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The cosine function satisfies its double-angle formula

$$(f(x))^2 = \frac{f(2x)+1}{2}$$

Now I was wondering if there are other continuous or even smooth (or other "sufficiently well behaved") functions $f: \mathbb K \to \mathbb K$ (with $\mathbb K = \mathbb R$ or $\mathbb C$) satisfying this equation.

My first attempt to solve this functional equation was differentiating it and applying techniques from solving differential equations, but unfortunately the $2x$ in the argument excludes standard methdos. For completeness sake, the equation would be

$$2f(x) f'(x) = f'(2x).$$

(With $f(0) = 1$ or $-0.5$ from the first equation.)

Are there any other methods to tackle this problem?

flawr
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    A trivial example is $f \equiv 1$ – geetha290krm Jul 05 '22 at 09:08
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    Or $f(x) = \cos(a x)$ for any $a \in \Bbb R$. – Martin R Jul 05 '22 at 09:10
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    Or $f(x)=\cosh x$. – Pavel R. Jul 05 '22 at 09:18
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    Great observations, thank you! I'm now curious to see whether there are any functions that are not from that family. – flawr Jul 05 '22 at 09:36
  • The functions $\cos(ax)$ (with arbitrary complex $a$) are solutions with $f(0) = 1$. Are there known solutions with $f(0) = -1/2$? – Martin R Jul 05 '22 at 09:48
  • Related: https://math.stackexchange.com/q/3524515/42969 – Martin R Jul 05 '22 at 09:51
  • The constant function $f \equiv -1/2$ would certainly work just like $f\equiv 1$ but it'd be interesting to see other solutions. – flawr Jul 05 '22 at 09:51
  • @MartinR Thanks, that is an dupe! – flawr Jul 05 '22 at 09:54
  • One can search for series in form $f(x)=a_0+a_1x+a_2x^2+a_3x^3+...$. For $a_0=1$ series has one independent parameter $a_2$. This solution corresponds to general formula $\frac{u^x+u^{-x}}{2}$, where $u$ can be real or complex number. For $a_0=-\frac{1}{2}$ all $a_i$ except $a_0$ must be zero, then there is only one solution with $f(0)=-\frac{1}{2}$ which is $f(x)=-\frac{1}{2}$. – Ivan Kaznacheyeu Jul 05 '22 at 14:48

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