I've noticed the derivative of $\sin^2 x$, $$\frac{d}{dx}\sin^2(x)=\sin 2x$$ and the same applies to $\sinh^2 x$, that is $$\frac{d}{dx}\sinh^2(x)=\sinh(2x)$$
I was wondering about the other functions of that property. (I posted a similar question about a week ago but I had an incorrect premise.)
So these functions must satisfy $$(f^2(x))'=f(2x)\rightarrow 2f'(x)f(x)=f(2x)$$ The only way I could think of solving these was the power series method which is something I've never been good with. $$f(x)=\sum_{n=0}^{\infty} C_nx^n,$$ $$f'(x)=\sum_{n=0}^{\infty}(n+1)C_{n+1}x^n$$ plugging this
$$2\sum_{n=0}^{\infty}(n+1)C_nC_{n+1}x^{2n}=\sum_{n=0}^{\infty}C_n(2x)^n.$$ and now I don't know what to do with that $x^{2n}$ term. Could someone help me find the general form of $f(x)$?
There seem to be 3 families of solutions: $$ \left( \begin{array}{cccc} Coeff & Solution\,1 & Solution\,2 & Solution\,3\\ a_0 & 0 & 0 & \frac{1}{8 a_2} \\ a_1 & 1 & 0 & \frac{1}{2} \\ a_2 & 0 & 0 & a_2 \\ a_3 & a_3 & 0 & \frac{4 a_2^2}{3} \\ a_4 & 0 & 0 & \frac{4 a_2^3}{3} \\ a_5 & \frac{3 a_3^2}{10} & 0 & \frac{16 a_2^4}{15} \\ a_6 & 0 & 0 & \frac{32 a_2^5}{45} \\ a_7 & \frac{3 a_3^3}{70} & 0 & \frac{128 a_2^6}{315} \\ a_8 & 0 & 0 & \frac{64 a_2^7}{315} \\ a_9 & \frac{a_3^4}{280} & 0 & \frac{256 a_2^8}{2835} \\ a_{10} & 0 & 0 & \frac{512 a_2^9}{14175} \\ a_{11} & \frac{3 a_3^5}{15400} & 0 & \frac{2048 a_2^{10}}{155925} \\ a_{12} & 0 & 0 & \frac{2048 a_2^{11}}{467775} \\ a_{13} & \frac{3 a_3^6}{400400} & 0 & \frac{8192 a_2^{12}}{6081075} \\ a_{14} & 0 & 0 & \frac{16384 a_2^{13}}{42567525} \\ \end{array} \right) $$ Solution 2 is clearly the constant $0$ function, Solution 1 seems to be: $$ f(x)=\sum_{n=0}^\infty\frac{2^{\frac{n}{2}-2} 3^{\frac{n}{2}-1} \left((-1)^n+1\right) a_3^{\frac{n}{2}-1}}{\Gamma (n)}x^n $$ Solution 3 seems to be of the type: $$ f(x)=\sum_{n=0}^\infty\frac{2^{2 n-5} a_2^{n-2}}{\Gamma (n)}x^n $$ This is far from a proof, but I don't think there are other families.