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Chapter 9: Problem 44

Find a power series that represents \(1 /(1+x)^{3}\) on the interval (-1,1)

Short Answer

Expert verified
The power series is \( \sum_{m=0}^{\infty} (m+1)(m+2)(-1)^m x^m \), valid for \( |x| < 1 \).

Step by step solution

01

Recognize a Standard Power Series

The function \( \frac{1}{1+x} \) can be represented by the geometric series \( \sum_{n=0}^{\infty} (-x)^n \) for \( |x| < 1 \). First, identify this standard series representation.
02

Differentiate the Power Series

To find the power series of \( \frac{1}{(1+x)^2} \), differentiate the series representation of \( \frac{1}{1+x} \) term-by-term: \( \sum_{n=0}^{\infty} (-1)^n x^n \). The derivative is \( \sum_{n=1}^{\infty} n(-1)^n x^{n-1} \).
03

Differentiate Again

Differentiate \( \frac{1}{(1+x)^2} \) to find the series for \( \frac{1}{(1+x)^3} \). Differentiate the series \( \sum_{n=1}^{\infty} n(-1)^n x^{n-1} \) term-by-term to get \( \sum_{n=1}^{\infty} n(n-1)(-1)^n x^{n-2} \).
04

Simplify the Power Series

Adjust the index of summation from \( n=1 \) to \( n=0 \) by substituting \( m = n-2 \), resulting in the series \( \sum_{m=0}^{\infty} (m+2)(m+1)(-1)^{m+2} x^m \). Simplifying gives \( \sum_{m=0}^{\infty} (m+1)(m+2)(-1)^m x^m \).
05

Conclude with the Interval of Convergence

Since the original series \( \sum_{n=0}^{\infty} (-x)^n \) converges for \( |x| < 1 \), the series found \( \sum_{m=0}^{\infty} (m+1)(m+2)(-1)^m x^m \) also converges on the same interval, \((-1,1)\).

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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 sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In mathematics, it can be expressed as: \[ S_n = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \]where "a" is the first term and "r" is the common ratio.
  • If the absolute value of the ratio is less than 1 (\(|r| < 1\)), the series converges.
  • A convergent geometric series can be used to represent certain functions as power series.
In the context of this problem, the function \( \frac{1}{1+x} \) resembles a geometric series with \( a = 1 \) and \( r = -x \). It expands to \( \sum_{n=0}^{\infty} (-x)^n \) for \(|x| < 1\), making it a foundational tool to solve the exercise.Understanding the geometric series allows us to perform operations like differentiation later on, thereby transforming basic series into more complex ones that model the function in question.
Term-by-term Differentiation
Term-by-term differentiation is a technique employed in calculus to differentiate an entire series by differentiating each term separately. This method presumes that the series converges uniformly, allowing differentiation to be performed on each term.When given a power series representing a function, like \( \sum_{n=0}^{\infty} (-x)^n \), each term \((-x)^n\) can be individually differentiated:\[ \text{Differentiating term-by-term} \rightarrow \sum_{n=1}^{\infty} n(-1)^n x^{n-1} \]This new series represents the derivative of the original function.Continuing this process allows us to find higher-order derivatives. For this problem, performing term-by-term differentiation twice allows us to transition from \( \frac{1}{1+x} \) to \( \frac{1}{(1+x)^3} \), ultimately transforming the function into the desired power series representation.
  • This step-by-step process ensures each derivative maintains the convergence properties of the original series.
  • It's crucial for ensuring the algebraic transformations accurately reflect the function's behavior.
Interval of Convergence
The interval of convergence refers to the set of all values of \(x\) for which a given power series converges, forming a vital part in understanding the series' behavior.For a power series \( \sum_{n=0}^{\infty} a_n x^n \), this involves finding the range of \(x\) such that the series sums to a finite number.
  • The interval of convergence is typically determined by applying the Ratio Test or Root Test.
  • Understanding this interval ensures the function is properly modeled where it makes sense mathematically.
In the given problem, the series \( \sum_{n=0}^{\infty} (-x)^n \) converges for \(|x| < 1\), derived from the behavior of a geometric series.After performing term-by-term differentiation multiple times, the interval of convergence remains \((-1, 1)\), ensuring the transformed series reliably represents the function within this interval.

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Most popular questions from this chapter

One can sometimes find a Maclaurin series by the method of equating coefficients. For example, let $$ \tan x=\frac{\sin x}{\cos x}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots $$ Then multiply by \(\cos x\) and replace \(\sin x\) and \(\cos x\) by their series to obtain $$ \begin{array}{c} x-\frac{x^{3}}{6}+\cdots=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right)\left(1-\frac{x^{2}}{2}+\cdots\right) \\ =a_{0}+a_{1} x+\left(a_{2}-\frac{a_{0}}{2}\right) x^{2}+\left(a_{3}-\frac{a_{1}}{2}\right) x^{3}+\cdots \end{array} $$ Thus. $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}-\frac{a_{0}}{2}=0, \quad a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}, \quad \cdots $$ SO $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}=0, \quad a_{3}=\frac{1}{3}, \quad \cdots $$ and therefore $$ \tan x=0+x+0+\frac{1}{3} x^{3}+\cdots $$ which agrees with Problem 1 . Use this method to find the terms through \(x^{4}\) in the series for \(\sec x\).

Find the Taylor series in \(x-a\) through the term \((x-a)^{3}\). $$ 1+x^{2}+x^{3}, a=1 $$

Let \(g(x)=p(x)+x^{n+1} f(x),\) where \(p(x)\) is a polynomial of degree at most \(n\) and \(f\) has derivatives through order \(n .\) Show that \(p(x)\) is the Maclaurin polynomial of order \(n\) for \(g\).

Find the radius of convergence of $$ \sum_{n=0}^{\infty} \frac{(p n) !}{(n !)^{p}} x^{n} $$ where \(p\) is a positive integer.

If an object of rest mass \(m_{0}\) has velocity \(v\), then (according to the theory of relativity) its mass \(m\) is given by \(m=\) \(m_{0} / \sqrt{1-v^{2} / c^{2}},\) where \(c\) is the velocity of light. Explain how physicists get the approximation $$ m \approx m_{0}+\frac{m_{0}}{2}\left(\frac{v}{c}\right)^{2} $$
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