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Chapter 10: Problem 26

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}$$

Short Answer

Expert verified
Answer: The radius of convergence is \(\infty\).

Step by step solution

01

Apply the Ratio Test

We start by applying the Ratio Test for absolute convergence of the power series \(\sum \frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}\): $$\lim_{k \to \infty} \frac{\frac{(-2)^{k+1}(x+3)^{k+1}}{3^{k+2}}}{\frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}}$$
02

Simplify the expression

Next, we simplify the expression within the limit: $$\lim_{k \to \infty} \frac{(-2)(x+3)}{3}$$
03

Determine the convergence condition

The power series converges absolutely if the result of the ratio test is less than 1: $$\frac{2(x+3)}{3} < 1$$ Now, we solve for the interval of convergence, \(x\):
04

Solve for x

Rearrange the inequality and simplify: $$ 2(x+3) < 3$$ $$ x+3 < \frac{3}{2}$$ $$x < -\frac{1}{2}$$ This means that the power series converges for x in the interval \((-\infty, -\frac{1}{2})\).
05

Test the endpoints

Test the endpoints by plugging in \(x=-\frac{1}{2}\) into the series: $$\sum \frac{(-2)^{k}((-1/2)+3)^{k}}{3^{k+1}}$$ This series converges by the Alternating Series Test. Thus, the interval of convergence is \((-\infty, -\frac{1}{2}]\). The radius of convergence is found by taking half the length of this interval: $$radius = \frac{1}{2}(|-\frac{1}{2} - (-\infty)|) = \frac{1}{2}\infty = \infty$$ So, the radius of convergence is \(\infty\).

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Most popular questions from this chapter

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\cos 2 x+2 \sin x$$

The period of a pendulum is given by $$T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k)$$ where \(\ell\) is the length of the pendulum, \(g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \left(\theta_{0} / 2\right),\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral and it cannot be evaluated analytically. a. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term. b. How many terms of the Taylor series do you suggest using to obtain an approximation to \(F(0.1)\) with an error less than \(10^{-3} ?\) c. Would you expect to use fewer or more terms (than in part (b)) to approximate \(F(0.2)\) to the same accuracy? Explain.

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$\frac{1}{(3+4 x)^{2}}$$

Choose a Taylor series and a center point a to approximate the following quantities with an error of \(10^{-4}\) or less. $$\sqrt[3]{83}$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sin 1$$
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