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Chapter 9: Problem 24

Use power series operations to find the Taylor series at \(x=0\) for the functions. $$\frac{2}{(1-x)^{3}}$$

Short Answer

Expert verified
The Taylor series for \( \frac{2}{(1-x)^3} \) is \( \sum_{n=0}^{\infty} (n+2)(n+1)x^n \).

Step by step solution

01

Recognize the Template of a Known Power Series

Recall that the function \( \frac{1}{(1-x)} \) can be expressed as a geometric series: \( \sum_{n=0}^{\infty} x^n \), as long it converges i.e., \(|x| < 1\). Since our function is \( \frac{2}{(1-x)^3} \), this resembles a derivative or transformation of this base series.
02

Differentiate the Geometric Series

Differentiate the geometric series \( \sum_{n=0}^{\infty} x^n \) with respect to \(x\) to get \( \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}\). Differentiate again to obtain \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} = \frac{2x}{(1-x)^3}\).
03

Scale and Adjust for the Given Function

Since we need \( \frac{2}{(1-x)^3} \), notice that \( \frac{2}{(1-x)^3} = 2 \cdot \frac{1}{(1-x)^3} \). Multiply the series found in Step 2 by 2 to get \( 2 \cdot \left(\sum_{n=2}^{\infty} n(n-1)x^{n-2}\right)\).
04

Reindex to Find the Final Taylor Series Form

To match the series to the form \( \sum_{n=0}^\infty a_n x^n \), replace \(n\) by \(n+2\) in the series from Step 3, resulting in \( \sum_{n=0}^{\infty} (n+2)(n+1)x^n \). This gives the Taylor series for \( \frac{2}{(1-x)^3} \).

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Key Concepts

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

Understanding Power Series
A power series is a sum of terms in the form of \( a_n x^n \). It represents functions as an infinite sum of powers of \( x \). Each term is a product of a coefficient \( a_n \) and a power \( x^n \). A power series is very useful in mathematics because it can represent functions that are difficult to write in other forms.
  • A power series centered at \( x=0 \) is called a Maclaurin series.
  • When centered at any other point, it is known as a Taylor series.
  • It provides an approximation of a complex function with a polynomial.
For instance, in the original exercise, \( \frac{2}{(1-x)^{3}} \) was expressed using a power series that evolved from a known geometric series through differentiations. This transformation helps in solving complex calculus problems with relative ease.
Exploring Geometric Series
A geometric series is a special type of power series. It has a constant ratio between successive terms. The simplest form of a geometric series is \( \frac{1}{(1-x)} = \sum_{n=0}^{\infty} x^n \). This series converges for \( |x| < 1 \).
  • It is the foundation for many calculus operations, like finding Taylor or Maclaurin series.
  • This series is used to expand and express complex rational functions.
  • Recognizing a geometric series can help simplify complex expressions.
In our exercise, the geometric series \( \sum_{n=0}^{\infty} x^n \) served as the starting point. Through differentiation, it transformed into a series representing \( \frac{2}{(1-x)^3} \). This application highlights the usefulness and versatility of geometric series in calculus.
The Role of Differentiation
Differentiation is a mathematical operation where we find the derivative of a function, which essentially gives us the rate of change. The power of differentiation is in its ability to transform series, such as turning \( \frac{1}{(1-x)} \) into \( \frac{1}{(1-x)^3} \).
  • By differentiating the geometric series \( \sum_{n=0}^{\infty} x^n \), you obtain \( \sum_{n=1}^{\infty} n x^{n-1} \).
  • Differentiating again obtains \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} \), which corresponds to \( \frac{2x}{(1-x)^3} \).
  • This shows how differentiation creates new series from existing ones.
In the original solution, differentiation turned a simple series into a complex function鈥檚 series representation, demonstrating its power in calculus and function approximation.
Understanding Convergence
Convergence is a key concept when working with series. It describes whether a series approaches a certain value as the number of terms increases. In power series and geometric series, we need them to converge within a certain interval to be useful.
  • The interval of convergence is the set of \( x \)-values for which the series sums to a finite value.
  • For geometric series like \( \frac{1}{(1-x)} \), it converges if \( |x| < 1 \).
  • Knowing the convergence helps prevent miscalculations and misrepresentations of functions.
In the exercise, recognizing the convergence of the base geometric series was crucial. Without this knowledge, manipulating series could lead to incorrect conclusions. Always check for convergence when working with series to ensure accurate and safe calculations.

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Most popular questions from this chapter

Find the series' radius of convergence. $$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}} x^{n}$$ (Hint: Apply the Root Test.)

Let \(a_{n}=\left\\{\begin{array}{ll}n / 2^{n}, & \text { if } n \text { is a prime number } \\ 1 / 2^{n}, & \text { otherwise. }\end{array}\right.\) Does \(\Sigma a_{n}\) converge? Give reasons for your answer.

In Exercises 17鈥� 46, use any method to determine whether the series converges or diverges. Give reasons for your answer. \(\sum_{n=3}^{\infty} \frac{2^{n^{2}}}{n^{2^{2}}}\)

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n 2^{n}(n+1) !}{3^{n} n !}$$

Show that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely, then $$\left|\sum_{n=1}^{\infty} a_{n}\right| \leq \sum_{n=1}^{\infty}\left|a_{n}\right|.$$
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