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Chapter 8: Problem 8

Express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{x}{1+x^{2}}\)

Short Answer

Expert verified
The power series is \(\sum_{n=0}^{\infty} (-1)^n x^{2n+1}\) with a radius of convergence \(R = 1\).

Step by step solution

01

Identify the series structure

We are given the function \(\frac{x}{1+x^2}\). We notice that this expression is of the form \(\frac{x}{1-(-x^2)}\), which resembles the geometric series \(\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}\). This suggests we can rewrite it using the geometric series framework.
02

Apply the geometric series formula

According to the geometric series formula, \(\frac{1}{1-(-x^2)}\) can be expanded as \(\sum_{n=0}^{\infty} (-x^2)^n\). Therefore, \(\frac{x}{1+x^2} = x \cdot \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n+1}\).
03

Verify the power series

The derived series is \(\sum_{n=0}^{\infty} (-1)^n x^{2n+1}\), which confirms that the function \(\frac{x}{1+x^2}\) has been expressed as a power series around the base point 0.
04

Determine the radius of convergence

For the geometric series \(\sum (-1)^n x^{2n}\), the series converges for \(|x^2| < 1\), which implies \(|x| < 1\). Thus, the radius of convergence \(R\) is \(1\).

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Key Concepts

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

Geometric Series
A geometric series is a type of infinite series where each term is a constant multiple of the previous one. It's represented mathematically as \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \] where \(a\) is the first term and \(r\) is the common ratio. This series is fundamental in mathematics because it's one of the simplest series with explicit solutions.
  • The series converges if the absolute value of the common ratio \(|r| < 1\).
  • To rewrite a function using a geometric series format, it must resemble \( \frac{1}{1-r} \) where \( r \) is the expression inside the series.
In the function \( \frac{x}{1+x^2} \), recognizing \(1+x^2\) as \(1-(-x^2)\) helps us identify the framework of a geometric series. This allows us to expand and express it in a power series form around a given point.
Radius of Convergence
The radius of convergence ensures the interval over which a power series converges. It's a crucial concept since it tells us the extent to which the series representation of a function is valid. The radius of convergence \( R \) is determined by analyzing the series and finding when it converges absolutely.
  • It can be calculated using the formula \( \frac{1}{R} = \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), known as the ratio test.
  • For geometric series, if the series term is \( (-x^2)^n \), it converges when \( |x^2| < 1 \).
Therefore, for \( \frac{x}{1+x^2} \), the radius of convergence is derived from this condition, leading to \(|x| < 1\). This validates that the series converges in the interval from \(-1\) to \(1\).
Convergence
Convergence of a series means that as we sum more and more terms, the total gets closer to some specific value. For a power series, understanding convergence helps us determine where the power series actually represents the original function.
  • For any power series, convergence depends on the variable \( x \) and its position relative to the base point where we expanded the series.
  • If \( |x| \) is within the radius of convergence, the series will converge to a finite value.
In our example of \( \frac{x}{1+x^2} \), convergence happens for \( |x| < 1 \), meaning the power series accurately represents the function within this interval. This guarantees that the series can be used for calculations and approximations inside this radius.
Base Point
The base point, often referred to as the center of a power series, is the value of \( x \) about which the series is expanded. It is crucial for defining the region where the series representation is valid.
  • Usually denoted by the letter \( x_0 \), it signifies the point around which the series elements are centered.
  • In many exercises, the base point is set to \(0\), simplifying calculations and keeping the series symmetric.
For our given function \( \frac{x}{1+x^2} \), the series is expanded around the base point \(0\). This means the power series is centered at the origin, making calculations straightforward when determining terms and convergence properties.

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Most popular questions from this chapter

Show that $$ 1 \cdot 3 \cdot 5 \cdot 7 \cdots(2 N-3)(2 N-1)=\frac{(2 N) !}{N ! 2^{N}} $$

In each of Exercises 91-94 a function \(f\), a base point \(c\), and a point \(x_{0}\) are given. Plot \(y=\left|f^{(3)}(t)\right|\) for \(t\) between \(c\) and \(x_{0}\). Use your plot to estimate the quantity \(M\) of Theorem 2 . Then use your value of \(M\) to obtain an upper bound for the absolute error \(\left|R_{2}\left(x_{0}\right)\right|=\left|f\left(x_{0}\right)-T_{2}\left(x_{0}\right)\right|\) that results when \(f\left(x_{0}\right)\) is approximated by the order 2 Taylor polynomial with base point \(c\). $$ f(x)=\ln \left(2 x^{2}-1\right) \quad c=1 \quad x_{0}=1.3 $$

Show that $$ \left(\begin{array}{c} -1 / 2 \\ n \end{array}\right)=\frac{(-1)^{n}(2 n) !}{2^{2 n}(n !)^{2}} $$ Using this formula, state the Maclaurin series of \((1+u)^{-1 / 2},\) and derive the Maclaurin series of \(\arcsin (x)\)

In each of Exercises \(45-60,\) determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}}\)

Determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n}\left(\frac{n^{2}+2 n+2}{3 n^{3}+7}\right)\)
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