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Chapter 8: Q. 18 (page 692)

Use an appropriate Maclaurin series to find the values of the series in Exercises 17鈥�22.
${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!}$

Short Answer

Expert verified

The required answer is ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!} = - 1$

Step by step solution

01

Step 1. Given Information

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!}$

02

Step 2. Explanation

The Maclaurin series for the function $f (x) = \cos x is \cos x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} x^{2 k}}{(2 k)!}$

So, the series ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!}$ is the maclaurin series for $\cos x at x = 蟺$

Since, ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} x^{2 k}}{(2 k)!} = \cos x$

Thus,

${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!} = 肠辞蝉蟺 {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k} 蟺^{2 k}}{(2 k)!} = - 1$

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Most popular questions from this chapter

In Exercises 49鈥�56 find the Taylor series for the specified function and the given value of $x_{0}$ . Note: These are the same functions and values as in Exercises 41鈥�48.

$50 . e^{x}, 1$

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}$

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

In Exercises 49鈥�56 find the Taylor series for the specified function and the given value of $x_{0}$ . Note: These are the same functions and values as in Exercises 41鈥�48.

role="math" localid="1649406173168" $49 . \cos (x), \frac{蟺}{2}$

Show that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ converges conditionally 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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