Is there any standard method for finding the function defined by a Taylor/Laurent series? Say you have a Taylor series defined by 
$$\sum_{n=0}^{\infty}a_nx^n$$
Is there any standard way to figure out what function is defined by the series? One option I see is just looking at the coefficients and finding some function that satisfies the values for $f(c)$, $f'(c)$, etc, where $c$ is the point the Taylor series is centered on, but it seems as though it would be hard to define a function with just some values in mind. You could say the same for a Laurent series, although I can imagine that would be much harder given that you would have to figure out all the nasty coefficients.
 A: Another way is through differential equations. Assuming your power series $f$ has radius of convergence $R>0$, you can differentiate it (at libidum) on $(-R,R)$, and doing so and rearranging the coefficients you may end up with a differential equation on $f$. If you can solve it to get an analytic form for the solutions, you will get an analytic form for $f$.
As a simple example, look at the power series defined by $f(x)=\sum_{n=0}^\infty x^n$, with radius of convergence $R=1$. Differentiating on $(-1,1)$, you get
$$\begin{align}
f^\prime(x) &= \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=1}^{\infty} n x^{n-1}
= \sum_{n=0}^{\infty} (n+1) x^{n} \\
&= \sum_{n=1}^{\infty} n x^{n} + \sum_{n=0}^{\infty} x^{n}
= x \sum_{n=1}^{\infty} n x^{n-1} + \sum_{n=0}^{\infty} x^{n} \\
&= x f^\prime(x) + f(x)
\end{align}$$
i.e. 
$$
f^\prime(x) = \frac{f(x)}{1-x}, \qquad x\in (-1,1)
$$
Solving this differential equation, you get $f(x) = \frac{f(0)}{1-x}$, and since $f(0) = 1$, you obtain $f$.
(of course, this particular example is not interesting, as the original series is easy to compute by other means; but it illustrates the technique)
