Working backwards from a Taylor Expansion Is there a common method for working backwards from an expansion to the expanded function?
Say, for example, I did not know that
$$
\ln{(x+1)} = x - \frac{x^2}{2} + \frac{x^3}{3} ...
$$
And was presented with the right hand side, which, due to context, I suspected was a taylor expansion of some kind. Is there a set of tools or approaches that could help in working backwards from an expansion?
Or would I just have to brute force it, ergo
Either f(x) is 0, or f(x) is x, then either f'(x) is 1, or ......until I recognize either a function or its derivatives? Or is there a better way to do this? Note that the $\ln(1 + x)$ is just an example, while hints for how this could be done in this specific case are appreciated, I am much more intrigued by the general problem of reverse engineering a function from a taylor expansion.
 A: You can try derivatives or integrals. For example, integrate (within the convergence interval, always) and hope you reach a recognizable function:
$$f(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}n\implies f'(x)=\sum_{n=1}^\infty(-1)^{n-1}x^{n-1}\stackrel{\text{geometric series!}}=\frac1{1+x}\implies$$
$$f(x)=\int\frac{dx}{1+x}=\log(1+x)\;,...etc.$$
A: Many basic Taylor expansions come from the geometric series:
$$\begin{align}x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots&=\int_0^x1-t+t^2-t^3+\dots\ dt\\&=\int_0^x\frac1{1+t}\ dt\\&=\ln(1+x)\end{align}$$
Other easily solved series expansions include the arctan integral (which gives a series expansion of $\pi$), noting symmetry by taking a few derivatives to set up a differential equation, etc.

$$\begin{align}x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\dots&=\int_0^x1-t^2+t^4-t^6+\dots\ dt\\&=\int_0^x\frac1{1+t^2}\ dt\\&=\arctan(x)\end{align}$$
set $x=1$ and you get
$$\frac\pi4=1-\frac13+\frac15-\frac17+\dots$$

$$y(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$
$$y''(x)=-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\right)$$
$$y''+y=0,\quad y(0)=0,y'(0)=1$$
$$\implies y(x)=\sin(x)$$
