Question

Use the series for \(f(x)=\frac{1}{1-x}\) on \(|x|<1\) to construct a series for \(\frac{1}{(1-x)(x-2)} .\) Determine the interval of convergence.

Short answer

The series for \( \frac{1}{(1-x)(x-2)} \) is constructed using partial fractions and converges for \( |x| < 1 \).

Step-by-step solution

Identify the Target Function

We want to construct a series for the function \( \frac{1}{(1-x)(x-2)} \). The first step is to rewrite this using partial fraction decomposition.

Perform Partial Fraction Decomposition

Express \( \frac{1}{(1-x)(x-2)} \) in terms of partial fractions: \( \frac{1}{(1-x)(x-2)} = \frac{A}{1-x} + \frac{B}{x-2} \). Solve for \( A \) and \( B \) by equating coefficients.

Solve for Coefficients A and B

To find \( A \) and \( B \), set \( 1 = A(x-2) + B(1-x) \). Solve this equation by setting convenient values for \( x \):1. Set \( x = 1 \): \( 1 = A(1-2) + B(1-1) \Rightarrow A = -1 \).2. Set \( x = 2 \): \( 1 = A(2-2) + B(1-2) \Rightarrow B = -1 \).

Rewrite the Function in Partial Fractions

Replace the original function with the found coefficients: \( \frac{1}{(1-x)(x-2)} = \frac{-1}{1-x} + \frac{-1}{x-2} \).

Express Each Fraction as a Series

Use the series \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \) for \( |x| < 1 \):- The first term: \( \frac{-1}{1-x} = -(1 + x + x^2 + x^3 + \cdots) \).- Rewrite the second term \( \frac{-1}{x-2} = \frac{-1}{-(2-x)} \).- Use \( \frac{1}{1-y} \) substitution where \( y = \frac{x}{2} \) for the second term: \( -\frac{1}{2}\left(1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \cdots \right) \).

Write the Combined Series

Combine the series into one expression:\[ \frac{1}{(1-x)(x-2)} = -(1 + x + x^2 + \cdots) - \frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4} + \cdots) \].Simplify the resulting series into a single expression.

Determine Interval of Convergence

The series for \( \frac{1}{1-x} \) converges for \( |x| < 1 \), and the series expanded from \( \frac{-1}{x-2} \) converges for \( |x/2| < 1 \) or \( |x| < 2 \). The intersection of these intervals is the convergence range, \( |x| < 1 \).

Key concepts

Interval of Convergence

The interval of convergence is a crucial concept when working with power series. It defines the set of values for which a series converges to a finite number. In other words, within this interval, the series behaves nicely, summing up to a specific value, while outside might diverge or become unmanageable.
To find the interval of convergence for the function \( \frac{1}{(1-x)(x-2)} \), we start by considering the intervals where its partial fraction decomposition gives converging series.

• Given \( \frac{1}{1-x} \), it converges for \(|x| < 1\).
• The term \( \frac{-1}{x-2} \) is rewritten using a shift, leading to its convergence when \(|x| < 2\).

When we combine these analyses, we look at the overlap of these intervals, as both series need to converge simultaneously for the overall function to converge. Therefore, the interval of convergence ends up being the common interval from \( |x| < 1 \). This means for all values \( x \) such that \( -1 < x < 1 \), the series will successfully sum up without any divergence issues.

Power Series

A power series is essentially an infinite sum of terms in the form of \( a_n(x-a)^n \), where \(a_n\) are coefficients, and \(a\) is the center of the series. Power series are handy for transforming complex functions into polynomials, making them easier to work with.
For instance, the function \( \frac{1}{1-x} \) can be expressed by the power series \( 1 + x + x^2 + x^3 + \cdots \). This series is considered centered at zero (since it has no offset \(a\)), allowing it to expand around \(x = 0\).
Power series can be found a lot in practical applications, particularly in solving differential equations, approximating functions, and analyzing convergence, where partial sums offer simpler approximations with increasing accuracy.
Using power series makes it feasible to understand intricate functions by observing them as infinitely long polynomials, giving a broader perspective and more comparability.

Series Expansion

Series expansion transforms functions into infinite polynomial series, which are generally called expansions. These expansions simplify complicated calculations by breaking down complex functions.
The primary goal here was to expand the function \( \frac{1}{(1-x)(x-2)} \) using partial fraction decomposition. By expressing the original function as \( \frac{-1}{1-x} + \frac{-1}{x-2} \), and then expanding each part using known series, we could achieve an expanded series representation that simplifies equating the function numerically

• Expanding \( \frac{-1}{1-x} \) using \( -(1 + x + x^2 + \cdots) \)
• And reformulating \( \frac{-1}{x-2} \) through a substitution

Once both parts were in series form, these were combined to create a unified expression, integrating all elements into a single series. Series expansions offer remarkable utility in providing easy, polynomial-like expressions even for complicated functions, aiding both understanding and practical computations.