Chapter 9: Problem 3
Find a power series representation for \(2 /(1-x)^{3}\).
Short Answer
Expert verified
The power series representation is \( \sum_{n=0}^{\infty} (n+2)(n+1) x^n \) for \(|x| < 1\).
Step by step solution
01
Identify the function form
Recognize that the function \( 2/(1-x)^3 \) is similar to the form of geometric series \( \frac{1}{1-x} \). However, it involves a third power which suggests a step involving derivatives of the geometric series.
02
Recall the geometric series
Recall that the geometric series \( \frac{1}{1-x} \) can be represented as an infinite series: \( \sum_{n=0}^{\infty} x^n \) for \(|x| < 1\). This is the base form from which we will work.
03
Differentiate the geometric series
Differentiate \( \frac{1}{1-x} \) with respect to \(x\) to replicate the denominator \( (1-x)^2 \). The derivative is \( \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} \).
04
Differentiate again
Differentiate \( \frac{1}{(1-x)^2} \) with respect to \(x\) to obtain \( \frac{1}{(1-x)^3} \). The double derivative is \( \frac{d^2}{dx^2} \frac{1}{1-x} = \frac{2}{(1-x)^3} \), which corresponds to \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} \).
05
Adjust the power series
Multiply the entire series from Step 4 by \(2\) to adjust for the original function, yielding: \( 2 \sum_{n=2}^{\infty} n(n-1) x^{n-2} \). Shift the index to start from \(n=0\) by letting \(n \to n+2\), resulting finally in \( \sum_{n=0}^{\infty} (n+2)(n+1) x^n \).
06
Combine insights
Combine all steps to form the power series representation for \( \frac{2}{(1-x)^3} \), which is: \( \sum_{n=0}^{\infty} (n+2)(n+1) x^n \) for \(|x| < 1\).
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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 one the simplest types of series to understand and work with. Essentially, a geometric series involves a constant ratio between consecutive terms. This can be expressed using the formula: \[ \sum_{n=0}^{fty} ar^n = a + ar + ar^2 + ar^3 + \cdots \]where \(a\) is the first term and \(r\) is the common ratio. A classic example of a geometric series is \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) for \(|x| < 1\). Here, \(a=1\) and \(r=x\). This represents an infinite series where each term is the previous term multiplied by \(x\).
- The geometric series is important as it forms the base for many more complex series.
- It must converge for calculations to be valid, which is why \(|x| < 1\) is a necessary condition.
Differentiation
Differentiation is a fundamental aspect of calculus that involves finding the rate at which a function changes. In the context of a power series, differentiation is used to transform one type of series into another, allowing us to handle more complex expressions. When differentiating the geometric series \( \sum_{n=0}^{\infty} x^n \), we derive new series representations. For instance:
- The first derivative of \( \frac{1}{1-x}\) is \( \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} \).
- The second derivative gives \( \frac{2}{(1-x)^3} \), leading to \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} \).
Convergence
Convergence is a critical concept when working with series, indicating if a series sums to a finite value. For power series representation, ensuring convergence guarantees that calculations reflect true values rather than diverging to infinity. When dealing with geometric series or its derivatives like \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), convergence is achieved if \(|x| < 1\). Here is why convergence is important:
- Without convergence, infinite series calculations can't yield meaningful results.
- Convergence ensures the series can be used in practical scenarios, such as physics or engineering calculations.