Q. 33 - Calculus

Question

Find the Maclaurin series for $e^{-x}$, and use it to approximate $\frac{1}{\sqrt{e}}$ to within 0.001 of its value. How many terms would you need to approximate $\frac{1}{\sqrt{e}}$ to within ${10}^{-6}$ of its value? 

Step-by-Step Solution

Answer

The Maclaurin series for $e^{-x}$ can be given as

$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-...$

The Maclaurin series for $e^{-\frac{1}{2}}$ can be given as

$e^{-\frac{1}{2}}=1-\frac{1}{2}+\frac{1}{8}-\frac{1}{48}+\frac{1}{384}-...$\

And first eight terms are required tro get value corrected up to ${10}^{-6}$

Step 1: Given information

We are given the Maclaurin series for $e^x$.

Step 2: Find the Maclaurin series for e - x

We know the Maclaurin series for $e^x$ which can be given as

$$\begin{gathered} e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+... \\ Replacing x by -x \\ {e}^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-... \end{gathered}$$

Is the required Maclaurin series.

Now substitute $x=\frac{1}{2}$, We get

$e^{-\frac{1}{2}}=1-\frac{1}{2}+\frac{1}{8}-\frac{1}{48}+\frac{1}{384}-...$

Also first 8 terms are reuqired to get values corrected upto ${10}^{-6}$