Polynomials are themselves Taylor expansions, correct?
ex. $4x+5x^2+3 = 3+4x+5x^2 +0x^3 +0x^4 + \dots$ I'm assuming has no closed form besides $\sum_{n=0}^{2}(3+n)x^n + \sum_{n=3}^{\infty}0x^n$ but we were lucky with the consecutive coefficients and powers.
But for example, we can get the taylor expansion of $e^x$ by differentiating and equating sides at a convenient value:
$e^x = a_0 +a_1x +a_2x^2+a_3x^3 + \dots\\ \implies e^0 =1= a_0.$
$\frac{d}{dx}e^x=e^x=a_1+2a_2x+3a_3x^2+\dots\\ \implies e^0=1=a_1$
$\frac{d^2}{dx^2}e^x=e^x=2a_2+(2*3)a_3x+\dots\\ \implies e^0=1=2a_2\implies a_2=\frac{1}{2}$
$\frac{d^3}{dx^3}e^x=e^x=(2*3)a_3+\dots\\ \implies e^0=1=(2*3)a_3\implies a_3=\frac{1}{2*3}=\frac{1}{3!}$
Gives $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\dots = \sum_{n=0}^{\infty}\frac{x^n}{n!}$.
Can the method find any taylor expansion? Do they all have a closed form? If not, what is the alternative?