Problem 7
Question
find the power series representation for \(f(x)\) and specify the radius of convergence. Each is somehow related to a geometric series. $$ f(x)=\frac{x^{2}}{1-x^{4}} $$
Step-by-Step Solution
Verified Answer
The power series is \( \sum_{n=0}^{\infty} x^{4n + 2} \), and the radius of convergence is 1.
1Step 1: Identify the function form
The given function is similar to the standard geometric series form \( \frac{1}{1-a} \). Here, \( a = x^4 \) instead of \( x \), and we are multiplying it by \( x^2 \).
2Step 2: Recognize the geometric series pattern
Recall the geometric series formula: \( \sum_{n=0}^{\infty} a^n = \frac{1}{1-a} \). By comparing this with the function \( f(x) = \frac{x^2}{1-x^4} \), it suggests that \( \sum_{n=0}^{\infty} (x^4)^n = \frac{1}{1-x^4} \).
3Step 3: Rewrite the function
Express \( f(x) \) using the geometric series: \[f(x) = x^2 \sum_{n=0}^{\infty} (x^4)^n = \sum_{n=0}^{\infty} x^2 \cdot x^{4n} = \sum_{n=0}^{\infty} x^{4n + 2}.\]
4Step 4: Determine the radius of convergence
For the geometric series \( \sum_{n=0}^{\infty} (x^4)^n \), it converges when \(|x^4| < 1\) or equivalently, when \(|x|^4 < 1\). Solving \(|x|^4 < 1\) gives \(|x| < 1\). Therefore, the radius of convergence is 1.
Key Concepts
Radius of ConvergenceGeometric SeriesCalculus
Radius of Convergence
Understanding the radius of convergence is crucial when working with power series, as it defines the interval where the series converges to a function. With our function \[f(x) = \frac{x^2}{1 - x^4}\]we have identified it as a geometric series. Geometric series converge when the absolute value of their terms is less than 1.
In this scenario, for the series \[\sum_{n=0}^{\infty} (x^4)^n\]this means the condition \(|x^4| < 1\).
By taking the fourth root, we find \(|x| < 1\), indicating that any value of \(x\) within this interval will lead to convergence. Thus, the radius of convergence, which is simply the distance from the center of the interval where the series converges, is \(1\).
This concept ensures that for any \(x\) in the interval \((-1, 1)\), the power series expansion of \(f(x)\) converges.
In this scenario, for the series \[\sum_{n=0}^{\infty} (x^4)^n\]this means the condition \(|x^4| < 1\).
By taking the fourth root, we find \(|x| < 1\), indicating that any value of \(x\) within this interval will lead to convergence. Thus, the radius of convergence, which is simply the distance from the center of the interval where the series converges, is \(1\).
This concept ensures that for any \(x\) in the interval \((-1, 1)\), the power series expansion of \(f(x)\) converges.
- The radius of convergence is determined by solving \(|x^4| < 1\).
- The solution \(|x| < 1\) indicates the interval of convergence.
- In this case, it is \(-1 < x < 1\).
Geometric Series
The geometric series is foundational in understanding power series representations. It is defined by a series where each term is a constant multiple of the previous one. This consistency is expressed in the formula:\[\sum_{n=0}^{\infty} a^n = \frac{1}{1-a}\]where \(a\) is the common ratio.
Our function \(f(x)\) mimics this form. It can be rewritten using a geometric series by expressing it as:\[f(x) = x^2 \sum_{n=0}^{\infty}(x^4)^n\]This transformation showcases the flexibility of geometric series in rewriting complex expressions into simpler, infinite series.
When utilized correctly, the geometric series concept allows us to manipulate and reframe functional equations into more manageable forms, significantly aiding in calculations and comprehension.
Our function \(f(x)\) mimics this form. It can be rewritten using a geometric series by expressing it as:\[f(x) = x^2 \sum_{n=0}^{\infty}(x^4)^n\]This transformation showcases the flexibility of geometric series in rewriting complex expressions into simpler, infinite series.
- Each term in the geometric series represents an increased power of the variable.
- It provides a powerful way to express functions as infinite sums.
When utilized correctly, the geometric series concept allows us to manipulate and reframe functional equations into more manageable forms, significantly aiding in calculations and comprehension.
Calculus
Calculus plays a pivotal role in understanding how series can represent functions and in determining their properties, like convergence. When we find the power series representation of a function, calculus allows us to manipulate and analyze this series further.
For example, in integrating or differentiating complex series like \(\frac{x^2}{1-x^4}\), calculus techniques let us explore rates of change and accumulation for functions expressed as infinite series.
In essence, calculus not only provides the groundwork for power series analysis but also ensures we understand the broader implications of these mathematical expressions, enhancing our ability to predict and model real-world phenomena.
For example, in integrating or differentiating complex series like \(\frac{x^2}{1-x^4}\), calculus techniques let us explore rates of change and accumulation for functions expressed as infinite series.
- Calculus bridges the gap between algebraic expressions and their numeric outcomes.
- It helps in understanding the behavior of functions within their convergence intervals.
- By using derivative and integral calculus, one can derive or integrate series term by term for additional insights.
In essence, calculus not only provides the groundwork for power series analysis but also ensures we understand the broader implications of these mathematical expressions, enhancing our ability to predict and model real-world phenomena.
Other exercises in this chapter
Problem 7
Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12) .\) $$ f(x)=\tan ^{-1} x $$
View solution Problem 7
Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications,
View solution Problem 7
Show that each series converges absolutely. $$ \sum_{n=1}^{\infty}\left(-\frac{3}{4}\right)^{n} $$
View solution Problem 7
Find the convergence set for the given power series. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n} $$
View solution