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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.

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    I'm pretty sure this problem is under-determined. While we have a general form for the Taylor series (i.e., $\sum_{k=0}^{\infty}\frac{y^{(k)}(0)(x-c)^k}{k!}$), there are an infinite number of functions the derivatives of which are equal to the sequence of derivatives--$\left\{y^{(k)}(0)\right\}_{k\in\mathbb{N}_{0}}$--necessary to complete the Taylor series. – 2017-01-29

2 Answers 2

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You can try derivatives or integrals. For example, integrate (within the convergence interval, always) and hope you reach a recognizable function:

$$f(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}n\implies f'(x)=\sum_{n=1}^\infty(-1)^{n-1}x^{n-1}\stackrel{\text{geometric series!}}=\frac1{1+x}\implies$$

$$f(x)=\int\frac{dx}{1+x}=\log(1+x)\;,...etc.$$

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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

$$\frac\pi4=1-\frac13+\frac15-\frac17+\dots$$

$$y(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$

$$y''(x)=-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\right)$$

$$y''+y=0,\quad y(0)=0,y'(0)=1$$

$$\implies y(x)=\sin(x)$$