How to identify this power series as $k\sin(k/x)$? In this question, a functional equation is solved for functions with a power series. We find a recursive formula: (copied from the answer by user achille hui)
\begin{align}
( 2^1 - 3 ) a_2 &= 0\\
( 2^2 - 4 ) a_3 &= 2 a_2 a_2 \iff 0 = 0 \text{ ( i.e. } a_3\;\text{ is arbitrary )}\\
( 2^3 - 5 ) a_4 &= 3 a_3 a_2 + 2 a_2 a_3\\
( 2^4 - 6 ) a_5 &= 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4\\
&\;\vdots
\end{align}
According to the answer, it turns out that $\sum a_i x^i$ is equal to $k \sin(x/k)$ or $k \sinh(x/k)$, depending on the sign of $a_3$. 
How does one see this? A messy inductive proof will probably prove equality, but that is not what I am looking for. What methods/tricks are there to find a closed form for the coefficents $a_i$ from the recursion, and then recognize it as the power series of $\sin$ or $\sinh$?
 A: 
What methods/tricks are there to find a closed form for the coefficents $a_i$ from the recursion, and then recognize it as the power series of sin or sinh?

The logic is the other way around.  If handed the recursion, standard generating function techniques will recognize it as an encoding of the differential equation in the other question, $f(2x) = 2f'(x)f(x)$ and further analysis would be based on the differential equation.  The problem is to find particular solutions.  You could do this by calculating $a_i$'s for the simplest cases, using the recursion, and looking for a pattern (power series with $a_n = \pm pq^n\frac{1}{(2n+1)!}$ will stand out quickly, or use OEIS), but things are easier than that.
It is not hard to "see" $\sin x$ as a solution, and the problem may have been inspired by the formula for $\sin (2x)$.   If you can somehow generate this solution, things get better. The inhomogeneity of the equation can be used: if $f(x)$ is a solution, so is $af(bx)$ for suitable $a$ and $b$, since both parameters will alter the ratio of the two sides in some way, and $b$ can be chosen to reverse the effect of $a$.  The condition happens to be $ab=1$, so we now have $k \sin (x/k)$ as a family of solutions. Four of the five solution types at the other question can be obtained from that by taking $k$ to be a complex number (it must be pure imaginary to have a real valued solution on $\mathbb{R}$) or letting it approach $0$ or $\infty$.   The $k$-deformation $kf(\frac{x}{k})$ leaves $f'(0)$ constant and by looking at solutions approximating a power function $px^q$ near $0$, one sees that there are very few possibilities.
The freezing of $f'(0)$ and the existence of an additional family of solutions, the $A \exp(Bx)$, with $f'(0)=1/2$, are somewhat unusual, and I suspect that if you relax the condition of analyticity near $0$ there could be huge families of solutions that imitate one of the analytic ones at points in a geometric progression $(2^k x_0), k \in \mathbb{Z}$ and interpolate almost arbitrarily in between, since the equation is of the delay-differential type.  
I'll offer the following as one approach to explaining why the solutions look as they do.   The analytic solutions satisfy a better functional equation 

$f(x+y)=f'(x)f(y)+f(x)f'(y) $

that specializes to the other one when $x=y$. I don't fully understand why to  expect the two functional equations to be equivalent for power series, nor could I have predicted the two variable relation without knowing it for the sine function (and therefore the other four sine-derived solutions) and checking that it works for the exponential solution.  But once it is there, solutions are seen to correspond to "half" group representations; the equation holds that for real $f$ and $x$,  the imaginary part of $g(x)= f'(x) + if(x)$ satisfies $\Im g(x+y)= \Im (g(x)g(y))$, or 

$(g(x+y) - g(x)g(y)) \in \mathbb{R}$ for real $x$ and $y$ 

I did not yet complete the analysis of this equation, but it is immediate from this way of writing things that if $g(x)$ is a solution, so is $g(kx)$, and that there are solutions $g(x)=Ae^x$ with $A-A^2 \in \mathbb{R}$.  If this is all solutions, it means $A$ real or on the famous line $1/2 + it$, with degenerate solutions for $A = 0,1$ and a special solution at $A=1/2$ where the two lines intersect.  These are five possiblities, and one thing to check is if they correspond to the five solution types of the original problem.
