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Chapter 8: Problem 17

Determine the radius and interval of convergence. $$\sum_{k=0}^{\infty} \frac{k}{4^{k}} x^{k}$$

Short Answer

Expert verified
The radius of convergence for the series is 4, and the interval of convergence is \(-4 < x < 4\).

Step by step solution

01

Finding the Radius of Convergence

To find the radius of convergence, we will use the Ratio Test, which involves calculating the limit as \(k\) approaches infinity of the absolute value of the ratio of the (k+1)th term and the \(k\)th term. For our power series, this is given by \n\n\[\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\left|\frac{\frac{k+1}{4^{k+1}}x^{k+1}}{\frac{k}{4^{k}}x^k}\right|\]\n\nWe simplify this expression to \n\n\[\lim_{k\to\infty}\left|\frac{k+1}{k}\times\frac{4^{k}}{4^{k+1}}\times\frac{x^{k+1}}{x^{k}}\right|\Rightarrow\lim_{k\to\infty}|\frac{k+1}{k}\times\frac{1}{4}\times x|\]\n\nAs \(k\) approaches infinity, this limit becomes \(\frac{1}{4}|x|\). For the series to converge, the Ratio Test requires this limit to be less than 1. Therefore, we have the inequality \(\frac{1}{4}|x| < 1\), which gives |x| < 4. This implies that the radius of convergence (R) is 4.
02

Finding the Interval of Convergence

Next, we determine the interval of convergence of the series. This can be found by setting the inequality obtained from the Ratio Test to less or equal to 1 and solving for \(x\). So, we solve.\n\n\[-4 鈮� x 鈮� 4\]\n\nTo confirm that this is indeed the interval of convergence, we need to check the end points of the interval: \(x = -4\) and \(x = 4\), by directly substituting these values into the series. We substitute and find that neither series converges. Thus, the interval of convergence for the series is \(-4 < x < 4\).

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Key Concepts

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

Ratio Test
The Ratio Test is a handy tool to determine the convergence of a series. If you have a series like \[ \sum_{k=0}^{\infty} a_k \]The Ratio Test tells you to look at the limit of the absolute value of the ratio of consecutive terms:\[ \lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right| \]If this limit is less than 1, the series converges absolutely. If it is greater than 1, the series diverges. However, if it equals 1, the test is inconclusive.In our example, after simplifying the series, we find the limit is \( \frac{1}{4} |x| \). To make sure the series converges, we check if this is less than 1, leading to the inequality \( |x| < 4 \). This gives us insights into the radius of convergence, as the series will converge for any \( x \) within this boundary.
Power Series
Power series are interesting mathematical objects that look like this:\[ \sum_{k=0}^{\infty} c_k x^k \]These series involve terms of \( x^k \) where \( k \) starts from 0 and goes to infinity. The \( c_k \) are coefficients that can be determined by the problem.An important aspect of power series is they form the basis for many functions in calculus and complex analysis. They are often used in approximating more complicated functions.In the given exercise, the power series is:\[ \sum_{k=0}^{\infty} \frac{k}{4^k} x^k \]Here, the coefficient \( c_k \) is given by \( \frac{k}{4^k} \). Understanding the structure of power series like this one is essential, as they help define functions over intervals, which can be analyzed and used in various applications.
Interval of Convergence
After applying the Ratio Test, when you find the radius of convergence \( R \), the next step is often determining the interval of convergence for which the series converges. This interval provides the values of \( x \) where the series behaves nicely and converges to a value.Given the inequality from the Ratio Test, \( |x| < 4 \), we have our starting point of the interval - The interval initially appears to be \(-4 < x < 4 \). - To find the exact interval, checking endpoints is crucial.For example, substituting \( x = -4 \) and \( x = 4 \) into the original series, we noticed both series do not converge at these endpoints. This observation clarifies our interval does not include the endpoints, hence the interval is marked as \(-4 < x < 4 \). This detailed analysis guarantees the interval of convergence for the power series remains accurate.

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Most popular questions from this chapter

The function \(\sin 8 \pi t\) represents a \(4-\mathrm{Hz}\) signal \((1 \mathrm{Hz} \text { equals } 1\) cycle per second) if \(t\) is measured in seconds. If you received this signal, your task might be to take your measurements of the signal and try to reconstruct the function. For example, if you measured three samples per second, you would have the data \(f(0)=0, f(1 / 3)=\sqrt{3} / 2, f(2 / 3)=-\sqrt{3} / 2\) and \(f(1)=0\) Knowing the signal is of the form \(A\) sin \(B t,\) you would use the data to try to solve for \(A\) and \(B\). In this case, you don't have enough information to guarantee getting the right values for A and \(B\). Prove this by finding several values of \(A\) and \(B\) with \(B \neq 8 \pi\) that match the data. A famous result of \(\mathrm{H}\). Nyquist from 1928 states that to reconstruct a signal of frequency \(f\) you need at least \(2 f\) samples.

Find all values of \(p\) such that the sequence \(a_{n}=\frac{1}{n^{p}}\) converges.

Show that a power series representation of \(f(x)=\ln \left(1+x^{2}\right)\) is given by \(\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k+2}}{k+1} .\) For the partial sums \(P_{n}(x)=\sum_{k=0}^{n}(-1)^{k} \frac{x^{2 k+2}}{k+1}, \quad\) compute \(\quad\left|f(0.9)-P_{n}(0.9)\right|\) for each of \(n=2,4,6 .\) Discuss the pattern. Then compute \(\left|f(1.1)-P_{n}(1.1)\right|\) for each of \(n=2,4,6 .\) Discuss the pattern. Discuss the relevance of the radius of convergence to these calculations.

Use long division to show that \(\frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots\).

Involve the binomial expansion. Show that the Maclaurin series for \((1+x)^{y}\) is \(1+\sum_{k=1}^{\infty} \frac{r(r-1) \cdots(r-k+1)}{k !} x^{k},\) for any constant \(r\)
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