Q. 67 - Calculus

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.

$e^{-0.3}$

Step-by-Step Solution

Answer

The approximate the value of $e^{-0.3}$ up to three decimal places is $0.741$

Also, to approximate the quantity $e^{-0.3}$ to within ${10}^{-6}$ of its value, the number of terms required is $7$

step 1: given information

Consider the function $e^{-03}$

Also, $f\left(x\right)=e^x$

step 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 $e^{-0.3}$, put $x=-0.3$

Therefore,

$f(x=-0.3)=\sum _{k=0}^{\infty }\frac{1}{k!}(-0.3{)}^k$

That is

$e^{-0.3}=\sum _{k=0}^{\infty }\frac{1}{k!}(-0.3{)}^2$

Now, to approximate the value of $e^{-03}$ up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$$\begin{gathered} e^{-0.3}=\sum _{k=0}^{\infty }\frac{1}{k!}(-0.3{)}^k \\ =1-0.3+\frac{1}{2!}(0.09)-\frac{1}{3!}(0.027)+\frac{1}{4!}(0.0081)-\cdots \\ =1-0.3+0.045-0.0045+0.0003375 \\ =0.74053 \end{gathered}$$

Therefore, the approximate the value of $e^{-0.3}$ up to three decimal places is $0.741$

Also, to approximate the quantity $e^{-0.3}$ to within ${10}^{-6}$ of its value, the number of terms required is $7$