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Chapter 8: Q 55. (page 670)

Explain why the series is not a power series in $x - x_{0}$ .Then use the ratio test for absolute convergence to find the values of $x$ for which the given series converge ${鈭�}_{k = 0}^{鈭�} \frac{{(\sin x)}^{k}}{k!}$

Short Answer

Expert verified

The value of $x$ for which the series ${鈭�}_{k = 0}^{鈭�} \frac{{(\sin x)}^{k}}{k!}$ converges when $x 鈭� R$ .

Step by step solution

01

Step 1. Given information.

The given power series is ${鈭�}_{k = 0}^{鈭�} \frac{{(\sin x)}^{k}}{k!}$

02

Step 2. Find the values of x for which the given series converge.

Since, the series consists of power of $S i n x$ . So the series in not power series in $x - x_{0}$ .

Now,

$b_{k} = \frac{{(\sin x)}^{k}}{k!}$ and $b_{k + 1} = \frac{{(\sin x)}^{k + 1}}{(k + 1)!}$

Thus,

$\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \underset{k 鈫� 鈭�}{l i m} |\frac{\frac{{(\sin x)}^{k + 1}}{(k + 1)!}}{\frac{{(\sin x)}^{k}}{k!}}| = \underset{k 鈫� 鈭�}{l i m} |\frac{\sin x}{(k + 1)}|$

So, for $k 鈫� 鈭�$ the value of limit will always be zero.

Therefore, the value of $x$ for which the series ${鈭�}_{k = 0}^{鈭�} \frac{{(\sin x)}^{k}}{k!}$ converges when $x 鈭� R$ .

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Most popular questions from this chapter

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{k + 1}{3^{k}} {(2 x - 5)}^{k}$

What is a Taylor polynomial for a function f at a point $x_{0}$ ?

What is $x_{0}$ if $(p, q)$ is the interval of convergence for the power series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} {(x 鈭� x_{0})}^{k}$ ?

Prove that if $(- p, p]$ is the interval of convergence for the series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} x^{k}$ , then the series converges conditionally at $p$ .

Show that the power series ${鈭�}_{k = 1}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{k^{2}} x^{k}$ converges absolutely when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

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