How to expand x² ? Is that possible?

Question

I have uploaded the steps in link provided. I don't understand what just happened to x² in solution attached. Is there a Taylor series expansion for x²? I don't think so. Can you please explain the expansion of x² in attached solution. Click here to see solution of above question

Answer 1

As it says in the solution, $x^2$ is already a Taylor series: every polynomial is. If you like, though, you can expand it in the usual way about $0$: if $f(x)=x^2$, then

$$f(x)=\sum_{n\ge 0}\frac{f^{(n)}(0)}{n!}x^n=\frac2{2!}x^2=x^2\,,$$

since all of the derivatives except the second are $0$ or evaluate to $0$ at $x=0$. If you insist on writing out every term of the series, you have

$$x^2=\sum_{n\ge 0}b_nx^n\,,$$

where

$$b_n=\begin{cases} 1,&\text{if }n=2\\ 0,&\text{otherwise.} \end{cases}$$

The solution then simply adds this Maclaurin series to the one for $e^x$:

$$\begin{align*} e^x+x^2&=\sum_{n\ge 0}\frac{x^n}{n!}+\sum_{n\ge 0}b_nx^n\\ &=\sum_{n\ge 0}\left(\frac1{n!}+b_n\right)x^n\\ &=1+x+\left(\frac1{2!}+1\right)x^2+\sum_{n\ge 3}\frac1{n!}x^n\\ &=\frac1{0!}+\frac1{1!}x+\frac3{2!}x^2+\sum_{n\ge 3}\frac1{n!}x^n\,, \end{align*}$$

so it is the exponential generating function for the sequence $\langle a_n:n\ge 0\rangle$, where $a_2=3$, and $a_n=1$ if $n\ne 2$.