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If the radius of convergence of the power series \(\Sigma_{n-0}^{\infty} c_{n} x^{n}\) is \(10,\) what is the radius of convergence of the series \(\sum_{n=1}^{\infty} n c_{n} x^{n-1} ?\) Why?

Short Answer

Expert verified
Both series have a radius of convergence of 10 because differentiation doesn't change it.

Step by step solution

01

Understanding the Radius of Convergence

The radius of convergence of a power series \(\sum_{n=0}^{\infty} c_n x^n\) is determined by the formula \( R = \frac{1}{\limsup_{n \to \infty} |c_n|^{1/n}} \). For this series, it is given as 10. This means \( R = 10 \). The terms \(c_n\) dictate the convergence behavior.
02

Relation to Derived Series

When a power series \(\sum_{n=0}^{\infty} c_n x^n\) undergoes term-by-term differentiation, the new series is \(\sum_{n=1}^{\infty} n c_n x^{n-1}\). The radius of convergence for both the original series and its derivative remains the same, provided they converge on the open interval defined by the radius of convergence.
03

Conclusion for Radius of Transformed Series

Given that differentiation does not affect the radius of convergence, the series \(\sum_{n=1}^{\infty} n c_n x^{n-1}\) retains the same radius of convergence as the original series, which is given as 10.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series
A power series is a series of the form \[ \sum_{n=0}^{\infty} c_n x^n \]where \(c_n\) represents the coefficients and \(x\) is the variable. Power series are used extensively in calculus because they provide a way to express functions as infinite polynomials.
Power series are defined for certain values of \(x\), which are determined by their radius of convergence. The radius of convergence, often denoted as \(R\), is the distance from the center of the series within which the series converges.
Understanding the concept of power series is crucial, as it lays the foundation for working with series in calculus, including tasks such as differentiation, integration, and finding convergence behaviors.
Differentiation of Series
When it comes to differentiating a power series, the process is straightforward and follows the pattern of term-by-term differentiation. If you have a series\[ \sum_{n=0}^{\infty} c_n x^n, \]its derivative can be found as\[ \sum_{n=1}^{\infty} n c_n x^{n-1}. \]
As you can see, each term’s exponent \(n\) is brought down as a coefficient, and the exponential power decreases by one. This works similarly to the power rule you may use in basic calculus assignments.
This makes differentiation of series a handy tool, especially when simplifying or finding the behavior of a function derived from a series! A key insight is that differentiating a power series does not change its radius of convergence, meaning it remains the same before and after the differentiation process.
Convergence Behavior
In the context of power series, convergence behavior refers to where and how the series converges to a finite value. Every power series has this behavior dictated by its radius of convergence. If the series converges within this radius, it means for these values of \(x\), as you sum more terms of the series, the total approaches a finite number.
Understanding the convergence behavior often entails finding the radius of convergence, \(R\), through the formula:\[ R = \frac{1}{\limsup_{n \to \infty} |c_n|^{1/n}}. \]
This is essential because, in many applications, you only need to focus on the interval \((-R, R)\), where the series behaves nicely and does not diverge.
  • The convergence is uniform within this interval.
  • Beyond this interval, the series may diverge.
  • At the endpoints, behavior can vary, and further testing may be needed.
Grasping convergence behavior is crucial for anyone dealing with infinite series and complex mathematical models where precision is required.

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Most popular questions from this chapter

Let $$ f(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} $$ Find the intervals of convergence for \(f, f^{\prime},\) and \(f^{\prime \prime}\)

If \(\Sigma a_{n}\) is convergent and \(\Sigma b_{n}\) is divergent, show that the series \(\Sigma\left(a_{n}+b_{n}\right)\) is divergent. [Hint: Argue by contradiction. \(]\)

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