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Chapter 9: Problem 22

Find the Maclaurin series for the functions. $$\frac{x^{2}}{x+1}$$

Short Answer

Expert verified
The Maclaurin series for \( \frac{x^2}{x+1} \) is \( \sum_{n=0}^{\infty} (-1)^n x^{n+2} \).

Step by step solution

01

Understanding the Maclaurin Series

The Maclaurin series is a Taylor series expansion of a function about 0. It is given by: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \). We need to find the derivatives of \( \frac{x^2}{x+1} \) and evaluate them at \( x = 0 \).
02

Simplify the Function

To expand \( \frac{x^2}{x+1} \), rewrite it as a geometric series. First, note that \( \frac{1}{x+1} = 1 - x + x^2 - x^3 + \dots \) (valid for \( |x| < 1 \)). Then, multiply this series by \( x^2 \): \( x^2(1 - x + x^2 - x^3 + \dots) = x^2 - x^3 + x^4 - x^5 + \dots \).
03

Write the Series Expansion

The simplified series becomes \( x^2 - x^3 + x^4 - x^5 + \dots \). Each subsequent term increases by a power of \( -x \) from the previous term. Therefore, the Maclaurin series for \( \frac{x^2}{x+1} \) is: \( \sum_{n=0}^{\infty} (-1)^n x^{n+2} \).
04

Validate by Initial Terms

To verify, substitute \( n = 0, 1, 2 \) into the series: \( n = 0 \), term is \( x^2 \); \( n = 1 \), term is \( -x^3 \); \( n = 2 \), term is \( x^4 \), and so on. This matches our series expansion \( x^2 - x^3 + x^4 - x^5 + \dots \).

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Key Concepts

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

Taylor series
The Taylor series is a powerful mathematical tool used to approximate functions. It expresses a function as an infinite sum of terms, calculated from the values of its derivatives at a specific point. The formula for a Taylor series centered around a point \( a \) is:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots \).
For the Maclaurin series, a special case of the Taylor series, \( a = 0 \), resulting in the simpler formula:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \).
This allows us, for example, to expand \( \frac{x^2}{x+1} \) at 0, providing an easy way to understand its behavior near this point.
Series Expansion
Series expansion involves representing a function as an infinite sum of simpler terms. It allows complex functions to be expressed in a form that can be easily managed and understood. In the context of a Taylor or Maclaurin series:
  • The function \( \frac{x^2}{x+1} \) is broken down into a series for \( |x| < 1 \).
  • We first express \( \frac{1}{x+1} \) as a geometric series.
  • Then, multiply by \( x^2 \) to reformat and generate terms like \( x^2 - x^3 + x^4 - x^5 + \dots \).
Series expansion aids in simplifying the process of finding derivatives and evaluating them at specific points.
Geometric series
A geometric series is a series of terms each of which is the product of the previous term and a fixed, non-zero number called the common ratio. For instance, \( 1 + r + r^2 + r^3 + \cdots \) is a simple geometric series. In our exercise:
  • We use the expansion \( \frac{1}{x+1} = 1 - x + x^2 - x^3 + \dots \) valid for \( |x| < 1 \).
  • This series is crucial because by multiplying each term by \( x^2 \), we construct an appropriate series for \( \frac{x^2}{x+1} \).
Understanding geometric series helps in making complex rational functions more approachable through series expansion techniques.
Derivatives
Derivatives measure how a function changes as its input changes, providing a way to analyze and approximate functions using series like Taylor and Maclaurin. The derivatives of \( \frac{x^2}{x+1} \) are found by applying standard differentiation rules:
  • The first derivative provides information on the slope of the curve.
  • Higher-order derivatives reveal more complex behaviors like curvature.
By evaluating these derivatives at \( x = 0 \), we can construct the Maclaurin series efficiently, leading to a simpler representation of the function near 0.
Polynomial approximation
Polynomial approximation involves approximating a complicated function with a polynomial, which is much easier to evaluate and understand. The purpose of using a Maclaurin series is to create such an approximation:
  • For \( \frac{x^2}{x+1} \), the expansion becomes \( x^2 - x^3 + x^4 - x^5 + \dots \), a polynomial-like series.
  • This approach simplifies calculations and helps us see patterns within the function.
Polynomial approximations are especially useful when dealing with functions that are difficult to compute directly, allowing for a clearer understanding of their behavior.

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Most popular questions from this chapter

Newton's method The following sequences come from the recursion formula for Newton's method, $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}.$$ Do the sequences converge? If so, to what value? In each case, begin by identifying the function \(f\) that generates the sequence. a. \(x_{0}=1, \quad x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}\) b. \(x_{0}=1, \quad x_{n+1}=x_{n}-\frac{\tan x_{n}-1}{\sec ^{2} x_{n}}\) c. \(x_{0}=1, \quad x_{n+1}=x_{n}-1\)

Find the first four nonzero terms in the Maclaurin series for the functions. $$\cos ^{2} x \cdot \sin x$$

Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{n}}{\left(2^{n}\right)^{2}}$$

Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\epsilon\) there corresponds an integer \(N\) such that for all \(m\) and \(n\), $$m>N \quad \text { and } \quad n>N \Rightarrow \quad\left|a_{m}-a_{n}\right|<\epsilon.$$

Determine if the sequence is monotonic and if it is bounded. $$a_{n}=\frac{(2 n+3) !}{(n+1) !}$$
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