Chapter 11: Problem 42
Suppose that the radius of convergence of the power series \(\sum c_{n} x^{n}\) is \(R .\) What is the radius of convergence of the power series \(\Sigma c_{n} x^{2 m} ?\)
Short Answer
Expert verified
The radius of convergence is \(\sqrt{R}\).
Step by step solution
01
Understand the Power Series
First, we have a power series \(\sum_{n=0}^{\infty} c_{n} x^{n}\), with a given radius of convergence \(R\). This means for \(|x| < R\), the series converges.
02
Identify the Pattern in the Given Series
Consider the series \(\sum_{m=0}^{\infty} c_{m} x^{2m}\), which contains terms \(x^{2m}\), effectively replacing \(n\) with \(2m\). This affects the convergence in terms of \(x^2\) instead of \(x\).
03
Relate the New Series to the Original Series
The transformation \(x^n\) to \(x^{2m}\) implies we are dealing with powers of \((x^2)^m\). Thus, the series behaves like \(\sum_{m=0}^{\infty} c_{m} (x^2)^{m}\).
04
Calculate the New Radius of Convergence
The original series has a radius \(R\). For the series \(\sum_{m=0}^{\infty} c_{m} (x^2)^{m}\), since it follows the power series format in terms of \(x^2\), the radius of convergence for \(x^2\) is still \(R\). Thus, the effective range for \(x\) to satisfy \(|x^2| < R\) is when \(|x| < \sqrt{R}\).
05
Conclusion
Therefore, the radius of convergence for the power series \(\sum_{m=0}^{\infty} c_{m} x^{2m}\) is \(\sqrt{R}\), due to the transformation affecting the convergence in terms of \(x^2\) instead of \(x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Convergence
In mathematics, when we talk about the radius of convergence of a power series, we're discussing the distance within which the series converges around its center. A power series is an infinite sum of terms in the form \( \sum_{n=0}^{\infty} c_{n} x^{n} \), where \( c_n \) are coefficients and \( x \) is a variable. The radius of convergence, denoted by \( R \), is crucial to determine where this series behaves well.
- For any point \( |x| < R \), the series converges, meaning the sum approaches a finite value.
- Conversely, if \( |x| > R \), the series diverges—this means the sum doesn't settle to a specific number.
Transformation of Series
Transforming a series means adjusting the form of the power series while examining how its properties, such as convergence, are affected. In this context, it refers to changing the power of \( x \) in the series from \( x^n \) to \( x^{2m} \). This substitution essentially squares the exponent of \( x \), altering the form of the convergence condition. For example, in the altered series given by \( \sum_{m=0}^{\infty} c_{m} x^{2m} \), the transformation implies that instead of looking at simply \( x \), we view the series in terms of \( (x^2) \).
- This means the variable we analyze for convergence changes from \( x \) to \( x^2 \).
- Every place you had \( x^n \) before, you now have \( (x^2)^{m} \), which impacts how you calculate convergence range.
Convergence of Series
The convergence of a series is a fundamental concept that tells us whether the series sums up to a finite value or not. Specifically, the convergence of a power series, like \( \sum_{n=0}^{\infty} c_{n} x^{n} \), depends heavily on the radius of convergence \( R \). The power series converges absolutely if the sum of the absolute values of its terms is finite.When you transform a series, such as changing \( x^n \) to \( x^{2m} \), the new form \( \sum_{m=0}^{\infty} c_{m} x^{2m} \) converges under a modified condition.
- The condition \( |x^2| < R \) arises, leading to the conclusion that \( |x| < \sqrt{R} \) is necessary for convergence.