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Chapter 9: Problem 26

Using a Power Series In Exercises 19-28, use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}, \quad|x|<1$$ to find a power series for the function, centered at \(0,\) and determine the interval of convergence. $$f(x)=\ln \left(x^{2}+1\right)$$

Short Answer

Expert verified
The function \(f(x)=\ln \left(x^{2}+1\right)\) can be represented by the power series \(\sum_{n=0}^{\infty} \frac{(-2)^{n} x^{n+1}}{n+1}\), and the interval of convergence is \(-1 < x \leq 1\).

Step by step solution

01

Express Function in Recognisable Form

The given function can be re-written as \(f(x) = 2\ln(|x| + 1)\) after taking out a factor of 2 from the logarithm. Now, we can express this as an integral: \(f(x) = 2 \int \frac{1}{1+x} dx\).
02

Substitute with Power Series

Now substitute \(\frac{1}{1+x}\) with its series representation \(\sum_{n=0}^{\infty}(-1)^{n} x^{n}\), which gives \(f(x) = 2 \int \sum_{n=0}^{\infty}(-1)^{n} x^{n} dx\). Using properties of series, the integral can be interchanged with the series to obtain \(f(x) = 2 \sum_{n=0}^{\infty}(-1)^{n} \int x^{n} dx\).
03

Integrate Term-by-Term

Evaluate the integral term-by-term to get: \(2 \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1}\). Therefore, the power series representation for \(f(x)\) centered at 0 is \(\sum_{n=0}^{\infty} \frac{(-2)^{n} x^{n+1}}{n+1}\).
04

Find Interval of Convergence

The interval of convergence is the same as that of the original series, ie., \(-1 < x < 1\). However, since this is a logarithmic function, the function is undefined at x = -1. Therefore, the correct interval of convergence is \(-1 < x \leq 1\).

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Key Concepts

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

Interval of Convergence
Understanding the interval of convergence is essential when dealing with power series. It represents the set of all real numbers for which the series converges to a finite value.

In the context of logarithmic functions represented by power series, our focus is on finding the range of values for our variable, such as 'x', within which the series remains convergent. It's crucial to determine this interval because outside of it, the series does not represent the function reliably.

To determine the interval of convergence for a given power series, one typically uses the ratio test or root test. Once identified, it should be checked at the endpoints specifically, as convergence at these points can change the type of interval (open or closed) considered.
Term-by-Term Integration
Term-by-term integration is a process that allows us to integrate power series by integrating each term separately. This technique is useful because it simplifies complex functions into manageable components.

When we apply term-by-term integration to the series representing a function, we operate under the assumption that the series converges uniformly within its interval of convergence. This enables us to interchange the integral sign with the summation sign, leading to a new series that represents the integral of the function.

It's important to understand that the interval of convergence for the integrated series often remains the same as that of the original power series. However, always re-evaluate at the endpoints, as the convergence could differ after integration.
Logarithm Function Properties
The logarithm function comes with a unique set of properties that play a critical role when we express it as a power series. One such property is the logarithm's ability to transform products into sums, which is particularly beneficial when managing series.

Other key properties include the natural logarithm's base (e), logarithms of positive real numbers are always defined, and the function is continuous and differentiable within its domain. These properties ensure that when we work with power series of logarithmic functions, we consider the function's behavior around points where it's undefined (such as 0 for the natural logarithm).
Series Representation
Series representation translates a function into a sum of its infinite components, which is an incredibly powerful tool in calculus. This form enables us to manipulate complex functions like logarithms more easily.

The goal is to find the coefficients that correspond to each term in the series such that, when summed, they approximate the function as closely as desired within its interval of convergence. For the logarithmic function, this approach helps us to understand its behavior near points of interest and to perform operations like integration or differentiation with respect to its series representation.

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Most popular questions from this chapter

Graphical Analysis In Exercises 3 and \(4,\) the figures show the graphs of the first 10 terms, and the graphs of the first 10 terms of the sequence of partial sums, of each series. (a) Identify the series in each figure. (b) Which series is a \(p\) -series? Does it converge or diverge? (c) For the series that are not p-series, how do the magnitudes of the terms compare with the magnitudes of the terms of the \(p\) -series? What conclusion can you draw about the convergence or divergence of the series? (d) Explain the relationship between the magnitudes of the terms of the series and the magnitudes of the terms of the partial sums. \(\sum_{n=1}^{\infty} \frac{6}{n^{3 / 2}}, \quad \sum_{n=1}^{\infty} \frac{6}{n^{3 / 2}+3^{\prime}}\) and $$\sum_{n=1}^{\infty} \frac{6}{n^{3 / 2}}, \quad \sum_{n=1}^{\infty} \frac{6}{n^{3 / 2}+3^{\prime}}$$

Finding the Interval of Convergence In Exercises \(15-38\) , find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1}$$

Differential Equation In Exercises \(59-64,\) show that the function represented by the power series is a solution of the differential equation. $$y=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}, \quad y^{\prime \prime}+y=0$$

Writing an Equivalent Series In Exercises \(45-48,\) write an equivalent series with the index of summation beginning at \(n=1 .\) $$\sum_{n=0}^{\infty}(-1)^{n+1}(n+1) x^{n}$$

Writing an Equivalent Series In Exercises \(45-48,\) write an equivalent series with the index of summation beginning at \(n=1 .\) $$\sum_{n=2}^{\infty} \frac{x^{n-1}}{(7 n-1) !}$$
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