Question

The exponential function ${e^x}$ can be defined as a power series as:
${e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...$
Can you use this definition to evaluate $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}}$ ?

Answer 1

In the above question, we are given the exponential function ${e^x}$ and its expansion in the form series. We have to use the power series to evaluate another series $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}}$ . Here, we can separate the function in two parts as ${e^{ - 0.2}}$ and $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$ so that we can easily evaluate $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$ using the given power series.

Complete step by step answer:
Given the power series of the exponential function ${e^x}$ is,
$\Rightarrow {e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...$
We have to use the above power series to evaluate the function,
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}}$
We can also write the above function as,
$\Rightarrow {e^{ - 0.2}}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$
This is the summation over $n$. So, taking out $e^{-0.2}$ out of the summation won’t change the value.
Now, consider $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$ .
Now since,
${e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...$
Using the above given expansion of ${e^x}$ in the expression $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$ , we can write,
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} = {e^{0.2}}$
Now, putting $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}} = {e^{0.2}}$ in the expression ${e^{ - 0.2}}\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}}}{{n!}}}$ we get,
$\Rightarrow {e^{ - 0.2}} \cdot {e^{0.2}}$
That gives,
$\Rightarrow \dfrac{{{e^{0.2}}}}{{{e^{0.2}}}}$
Hence,
$\Rightarrow 1$
That is the required value of the above given function $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}}$ .
Therefore, the value of the series $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {0.2} \right)}^n}{e^{ - 0.2}}}}{{n!}}}$ is $1$.

Note:
The number $e$ given in the above question is known as the Euler’s number. The function of the Euler’s number of the type ${e^x}$ with the independent variable $x$ ranging over the entire real number line as the exponent of the positive constant number $e$ . It is known as a natural exponential function or just as an exponential function.
The Euler’s number $e$ is a constant having a value of nearly $e = 2.718281828...$
The inverse function of the natural exponential function ${e^x}$ is ${\log _e}x$ .
This is known as the natural logarithmic function or just as the logarithmic function, also denoted by $\ln x$ .