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Find the interval of convergence of the (nonpower) series and the corresponding series of derivatives. $$\sum_{k=0}^{\infty} e^{-2 k x}$$

Short Answer

Expert verified
The interval of convergence for both the given series and its derivative series is \(x > 0\).

Step by step solution

01

Analyze the given series

Firstly, we can see that the series is in the form of a geometric series: \(e^{-2kx}\) can be rewritten as \((e^{-2x})^k\). In a geometric series of the form \(a*r^k\) where 'a' is the first term, 'r' is the common ratio, and 'k' goes from 0 to infinity, the series converges if \(|r| < 1\).
02

Find the interval of convergence

For the series to converge, \((e^{-2x})^k\), we need \(|e^{-2x}| < 1\). This gives us two inequalities: \(e^{-2x} < 1\) and \(-e^{-2x} < 1\). Solving these, we get \(x > 0\). So, the interval of convergence for the given non-power series is \(x > 0\).
03

Find the convergence of the series of derivatives

Differentiate the series term by term, we get \(\sum_{k=0}^{\infty} (-2k)e^{-2kx}\). As this series is still a geometric series with common ratio \(e^{-2x}\), it will converge for the same conditions, and thus the interval of convergence is the same: \(x > 0\).

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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 a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. Mathematically, this is represented as \(a, ar, ar^2, ar^3, ...\), where \(a\) is the first term and \(r\) is the common ratio. This kind of series is encountered frequently in algebra and calculus.

When attempting to sum a geometric series, we use the formula for the sum of an infinite geometric series, \( S = \frac{a}{1-r} \), but this formula is only valid if the absolute value of the common ratio is less than one, that is \(|r| < 1\). Applying this to our exercise, we identified that the series \(\sum_{k=0}^{\infty} e^{-2 k x}\) is a geometric series with \(a = 1\) and the common ratio \(r = e^{-2x}\).

We can see the geometric nature by rewriting the term as \(e^{-2kx} = (e^{-2x})^k\), which clearly exhibits the structure of a geometric series. This understanding allows us to proceed with finding the interval of convergence by analyzing the common ratio.
Series of Derivatives
When dealing with infinite series, particularly in calculus, one might investigate the behavior of the derivatives of the terms in the series. The series of derivatives refers to taking the derivative of each term in the original series, which can often provide insight into the convergence or behavior of the series for different intervals.

In the case of a geometric series, the process of finding the series of derivatives is systematic. For the function \(e^{-2kx}\), deriving term by term with respect to \(x\) yields \(\sum_{k=0}^{\infty} (-2k)e^{-2kx}\), which introduces a factor of \(k\) to each term of the series. Despite this change, the series retains its geometric character because the derived terms still involve a common ratio of \(e^{-2x}\), just multiplied by \(k\). This means that the series of derivatives will converge under the same condition as the original series: when \(|e^{-2x}| < 1\).

Importance of Series of Derivatives

Why is this important? Because understanding the convergence of a series of derivatives helps us analyze the behavior of functions represented by series. If the original series represents a function, then knowing where its derivative series converges gives us information about the function's continuity and differentiability within certain intervals.
Convergence Criteria
The convergence of an infinite series is not guaranteed, and various tests and criteria are used in mathematics to determine whether or not a series converges. The convergence criteria for geometric series claim that such a series converges if the common ratio's absolute value is less than one, \(|r| < 1\).

Applying this criterion to our original problem, we examine the common ratio \(e^{-2x}\). For convergence, we require \(\left|e^{-2x}\right| < 1\). We analyze this condition by breaking it into two parts: \(e^{-2x} < 1\) and \(e^{-2x} > -1\). The second inequality is always true, since \(e^{-2x}\) is always positive. The first inequality simplifies to \(x > 0\). This tells us the interval of convergence for the original series is for all \(x\) greater than zero.

Why Interval Matters

The interval of convergence is crucial as it pinpoints where the function, represented by a series, behaves predictably and where it has a real sum. Anything outside this interval results in the series diverging, meaning it does not approach a finite value as \(k\) goes to infinity. For functions represented by series and their applications in real-world scenarios, understanding where they converge is essential for accurate modeling and analysis.

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Most popular questions from this chapter

Determine the radius and interval of convergence. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} x^{k}}{\sqrt{k}}$$

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

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