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Chapter 9: Problem 3

In Problems 1-10, find the power series representation for \(f(x)\) and specify the radius of convergence. Each is somehow related to a geometric series (see Examples 1 and 2). $$ f(x)=\frac{1}{(1-x)^{3}} $$

Short Answer

Expert verified
Power series: \( \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2} x^n \) with radius of convergence \( |x| < 1 \).

Step by step solution

01

Recall the Geometric Series Formula

We start by recalling the formula for the sum of a geometric series. The infinite geometric series: \( 1 + r + r^2 + r^3 + \dots \) can be written as \( \frac{1}{1-r} \) for \( |r| < 1 \).
02

Differentiate the Geometric Series Sum

In order to find the power series representation of \( f(x) = \frac{1}{(1-x)^3} \), first observe that it is related to the cube of a geometric series sum. Begin by differentiating \( \frac{1}{1-x} \) with respect to \( x \) to find a related expression. The derivative of \( \frac{1}{1-x} \) with respect to \( x \) is \( \frac{1}{(1-x)^2} \).
03

Differentiate Again

Since we need \( \frac{1}{(1-x)^3} \), differentiate \( \frac{1}{(1-x)^2} \) with respect to \( x \) to further reduce it to \( \frac{1}{(1-x)^3} \). The second derivative of \( \frac{1}{1-x} \) with respect to \( x \) is \( \frac{2}{(1-x)^3} \).
04

Adjust the Expression

Notice that we have found \( 2 \cdot \frac{1}{(1-x)^3} \), but we want an expression for \( \frac{1}{(1-x)^3} \) alone. Hence, divide the expression by 2 to obtain \( \frac{1}{(1-x)^3} = \frac{1}{2} \cdot \frac{d^2}{dx^2}\left(\frac{1}{1-x}\right) \).
05

Derive the Power Series Expression

Now, we expand \( \frac{1}{1-x} \) as a geometric series: \( 1 + x + x^2 + x^3 + \cdots \). Taking two derivatives, we get the series for \( \frac{1}{(1-x)^3} \).- The first derivative gives: \( 1 + 2x + 3x^2 + 4x^3 + \cdots \)- The second derivative gives: \( 2(1 + 3x + 6x^2 + 10x^3 + \cdots) \).Dividing by 2, the power series for \( \frac{1}{(1-x)^3} \) becomes: \( \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2} x^n \).
06

Determine the Radius of Convergence

Based on the original geometric series \( \frac{1}{1-x} \), the series converges for \( |x| < 1 \). Since differentiation does not change the radius of convergence, the radius of convergence for \( \frac{1}{(1-x)^3} \) remains the same: \( |x| < 1 \).

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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 an essential type of infinite series. It sums a sequence of terms where each term after the first is the product of the previous one and a fixed value, known as the common ratio. The formula for the sum of an infinite geometric series is uniquely simple:
  • The sum is given by \( S = \frac{1}{1 - r} \) where \( |r| < 1 \).
This implies that the series converges only when the absolute value of the common ratio \( r \) is less than 1. In this exercise, the function \( f(x) = \frac{1}{1-x} \) is a classic geometric series with common ratio \( x \). Understanding how this series operates helps in recognizing how the expression can be differentiated or manipulated in more complex problems.
Radius of Convergence
The radius of convergence is an important term that describes where a power series converges. Simply put, it tells you the interval around the center where the series sums up to a finite value. For a series expressed normally as \( \sum a_n x^n \), the convergence is determined by the ratio test or directly from representation.
  • For the geometric series \( \frac{1}{1 - x} \), the radius of convergence is \( |x| < 1 \).
  • This means the series converges as long as the absolute value of \( x \) is less than 1.
For any changes made to the series, such as differentiation, the radius of convergence will typically remain unchanged, as illustrated in the given exercise. It stays consistent at \( |x| < 1 \) when derived further for transformations like converting \( \frac{1}{1-x} \) to \( \frac{1}{(1-x)^3} \).
Differentiation
Differentiation is a central process in calculus that deals with finding the derivative of a function. It involves calculating how a function changes as its input changes. In the context of power series and geometric series, differentiation can be applied to manipulate the series for new purposes.
  • When differentiating \( \frac{1}{1-x} \), you obtain the series \( \frac{1}{(1-x)^2} \).
  • Further differentiating provides \( \frac{2}{(1-x)^3} \).
  • By dividing by 2, we isolate to get \( \frac{1}{(1-x)^3} \).
  • The power series expansion then becomes \( \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2} x^n \).
This series showcases how calculus techniques like differentiation simplify or transform basic series into more intricate forms.
Calculus
Calculus is a branch of mathematics that helps us understand changes through differentiation and integration. In this exercise, calculus principles are employed to expand and manipulate geometric series into the series of interest.
  • Initially, by differentiating a simple function \( \frac{1}{1-x} \), it models changes and computes new related functions \( \frac{1}{(1-x)^3} \).
  • Through these calculations, advanced expressions for functions emerge, showcasing the power of calculus.
Overall, calculus allows us to derive properties, such as power series expansions and convergence, lending a systematic approach to understanding change and area under curves in various contexts.

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