Question

The expansion of $ {{\text{e}}^{\text{x}}} $ is which among the following?
 $
  {\text{A}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{\text{r}}}} \\
  {\text{B}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}} \\
  {\text{C}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{{\text{r + 1}}}}}}{{{\text{r + 1}}}}} \\
  {\text{D}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{{\text{r + 1}}}}}}{{\left( {{\text{r + 1}}} \right)!}}} \\
 $

Answer 1

Hint: To determine the expansion of $ {{\text{e}}^{\text{x}}} $ , we observe the Taylor series expansion for the given term and then evaluate which among the given options is a valid general term for the expansion. The general term should hold true for all the individual terms of the expansion.


Complete step-by-step answer:
Given Data, $ {{\text{e}}^{\text{x}}} $
The Taylor series expansion of the term $ {{\text{e}}^{\text{x}}} $ is given as
 $ {{\text{e}}^{\text{x}}} $ = $ 1 + {\text{x + }}\dfrac{{{{\text{x}}^2}}}{{2!}} + \dfrac{{{{\text{x}}^3}}}{{3!}} + \dfrac{{{{\text{x}}^4}}}{{4!}} + ........\infty $
The term $ \sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}} $ holds true for every term of the expansion of $ {{\text{e}}^{\text{x}}} $ .
For the first term of $ {{\text{e}}^{\text{x}}} $ , i.e. r = 0
 $ \sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}} $ = $ \sum\limits_{{\text{r = 0}}}^0 {\dfrac{{{{\text{x}}^0}}}{{{\text{0!}}}} = 1} $ --- (0! = 1 and any number to the power 0 = 1)
It also holds true for all the other terms of the equation.
Hence the expansion of $ {{\text{e}}^{\text{x}}} $ is $ \sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}} $ .
Option B is the correct answer.

Note – In order to solve this type of questions the key is to understand the concept of ‘e’ also known as the Napier’s constant used for the natural logarithm and it has a value of 2.71828. The Taylor series of a polynomial is a polynomial itself, $ {{\text{e}}^{\text{x}}} $ expansion holds because the derivative of $ {{\text{e}}^{\text{x}}} $ is $ {{\text{e}}^{\text{x}}} $ itself and $ {{\text{e}}^0} $ equals to 1.