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Chapter 10: Problem 12

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

Short Answer

Expert verified
Answer: The radius of convergence is \(\infty\) and the interval of convergence is \((-\infty, \infty)\).

Step by step solution

01

Apply the Ratio Test

First, apply the Ratio Test to the given power series by finding the limit as \(k\) goes to infinity of the absolute value of the ratio between the \((k+1)\)-th term and the \(k\)-th term: $$\lim_{k\to\infty} \left|\frac{\frac{(x-1)^{(k+1)}}{(k+1) !}}{\frac{(x-1)^{k}}{k !}}\right|$$
02

Simplify the expression

Simplify the expression inside the limit: $$\lim_{k\to\infty} \frac{(x-1)^{(k+1)}k!}{(k+1)!(x-1)^k}$$ We can see that \((x-1)^k\) appears in both the numerator and denominator. Cancel out these common factors: $$\lim_{k\to\infty} \frac{(x-1)k!}{(k+1)!}$$
03

Rewrite \((k+1)!\) in the denominator

Rewrite \((k+1)!\) as \((k+1)k!\): $$\lim_{k\to\infty} \frac{(x-1)k!}{(k+1)k!}$$ Then cancel out the common factors of \(k!\): $$\lim_{k\to\infty} \frac{x-1}{k+1}$$
04

Find the limit

As \(k\) approaches infinity, the denominator goes to infinity while the numerator \((x-1)\) remains constant. This means that the limit is zero: $$\lim_{k\to\infty} \frac{x-1}{k+1} = 0$$ Since the limit is 0, the Ratio Test tells us that the power series converges for all \(x\). Hence, the radius of convergence is \(\infty\).
05

Test the endpoints

Since the radius of convergence is infinity, the interval of convergence is \((-\infty, \infty)\). The power series converges for all values of \(x\).

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Most popular questions from this chapter

Tangent line is \(p_{1}\) Let \(f\) be differentiable at \(x=a\) a. Find the equation of the line tangent to the curve \(y=f(x)\) at \((a, f(a))\) b. Find the Taylor polynomial \(p_{1}\) centered at \(a\) and confirm that it describes the tangent line found in part (a).

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty} \frac{x^{k}}{2^{k}}$$

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int_{0}^{2} \frac{d x}{1-x},\) one could expand the integrand in a Taylor series and integrate term by term. b. To approximate \(\pi / 3,\) one could substitute \(x=\sqrt{3}\) into the Taylor series for \(\tan ^{-1} x\) c. \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}=2\)

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.

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{k}}{3^{k}}$$
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