Chapter 9, Problem 9RWDT

To determine

To find: If the function $g$ has Maclaurin series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n$ must it be the case that $g\left(x\right)=e^n$ .

Explanation of Solution

Given information:

The function $g$ has Maclaurin series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n$ .

Expand the given series gives:

  ${\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n=1+a_{1}x+a_2{x}^2+a_3{x}^3+\mathrm{....}$

In order to prove if function g has McLaurin series then first we check whether the series converges.

Suppose ${\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n$ is a power series. Also suppose that $n\to \infty \left|\frac{a_{n+1}}{a_n}\right|=0$ . Then radius of convergence of the power series is $\infty$ and the power series converges for all real number $x$ .

The exponential series is of the form $1+{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n!}}$ is power series. Here, $a_0=1$ and $a_n=n!$ for all $n\in N$

  $\begin{array}{l}n\to \infty \left|\frac{a_{n+1}}{a_n}\right|=n\to \infty \frac{n!}{(n+1)!}\\ n\to \infty \frac{n!}{(n+1)!}=n\to \infty \frac{1}{n+1}\\ n\to \infty \frac{n!}{(n+1)!}=0\end{array}$

Hence the exponential series converges for all $x\in R$ .

This gives the idea that exponential series converges in the interval $(-\infty ,\infty )$ .

Therefore if function g has McLaurin series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n$ it must be exponential.

Now to find the coefficients of Maclaurin series it must evaluate ${\left(\frac{d^k}{d{x}^k}f\left(x\right)\right)}_{x=0}$ for $k=1,2,3,\mathrm{4.....}$

If the function is exponential that is $g=e^x$ then first draw the graph of $g=e^x$ .

  

It can be seen from the graph that all the derivative of $g\left(x\right)$ at $x=0$ is equal to 1.

Maclaurin series coefficients, $a_k$ are always calculated using the formula $a_k=\frac{g^k\left(0\right)}{k!}$ .

For $k=1$

  $a_1=\frac{g^1\left(0\right)}{1!}$

  $\begin{array}{l}\frac{d}{dx}g(x)=\frac{d}{dx}{e}^x\\ \frac{d}{dx}g(x)=e^x\\ {\left(\frac{d}{dx}g(x)\right)}_{x=0}=e^0\\ {\left(\frac{d}{dx}g(x)\right)}_{x=0}=1\end{array}$

For $k=2$

  $a_2=\frac{g^2\left(0\right)}{1!}$

  $\begin{array}{l}\frac{d^2}{d{x}^2}g(x)=\frac{d}{dx}{e}^x\\ g^2(x)=e^x\\ {\left(g^2(x)\right)}_{x=0}=e^0\\ {\left(g^2(x)\right)}_{x=0}=1\end{array}$

Similarly more coefficient at $x=0$ can be found which all are equal to 1.

  $g$ has all order of derivatives at $x=0$

  $g^{(n)}\left(0\right)=1$

The series will be in the form:

  $e^x={\displaystyle \sum _{k=0}^{\infty }\frac{x^k}{k!}}=1+x+\frac{x^2}{2!}+\mathrm{...}$