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Chapter 10: Problem 55

Find the radius of convergence of \(\sum \frac{k ! x^{k}}{k^{k}}\)

Short Answer

Expert verified
Answer: The radius of convergence for the power series is \(e\).

Step by step solution

01

Identify the sequence terms

In this power series \(\sum \frac{k ! x^{k}}{k^{k}}\), we can identify the sequence terms \(a_k = \frac{k ! x^{k}}{k^{k}}\).
02

Apply the Ratio Test

To use the Ratio Test, we must first find the ratio of consecutive terms \(\frac{a_{k+1}}{a_k}\). Note that \(a_{k+1} = \frac{(k+1) ! x^{k+1}}{(k+1)^{k+1}}\). Divide \(a_{k+1}\) by \(a_k\): \(\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1) ! x^{k+1}}{(k+1)^{k+1}}}{\frac{k ! x^{k}}{k^{k}}} = \frac{(k+1) ! x^{k+1} k^{k}}{k ! x^{k} (k+1)^{k+1}}\)
03

Simplify the Ratio Expression

Now we simplify the expression: \(\frac{a_{k+1}}{a_k} = \frac{(k+1) k ! x^{k} k^{k} x}{k ! (k+1)^{k} (k+1) x^{k}}\) Since we have \((k+1) k ! = (k+1) !\), this expression simplifies to: \(\frac{a_{k+1}}{a_k} = \frac{x k^{k}}{(k+1)^{k}}\)
04

Find the Limit

Now, we find the limit as \(k \to \infty\): \(L = \lim_{k \to \infty} \frac{x k^{k}}{(k+1)^{k}} = x \lim_{k \to \infty} \frac{k^{k}}{(k+1)^{k}}\) Consider the expression inside the limit: \(\lim_{k \to \infty} \frac{k^{k}}{(k+1)^{k}} = \lim_{k \to \infty} \left(\frac{k}{k+1}\right)^{k}\) This is a well-known limit that approaches \(\frac{1}{e}\). Therefore: \(L = x \cdot \frac{1}{e}\)
05

Determine Convergence

To ensure convergence, \(L\) must be less than 1: \(x \cdot \frac{1}{e} < 1\) Solving for \(x\), we have: \(|x| < e\)
06

Find the Radius of Convergence

Now we can find the radius of convergence \(R\): \(R = e\) So, the radius of convergence for the power series \(\sum \frac{k ! x^{k}}{k^{k}}\) is \(e\).

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Most popular questions from this chapter

The expected (average) number of tosses of a fair coin required to obtain the first head is \(\sum_{k=1}^{\infty} k\left(\frac{1}{2}\right)^{k} .\) Evaluate this series and determine the expected number of tosses. (Hint: Differentiate a geometric series.)

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. $$f(x)=\frac{e^{x}+e^{-x}}{2}$$

Compute the coefficients for the Taylor series for the following functions about the given point a and then use the first four terms of the series to approximate the given number. $$f(x)=1 / \sqrt{x} \text { with } a=4 ; \text { approximate } 1 / \sqrt{3}$$

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.

Let $$f(x)=\sum_{k=0}^{\infty} c_{k} x^{k} \quad \text { and } \quad g(x)=\sum_{k=0}^{\infty} d_{k} x^{k}$$ a. Multiply the power series together as if they were polynomials, collecting all terms that are multiples of \(1, x,\) and \(x^{2} .\) Write the first three terms of the product \(f(x) g(x)\) b. Find a general expression for the coefficient of \(x^{n}\) in the product series, for \(n=0,1,2, \ldots\)
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