/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 \(23-26\) Evaluate the indefinit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 24

\(23-26\) Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int \frac{\ln (1-t)}{t} d t $$

Short Answer

Expert verified
The integral is \(-\sum_{n=1}^{\infty} \frac{t^n}{n^2} + C\), with radius of convergence \(|t| < 1\).

Step by step solution

01

Understand the Integrand

We need to integrate \( \frac{\ln(1-t)}{t} \) with respect to \( t \). The function \( \ln(1-t) \) can be expanded using the Taylor series for \( |t| < 1 \) as \( \ln(1-t) = -\sum_{n=1}^{\infty} \frac{t^n}{n} \).
02

Substitute Series Expansion

Substitute the series expansion into the integrand:\[ \int \frac{\ln(1-t)}{t} \ d t = \int \frac{-\sum_{n=1}^{\infty} \frac{t^n}{n}}{t} \ d t \]This simplifies to:\[ -\int \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ d t = -\int \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \ d t \]
03

Integrate Term by Term

Now integrate each term of the series:\[ -\int \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \ d t = -\sum_{n=1}^{\infty} \int \frac{t^{n-1}}{n} \ d t = -\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{t^n}{n} + C \]This results in:\[ -\sum_{n=1}^{\infty} \frac{t^n}{n^2} + C \]
04

Determine Radius of Convergence

The radius of convergence for the series \( \ln(1-t) = -\sum_{n=1}^{\infty} \frac{t^n}{n} \) is determined by the convergence of the geometric series \( t^n \). Thus, the radius of convergence is \( |t| < 1 \), as it is based on the power series expansion of \( \ln(1-t) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Power Series
A power series is a way of representing a function as an infinite sum of terms calculated from the values of its derivatives. Each term in a power series is of the form \( a_n (x - c)^n \), where \( a_n \) are coefficients, \( x \) is a variable, and \( c \) is the center of the series. This representation is incredibly useful in calculus because it allows us to work with complex functions in a more manageable way.
  • Power series can converge to define a function within a certain interval.
  • When we integrate or differentiate a power series term-by-term, the resulting series still converges within this interval.
The exercise involves rewriting a given function, \( \frac{\ln(1-t)}{t} \), as a power series so that we can easily integrate it by integrating each term individually. This method simplifies the problem by breaking it into smaller, more manageable tasks.
Radius of Convergence
The radius of convergence is a measure of the interval in which a power series converges to a function. It determines how far from the center of the series the convergence holds true. For a series such as \( \sum_{n=0}^{\infty} a_n (x-c)^n \), the radius of convergence \( R \) can be found using the ratio test or the root test.
  • If the series \( |t| < R \) converges, then the series will approximate the function within this radius.
  • Outside of this interval, the series may diverge, meaning it won't accurately represent our function.
In our exercise, the power series expansion of \( \ln(1-t) \) has a radius of convergence \( |t| < 1 \). This is crucial for ensuring that our integration results in a function that converges everywhere within this interval.
Taylor Series
A Taylor series is a special type of power series that aims to approximate a function using its derivatives at a single point. It is expressed as \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \). This series can be particularly useful for approximating complicated functions with polynomials, which are easier to work with.
  • A Taylor series centered at zero is also known as a Maclaurin series.
  • Taylor series are valuable in both theoretical math and practical applications, such as engineering and physics.
In the provided solution, the Taylor series expansion for \( \ln(1-t) \) is utilized. This allows each part of the function to be represented as a series of simpler terms, making integration straightforward.
Geometric Series
A geometric series is a series of the form \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term and \( r \) is the common ratio. It is one of the simplest types of series, which can be summed up easily if \(|r| < 1\). The sum of the first \( n \) terms is given by: \[ S_n = a \frac{1-r^n}{1-r} \]For an infinite series where \(|r| < 1\), the sum is:\[ S = \frac{a}{1-r} \]In calculus, recognizing and using geometric series is a powerful tool for solving or simplifying problems, as they have clear convergence properties.
  • The Taylor series for \( \ln(1-t) \) leverages the familiar attributes of a geometric series.
  • The recognition of the geometric series pattern in the integrand \( \ln(1-t) \) is what allows us to determine the radius of convergence for \( |t| < 1 \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that if \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\left\\{b_{n}\right\\}\) is bounded, then \(\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=0\)

Suppose that the radius of convergence of the power series \(\sum c_{n} x^{n}\) is \(R .\) What is the radius of convergence of the power series \(\Sigma c_{n} x^{2 m} ?\)

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \frac{1}{n+n \cos ^{2} n}$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sinh 2 x, \quad a=0, \quad n=5, \quad-1 \leqslant x \leqslant 1$$

Test the series for convergence or divergence. $$\sum_{k=1}^{\infty} \frac{k \ln k}{(k+1)^{3}}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.