Question

Use appropriate Maclaurin series to express the quantities in Exercises $67-76$as alternating series. Then use Theorem 7.38 to approximate the value of the specified quantities to within $0.001$of their actual value. How many terms in each series would be needed to approximate the given quantity to within ${10}^{-6}$of its value? In Exercises $73-76$be sure to convert to radian measure first.

$e^{-2}$

Short answer

Therefore, the approximate the value of $e^{-2}$ up to three decimal places is $0.135$

Step-by-step solution

Given information

Consider the function $e^{-2}$

:calculation

The Maclaurin series for the function $f\left(x\right)=e^x$is

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$

So, to find the Maclaurin series for the function $e^{-2}$, put $x=-2$Therefore,

$f(x=-2)=\sum _{k=0}^{\infty }\frac{1}{k!}(-2{)}^k$

That is

$e^{-2}=\sum _{\ell =0}^{\infty }\frac{1}{k!}(-2{)}^k$

Now, to approximate the value of $e^{-2}$up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$e^{-2}=\sum _{k=0}^{\infty }\frac{1}{k!}(-2{)}^k$

$=1-2+\frac{1}{2!}\left(4\right)-\frac{1}{3!}\left(8\right)+\frac{1}{4!}\left(16\right)-\frac{1}{5!}\left(32\right)$

$=1-2+2-1.33333+0.66666-0.26666$

$=0.13533$

Therefore, the approximate the value of $e^{-2}$up to three decimal places is $0.135$

Also, to approximate the quantity $e^{-2}$to within ${10}^{-6}$of its value, the number of terms required is 7