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We know that $f(x) = x - \frac{1}{x}$ , for $x \neq 0$ , is a particular solution of the equation, but what would be the method to arrive at this solution?

JaberMac
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    You can't arrive at the solution (unless you impose an extra condition on $f$ such as continuity). Observe that gluing $f$ defined on the exponential classes ${\pm c^k\mid k\in\mathbb{Z}},c>0$ always gives a solution to your equation. – Moonshine Jul 07 '23 at 16:34
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    How about: prove, if the domain includes $0$, then there is no real-valued function satisfying this? – GEdgar Jul 07 '23 at 16:58
  • No, exponential classes {c^k | k ∈ Z} are not solutions to this equation. This is easily verified when we make the substitution in the equation. – JaberMac Jul 07 '23 at 17:02
  • Assuming a Laurent series expansion? – Bob Dobbs Jul 07 '23 at 17:06
  • @Moonshine means: For each $c>0$, the equation $f(x^2) + (f(x))^2 = 2x^2 - 2$ is a recurrence for a function with domain ${\pm c^k\mid k\in\mathbb{Z}}$. Combine these solutions to get a solution on $\mathbb R\setminus{0}$. But instead it should be sets $${\pm c^{2^j}\mid j\in\mathbb{Z}}$$ – GEdgar Jul 07 '23 at 17:12
  • @GEdgar - it seems that showing nonexistence for domain containing $0$ would be too easy. Plugging $x=0$, we get $f(0)+f(0)^2=-2$ and this obviously impossible. – Moonshine Jul 07 '23 at 17:24
  • The reverse engineering method is pretty cool and I've used it to solve similar problems. But I think this method is very intuitive, since, when we already know the solution, we try to inducedly convert one part of the equation to be similar to the other part and then discover a mirror equality. Would only this method be the most efficient for this type of equation? It sure is very practical. – JaberMac Jul 07 '23 at 21:16

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Note that you can find values immediately for $x = 0,1$ $$ [f(0)]^2 + f(0) + 2 = 0 \implies f(0) = \frac{-1 \pm i\sqrt{3}}{2}. $$ Therefore, if you are only interested in $f: \mathbb R \to \mathbb R$ you need to restrict yourself to $x \neq 0$. For $f(1)$ we see $$ [f(1)]^2 + f(1) = 0 \implies f(1) = 0,-1. $$ Look at the action from $x \to -x$, which yields the relation $f(x) = \pm f(-x) $ (hence the function is either even or odd.

Assuming the function $f$ is differentiable, we obtain $$ xf'(x^2) + f(x)f'(x) = 2x. $$ Evaluating this at $x = 1$, $$ [1 + f(1)]f'(1) = 2 $$ and we see that if $f(1) = -1$, then the equation cannot be satisfied and therefore if our function did satisfy $f(1) = -1$, then it can't be differentiable at $x = 1$.

Lets try a series solution (noting that it must be even or odd, lets try odd first) $f(x) = \sum_n a_{2n+1} x^{2n+1}$, note that this is differentiable everywhere and so $f(1) = 0 = \sum_n a_{2n+1} = 0$ and $f'(1) = 2 = \sum_n (2n+1) a_{2n+1}$. Plugging this into the functional equation \begin{align} \sum_n a_{2n+1} x^{2(2n+1)} + \sum_n \sum_m a_{2n+1} a_{2m+1} x^{2(n+m+1)} = 2x^2 - 2 \end{align} Equating powers shows that $$ a_{2n+1} + \sum_k a_{2k+1} a_{2(2n-k)+1} = 2 \delta_{n0} \tag{1}$$ and $$ \sum_k a_{2k+1} a_{2(2n-k)-1} = -2 \delta_{n0} \tag{2}$$ for all $n$. At this point, I am not sure if there is a systematic way to proceed here, but you can show certain properties if the solution is of this form. First, note that you require at least two of the $a$'s to be nonzero. Searching for a solution of this type implies that it must be the pair $a_{2p+1},a_{-(2p+1)}$ to satisfy the second equality with $n = 0$, $a_{2p+1} a_{-(2p+1)} = -1$. With this, the first equality at $n = 0$, leads to $a_1^2 + a_1 = 2$ (the other terms drop out because the products are all 0 if $k \neq 0$). Solutions for $a_1 = 1,-2$. This also shows that $p = 0$ and so $a_{-1} = -1,1/2$.

Checking the case in equality (1) where $n \neq 0$: $$ 0 = a_{2n+1} + \sum_k a_{2k+1} a_{2(2n-k)+1} = a_{2n+1} + a_{2n+1}^2 \implies a_{2n+1} = 0, -1. $$ So all is fine (equality (2) is trivially satisfied when $n \neq 0$). Finally, the condition $f(1) = 0 = a_1 + a_{-1}$ implies that we choose the solution $a_{\pm 1} = \pm 1$ hence $f(x) = x - \frac{1}{x}$ is the only power series solution with two terms.

I am not sure if much more is possible.

Gregory
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