Chapter 5: Problem 8
\(\frac{1-x}{2+x}=\frac{1}{2}-\frac{3}{4} x+\frac{3}{8} x^{2}-\frac{3}{16} x^{3}+\cdots\) since the function is undefined at \(x=-2,\) the series converges on (-2,2).
Short Answer
Expert verified
Perform a series expansion of \(\frac{1-x}{2+x}\) to match the provided series. Validate its convergence within \((-2,2)\).
Step by step solution
01
Identify the Function to Expand
Notice that the left-hand side of the equation is a fraction, \( \frac{1-x}{2+x} \). We need to express this as a series that matches the given series on the right-hand side.
02
Manipulate the Denominator
We can rewrite the denominator \(2 + x\) as \(2(1 + \frac{x}{2})\). This will help us in finding a suitable series expansion.
03
Perform Long Division or Partial Fractions
Perform long division on \(\frac{1-x}{2+x}\) to find the series expansion. This involves dividing the numerator by the denominator term by term until we notice a pattern or repeat terms, which leads to the series.
04
Applying Geometric or Binomial Expansion
At this stage, use the geometric series expansion formula \((1 - z)^{-1} = 1 + z + z^2 + z^3 + \cdots\) for \(|z| < 1\). Here, consider \(z = -\frac{x}{2}\) and multiply the series found if necessary to match the form \(\frac{1-x}{2+x}\).
05
Matching with Given Series
Adjust any coefficients by comparing the terms derived from your expansion with the given series \(\frac{1}{2} - \frac{3}{4}x + \frac{3}{8}x^{2} - \frac{3}{16}x^{3} + \cdots\). Ensure that each term's coefficients from your expansion match with the given series.
06
Convergence Considerations
Since the expression is undefined at \(x = -2\), consider the interval of convergence for the given series, which is \(-2 < x < 2\). Verify that the steps and final series are valid within this interval.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fractions
Partial Fractions is a technique used in mathematics to simplify complex rational expressions. It involves decomposing a rational function into simpler fractions, which can then be used for easier integration or computation. Imagine you have a fraction with polynomials in the numerator and denominator. The idea is to express this as a sum of simpler fractions, making calculations much more manageable.
For our given function \( \frac{1-x}{2+x} \), using partial fractions might involve breaking it into forms that are easier to expand into a series.
For our given function \( \frac{1-x}{2+x} \), using partial fractions might involve breaking it into forms that are easier to expand into a series.
- The purpose of partial fraction decomposition is to work with expressions that can be more straightforwardly summed or integrated.
- Typically, this method is used in integration, but in series expansions, it helps to isolate terms in ways that suggest simple series forms.
- Partial fractions might not always directly apply, as our expression \( \frac{1-x}{2+x} \) primarily requires series expansion through other methods like geometric series expansion or long division.
Geometric Series
A geometric series is a crucial mathematical concept where each term after the first is obtained by multiplying the preceding term by a fixed, non-zero number called the common ratio.
This series looks like \( a + ar + ar^2 + ar^3 + \cdots \), where \( a \) is the first term and \( r \) is the ratio. For the series to converge, \(|r| < 1\).
In our exercise, the transformation of the denominator in \( \frac{1-x}{2+x} \) allowed us to find an expression matching the geometric series form.
This series looks like \( a + ar + ar^2 + ar^3 + \cdots \), where \( a \) is the first term and \( r \) is the ratio. For the series to converge, \(|r| < 1\).
In our exercise, the transformation of the denominator in \( \frac{1-x}{2+x} \) allowed us to find an expression matching the geometric series form.
- By factorizing the denominator as \( 2(1 + \frac{x}{2})\), it aligns with the geometric series form \((1 - z)^{-1}\), setting the stage for expansion.
- In the expansion of \( (1 - z)^{-1} = 1 + z + z^2 + z^3 + \cdots \), we substitute \( z = -\frac{x}{2} \).
- This gives us a reasonable approach to derive or verify the series form mentioned in the exercise.
Binomial Expansion
Binomial expansion provides a way to expand expressions that are raised to any power. It is based on the binomial theorem, which is represented as \((a + b)^n = \sum_{k=0}^n {n \choose k} a^{n-k} b^k\).
This concept can also be applied when each term of a series involves powers that follow this form. In our case, we did not directly use a binomial expansion; however, understanding the principle helps link how certain transformations in a series are approached.
This concept can also be applied when each term of a series involves powers that follow this form. In our case, we did not directly use a binomial expansion; however, understanding the principle helps link how certain transformations in a series are approached.
- In some series, binomial expansions help express polynomial power sums in a concise and understandable form.
- If a quotient can be restructured using powers like \((1 + x)^n\), binomial expansion simplifies finding each term's coefficient.
- Despite not being directly used here, the binomial theorem remains foundational in series methods, enhancing skills in manipulating series-related expressions.
Interval of Convergence
The interval of convergence is the range of values for which a series converges to a finite result. Knowing this range is essential in determining where a series-based function behaves appropriately.
For our series \( \frac{1-x}{2+x} = \frac{1}{2} - \frac{3}{4} x + \frac{3}{8} x^{2} - \frac{3}{16} x^{3} + \cdots \), the interval of convergence is given as \(-2 < x < 2\).
For our series \( \frac{1-x}{2+x} = \frac{1}{2} - \frac{3}{4} x + \frac{3}{8} x^{2} - \frac{3}{16} x^{3} + \cdots \), the interval of convergence is given as \(-2 < x < 2\).
- This interval comes from analyzing where the series remains defined, especially since the original function is undefined at \(x = -2\).
- Understanding the interval involves determining where associated transformations, like our geometric series derivation, stay valid under series rules.
- Working within the interval ensures convergence, preventing divergence that renders a series sum infinite or undefined.