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Chapter 8: Q. 22 (page 692)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given Information

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

02

Step 2. Explanation

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

So, the given series is the maclaurin series for $\frac{1}{1 - x} a t x = - \frac{1}{蟺}$

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

Thus,

${鈭�}_{k = 0}^{鈭�} {(- \frac{1}{蟺})}^{k} = \frac{1}{1 - (- \frac{1}{蟺})} {鈭�}_{k = 0}^{鈭�} {(- \frac{1}{蟺})}^{k} = \frac{蟺}{蟺 + 1}$

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Most popular questions from this chapter

In Exercises 41鈥�48 find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$

$46 . \sqrt[3]{x}, 1$

If m is a positive integer, how can we find the Maclaurin series for the function $c x^{m} f (x)$ if we already know the Maclaurin series for the function f(x)? How do you find the interval of convergence for the new series?

What is the relationship between a Maclaurin series and a power series in x?

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

The second-order differential equation

$x^{2} y^{''} + x y^{'} + (x^{2} - p^{2}) = 0$

where p is a nonnegative integer, arises in many applications in physics and engineering, including one model for the vibration of a beaten drum. The solution of the differential equation is called the Bessel function of order p, denoted by $J_{p} (x)$ . It may be shown that $J_{p} (x)$ is given by the following power series in x:

$J_{p} (x) = {鈭�}_{k = 0}^{鈭�} \frac{(- 1)^{k}}{k! (k + p)! 2^{2 k + p}} x^{2 k + p}$

Graph the first four terms in the sequence of partial sums of $J_{1} (x)$ .

See all solutions

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