Question

How do I find a power series representation for ${{e}^{x}}$ and what is the radius of convergence?

Answer 1

Hint: A power series about $x=a$ is a series of the form $\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{\left( x-a \right)}^{n}}={{c}_{0}}+{{c}_{1}}\left( x-a \right)+{{c}_{2}}{{\left( x-a \right)}^{2}}+...+{{c}_{n}}{{\left( x-a \right)}^{n}}+...}$ in which the center $a$ and the coefficients ${{c}_{0}},{{c}_{1}},...{{c}_{n}},...$ are constants. Half of the length of the interval of convergence is called the radius of convergence.

Complete step by step answer:
Let us consider the given function ${{e}^{x}}.$
We are asked to find a power series representation of this function. Also, we need to find the radius of convergence of this function.
Let $f\left( x \right)={{e}^{x}}.$
Let us first find the power series representation of the given function.
We know that a power series about $x=a$ is defined as a series of the form $\sum\limits_{n=0}^{\infty }{{{c}_{n}}{{\left( x-a \right)}^{n}}={{c}_{0}}+{{c}_{1}}\left( x-a \right)+{{c}_{2}}{{\left( x-a \right)}^{2}}+...+{{c}_{n}}{{\left( x-a \right)}^{n}}+...}$ in which the center $a$ and the coefficients ${{c}_{0}},{{c}_{1}},...{{c}_{n}},...$ are constants.
In this case, $a=0.$
Let us find the series coefficients by ${{\left( \dfrac{{{d}^{k}}}{d{{x}^{k}}}\left( f\left( x \right) \right) \right)}_{x=0}}$ for $k=0,1,2,...$
Let us apply $k=0$ in the above derivative, we will get $f\left( 0 \right)={{e}^{0}}=1.$
Now, we will apply $k=1$ in the derivative to get ${{\left( \dfrac{df\left( x \right)}{dx} \right)}_{x=0}}={{\left( \dfrac{d{{e}^{x}}}{dx} \right)}_{x=0}}={{\left( {{e}^{x}} \right)}_{x=0}}={{e}^{0}}=1.$
We will get ${{\left( \dfrac{{{d}^{k}}}{d{{x}^{k}}}\left( f\left( x \right) \right) \right)}_{x=0}}={{\left( \dfrac{{{d}^{k}}{{e}^{x}}}{d{{x}^{k}}} \right)}_{x=0}}={{\left( {{e}^{x}} \right)}_{x=0}}={{e}^{0}}=1.$
So, the coefficients are ${{c}_{k}}=\dfrac{{{f}^{k}}\left( 0 \right)}{k!}=\dfrac{1}{k!}{{\left( \dfrac{{{d}^{k}}}{d{{x}^{k}}}\left( f\left( x \right) \right) \right)}_{x=0}}.$
Therefore, we will get the power series representation as
\[\Rightarrow \sum\limits_{k=0}^{\infty }{\dfrac{{{f}^{k}}\left( 0 \right)}{k!}{{\left( x-0 \right)}^{k}}=}\dfrac{{{x}^{0}}}{0!}+\dfrac{{{x}^{1}}}{1!}+\dfrac{{{x}^{2}}}{2!}+...=\sum\limits_{k=0}^{\infty }{\dfrac{{{x}^{k}}}{k!}.}\]
Let us find the radius of convergence.
We will use the ratio test to find the radius of convergence.
So, we will get $\displaystyle \lim_{k \to \infty }\left| \dfrac{\dfrac{{{x}^{k+1}}}{\left( k+1 \right)!}}{\dfrac{{{x}^{k}}}{k!}} \right|=\displaystyle \lim_{k \to \infty }\left| \dfrac{{{x}^{k+1}}}{\left( k+1 \right)!}\dfrac{k!}{{{x}^{k}}} \right|=\displaystyle \lim_{k \to \infty }\left| \dfrac{x}{\left( k+1 \right)} \right|=0.$
By the ratio test, the series converges for all value of $x.$ Therefore, the interval of convergence is $\left( -\infty ,\infty \right).$ Therefore, the radius of convergence is $\infty .$
Hence the power series representation of the given function is \[\sum\limits_{k=0}^{\infty }{\dfrac{{{x}^{k}}}{k!}}\] and the radius of convergence is $\infty .$

Note:
If the series is convergent for all values of $x,$ then the radius of convergence is $\infty .$ If the series is convergent only for $x=a,$ then the radius of convergence is $0.$ By the ratio test, if the limit is less than $1,$ then the series is convergent. If the limit is greater than $1$ or $\infty ,$ then the series is divergent.