Suppose $f(x) = \sum_{n=0}^\infty c_nx^n$ is a function on $(-R, R)$ for positive radius of convergence $R>0$. Then $f(0) = c_0$. Additionally:
$$ f'(x) = \sum_{n=0}^\infty (n+1)c_{n+1}x^n, \text{ so } f'(0) = 1\cdot c_1 $$
$$ f''(x) = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2}x^n, \text{ so } f''(0) = 2\cdot1\cdot c_2 $$
$$ f'''(x) = \sum_{n=0}^\infty (n+3)(n+2)(n+1)c_{n+3}x^n, \text{ so } f'''(0) = 3\cdot2\cdot1\cdot c_2 $$
A pattern emerges, and we find that:
$$ f^{(k)}(0) = k! c_k \text{ for all } k $$
$f^{(k)}(0)$ is the $k$th derivative of $f$ evaluated at $x=0$.
The Taylor series of an infinitely-differentiable function $f(x)$ at $x = a$ is:
$$ \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n $$
<aside>
What is the Taylor series of $e^x$ centered at $x=0$?
</aside>
All derivatives of $e^x$ are simply $e^x$ itself, so $f^{(n)}(x) = e^x$. This means that $f^{(n)}(a) = f^{(n)}(0) = e^0 = 1$. Therefore, the Taylor series is:
$$ \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n = \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^\infty\frac{1}{n!}x^n = \sum_{n=0}^\infty\frac{x^n}{n!} $$
<aside>
What is the Taylor series of $f(x) = \sin(x)$ centered at $x=0$?
</aside>
The derivatives of $f(x) = \sin(x)$ follow a pattern: