Chapter 8: Q. 17 (page 692)

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.
$\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!}$

Short Answer

Expert verified

The required answer is $\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!} = e^{- π}$

Step by step solution

Step 1. Given Information

The given series is $\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!}$

Step 2. Explanation

The maclaurin series for the function $f (x) = e^{x} is e^{x} = \sum_{k = 0}^{\infty} \frac{1}{k!} x^{k}$

So, the given series is the maclaurin series for $e^{x} at x = - π$

Since, $\sum_{k = 0}^{\infty} \frac{1}{k!} e^{k} = e^{x}$

Thus, $\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!} = e^{- π}$

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Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.
$\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.∑k=0∞-πkk!

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.
$\sum_{k = 0}^{\infty} \frac{{(- π)}^{k}}{k!}$