ratio Test
The Ratio Test is a method used in calculus to determine the absolute convergence of an infinite series. Specifically, in the context of power series, it helps in identifying the radius of convergence. To apply the Ratio Test, you take the limit of the absolute value of the ratio of successive terms of the series.

The formal rule for the Ratio Test says that for a series \( \sum a_n\), if \[ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L < 1 \.\] then the series converges absolutely. If \(L > 1\) or if the limit does not exist, then the series diverges. And if \(L = 1\), the test is inconclusive.

In our example, we utilize the Ratio Test to find the limit as \(n\) approaches infinity for the series' terms. The outcome of this limit, as per the Ratio Test, determines whether the series converges or diverges, and within what range of \(x\) values the convergence occurs. This range, or interval, is symmetric about \(x_0\) and defines the series' radius of convergence.
Limit of a Sequence
The concept of a limit in the context of a sequence is about what value the sequence approaches as the number of terms grows indefinitely large. Formally, the limit of a sequence \(a_n\) as \(n\) goes to infinity, denoted by \(\lim_{n \to \infty} a_n\), is the value that the terms of the sequence get closer to as \(n\) increases without bound.

In our exercise, we compute the limit of the ratio of the power series' consecutive terms to apply the Ratio Test. This limit is a way to assess how the terms of the series behave as they progress towards infinity, which is an essential step in determining the series' convergence.

In this specific example, as \(n\) becomes very large, the ratio of \(\frac{n}{n+1}\) approaches 1. Hence, we find that the limit of the absolute value of \(\left|\frac{(x-x_0)n}{n+1}\right|\) simplifies to \(\left|x-x_0\right|\). It means that as long as \(x\) is within 1 unit of \(x_0\), the terms of the power series will get increasingly closer to zero, which implies convergence within that interval, establishing a radius of convergence equal to 1.

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Most popular questions from this chapter

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that the I.egendre equation can also be written as $$ \left[\left(1-x^{2}\right) y^{\prime}\right]=-\alpha(\alpha+1) y $$ Then it follows that \(\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime}=-n(n+1) P_{n}(x)\) and \(\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}=\) \(-m(m+1) P_{m}(x) .\) By multiplying the first equation by \(P_{m}(x)\) and the second equation by \(P_{n}(x),\) and then integrating by parts, show that $$ \int_{-1}^{1} P_{n}(x) P_{m}(x) d x=0 \quad \text { if } \quad n \neq m $$ This property of the Legendre polynomials is known as the orthogonality property. If \(m=n,\) it can be shown that the value of the preceding integral is \(2 /(2 n+1) .\)

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x(3-x) y^{\prime \prime}+(x+1) y^{\prime}-2 y=0\)

It can be shown that \(J_{0}\) has infinitely many zeros for \(x>0 .\) In particular, the first three zeros are approximately \(2.405,5.520, \text { and } 8.653 \text { (see figure } 5.8 .1) .\) Let \(\lambda_{j}, j=1,2,3, \ldots,\) denote the zeros of \(J_{0}\) it follows that $$ J_{0}\left(\lambda_{j} x\right)=\left\\{\begin{array}{ll}{1,} & {x=0} \\ {0,} & {x=1}\end{array}\right. $$ Verify that \(y=J_{0}(\lambda, x)\) satisfies the differential equation $$ y^{\prime \prime}+\frac{1}{x} y^{\prime}+\lambda_{j}^{2} y=0, \quad x>0 $$ Ilence show that $$ \int_{0}^{1} x J_{0}\left(\lambda_{i} x\right) J_{0}\left(\lambda_{j} x\right) d x=0 \quad \text { if } \quad \lambda_{i} \neq \lambda_{j} $$ This important property of \(J_{0}\left(\lambda_{i} x\right),\) known as the orthogonality property, is useful in solving boundary value problems. Hint: Write the differential equation for \(J_{0}(\lambda, x)\). Multiply it by \(x J_{0}\left(\lambda_{y} x\right)\) and subtract it from \(x J_{0}\left(\lambda_{t} x\right)\) times the differential equation for \(J_{0}(\lambda, x)\). Then integrate from 0 to \(1 .\)

In several problems in mathematical physics (for example, the Schrödinger equation for a hydrogen atom) it is necessary to study the differential equation $$ x(1-x) y^{\prime \prime}+[\gamma-(1+\alpha+\beta) x] y^{\prime}-\alpha \beta y=0 $$ where \(\alpha, \beta,\) and \(\gamma\) are constants. This equation is known as the hypergeometric equation. (a) Show that \(x=0\) is a regular singular point, and that the roots of the indicial equation are 0 and \(1-\gamma\). (b) Show that \(x=1\) is a regular singular point, and that the roots of the indicial equation are 0 and \(\gamma-\alpha-\beta .\) (c) Assuming that \(1-\gamma\) is not a positive integer, show that in the neighborhood of \(x=0\) one solution of (i) is $$ y_{1}(x)=1+\frac{\alpha \beta}{\gamma \cdot 1 !} x+\frac{\alpha(\alpha+1) \beta(\beta+1)}{\gamma(\gamma+1) 2 !} x^{2}+\cdots $$ What would you expect the radius of convergence of this series to be? (d) Assuming that \(1-\gamma\) is not an integer or zero, show that a second solution for \(0

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