
Find the value of ${{e}^{-\dfrac{1}{5}}}$ correct to four places of decimal.
Answer
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Hint: For solving this problem, the expansion of ${{e}^{x}}$ is required in terms of x. By using this expansion and putting x as $-\dfrac{1}{5}$, we can evaluate the required value correctly up to 4 decimal places.
Complete step-by-step answer:
In our question, we are required to evaluate the value of ${{e}^{-\dfrac{1}{5}}}$ correct to four decimal places. First, we write the expansion of ${{e}^{x}}$ in terms of x up to infinity and then put the value of x in that expression to calculate the value ${{e}^{-\dfrac{1}{5}}}$ correct up to 4 decimal places. So, the expansion of is ${{e}^{x}}$:
${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+......\text{ }to\text{ }\infty \text{ }\ldots (1)$
Now, putting the value of x in equation (1) as $x=-\dfrac{1}{5}$, we get
${{e}^{-\dfrac{1}{5}}}=1-\dfrac{1}{5}+\dfrac{1}{2!}\cdot {{\left( -\dfrac{1}{5} \right)}^{2}}+\dfrac{1}{3!}\cdot {{\left( -\dfrac{1}{5} \right)}^{3}}+\dfrac{1}{4!}\cdot {{\left( -\dfrac{1}{5} \right)}^{4}}.......\infty $
Now, simplifying each term by multiplication, we get
$\begin{align}
& {{e}^{-\dfrac{1}{5}}}=1-\dfrac{1}{5}+\dfrac{1}{50}-\dfrac{1}{750}+\dfrac{1}{15000}.......\infty \\
& {{e}^{-\dfrac{1}{5}}}=1-0.2+0.02-0.001333+0.0000066 \\
& {{e}^{-\dfrac{1}{5}}}=0.8187 \\
\end{align}$
So, the correct value of ${{e}^{-\dfrac{1}{5}}}$ to four decimal places is 0.8187.
Note: The key step in solving this problem is the knowledge of series expansion of various functions of x. Students must be careful while putting the value of x in the expansion. They must take care of the negative sign and power associated with sign to avoid errors in calculation.
Complete step-by-step answer:
In our question, we are required to evaluate the value of ${{e}^{-\dfrac{1}{5}}}$ correct to four decimal places. First, we write the expansion of ${{e}^{x}}$ in terms of x up to infinity and then put the value of x in that expression to calculate the value ${{e}^{-\dfrac{1}{5}}}$ correct up to 4 decimal places. So, the expansion of is ${{e}^{x}}$:
${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+......\text{ }to\text{ }\infty \text{ }\ldots (1)$
Now, putting the value of x in equation (1) as $x=-\dfrac{1}{5}$, we get
${{e}^{-\dfrac{1}{5}}}=1-\dfrac{1}{5}+\dfrac{1}{2!}\cdot {{\left( -\dfrac{1}{5} \right)}^{2}}+\dfrac{1}{3!}\cdot {{\left( -\dfrac{1}{5} \right)}^{3}}+\dfrac{1}{4!}\cdot {{\left( -\dfrac{1}{5} \right)}^{4}}.......\infty $
Now, simplifying each term by multiplication, we get
$\begin{align}
& {{e}^{-\dfrac{1}{5}}}=1-\dfrac{1}{5}+\dfrac{1}{50}-\dfrac{1}{750}+\dfrac{1}{15000}.......\infty \\
& {{e}^{-\dfrac{1}{5}}}=1-0.2+0.02-0.001333+0.0000066 \\
& {{e}^{-\dfrac{1}{5}}}=0.8187 \\
\end{align}$
So, the correct value of ${{e}^{-\dfrac{1}{5}}}$ to four decimal places is 0.8187.
Note: The key step in solving this problem is the knowledge of series expansion of various functions of x. Students must be careful while putting the value of x in the expansion. They must take care of the negative sign and power associated with sign to avoid errors in calculation.
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