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Question: How do I find a power series representation for \({{e}^{x}}\) and what is the radius of convergence?...

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

Explanation

Solution

A power series about x=ax=a is a series of the form ∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+...+cn(x−a)n+...\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 aa and the coefficients c0,c1,...cn,...{{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 ex.{{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(x)=ex.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=ax=a is defined as a series of the form ∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+...+cn(x−a)n+...\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 aa and the coefficients c0,c1,...cn,...{{c}_{0}},{{c}_{1}},...{{c}_{n}},... are constants.
In this case, a=0.a=0.
Let us find the series coefficients by (dkdxk(f(x)))x=0{{\left( \dfrac{{{d}^{k}}}{d{{x}^{k}}}\left( f\left( x \right) \right) \right)}_{x=0}} for k=0,1,2,...k=0,1,2,...
Let us apply k=0k=0 in the above derivative, we will get f(0)=e0=1.f\left( 0 \right)={{e}^{0}}=1.
Now, we will apply k=1k=1 in the derivative to get (df(x)dx)x=0=(dexdx)x=0=(ex)x=0=e0=1.{{\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 (dkdxk(f(x)))x=0=(dkexdxk)x=0=(ex)x=0=e0=1.{{\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 ck=fk(0)k!=1k!(dkdxk(f(x)))x=0.{{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
⇒∑k=0∞fk(0)k!(x−0)k=x00!+x11!+x22!+...=∑k=0∞xkk!.\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 lim⁡k→∞∣xk+1(k+1)!xkk!∣=lim⁡k→∞∣xk+1(k+1)!k!xk∣=lim⁡k→∞∣x(k+1)∣=0.\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.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 ∑k=0∞xkk!\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,x, then the radius of convergence is ∞.\infty . If the series is convergent only for x=a,x=a, then the radius of convergence is 0.0. By the ratio test, if the limit is less than 1,1, then the series is convergent. If the limit is greater than 11 or ∞,\infty , then the series is divergent.