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Chapter 8: Q. 57 (page 659)

Find the Maclaurin series for the functions in Exercises 51鈥�60
by substituting into a known Maclaurin series. Also, give the
interval of convergence for the series.
$\frac{e^{x} + e^{- x}}{2}$

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given Information

Consider the function $\frac{e^{x} + e^{- x}}{2}$

02

Step 2

We know that the Maclaurin series for the function $g (x) = e^{x}$ is $e^{x} = {鈭�}_{k - 0}^{鈭�} \frac{1}{k!} x^{k}$

So, the series for $h (x) = e^{- x}$ can be found by substituting $x$ by $- x$

That is $e^{- x} = {鈭�}_{k = 0}^{鈭�} \frac{1}{k!} {(- x)}^{k}$

Finally, to find the result for the function role="math" localid="1649869343697" $f (x) = \frac{e^{x} + e^{- x}}{2}$ we add the series for $e^{x}$ and $e^{- x}$ and then divide by $2$

Thus,

$\frac{e^{x} + e^{- x}}{2} = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + . . .$ implies that,

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

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Most popular questions from this chapter

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: ${鈭�}_{k = 0}^{鈭�} \frac{1}{k!} x^{k}$

What is Lagrange鈥檚 form for the remainder? Why is Lagrange鈥檚 form usually more useful for analyzing the remainder than the definition of the remainder or the integral provided by Taylor theorem?

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

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 $p$ .

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