Find all $f: \mathbb R\longrightarrow \mathbb R $ such that $f $ is a continuous function at $0$ and satisfies $$\;\forall \:x \in \mathbb R,\; f\left(2x\right) = f\left(x\right)\cos x $$
My try: I just found the $f (x)$ is periodic, i.e. $f (2\pi / 2)= f (\pi/2) \cos (\pi /2) $
And$ f (\pi)=f (3\pi)$ ... and so on, Best I came up with is $$f (2^n x) = f (x) \cos (x) \cos (2x) ... \cos (2^{n-1} x)$$
On
To elaborate on the comment by @JimmyK4542, on possible answer is $\;f\left(x\right) := \dfrac{\sin x}x\,$ for all $\,x\neq 0,\,$ so that $$ f\left(2x\right) = \dfrac{\sin 2x}{2x} = \dfrac{2\sin x\cos x}{2x} = \dfrac{\sin x}{x}\,\cos x =f\left(x\right)\cos x $$
Alternatively, you can find solution in terms of Taylor series, assuming $\,f\,$ is sufficiently smooth.
Let $\;f\left(x\right) = \sum a_n\,x^n,\,$ and $\;f\left(2x\right) = \sum a_n\,\left(2x\right)^n = \sum 2^na_n\,x^n.\,$ Recall $\displaystyle\,\cos x = \sum_{n=0}^{\infty} \dfrac{\left(-1\right)^{n}}{\left(2n\right)!}\,x^{2n},\,$ then
\begin{align} f\left(2x\right) &= f\left(x\right)\cos x &\iff&& \sum 2^na_n\,x^n & = \left(\sum_{} a_n\,x^n\right) \left(\sum_{} \dfrac{\left(-1\right)^{n}}{\left(2n\right)!}\,x^{2n}\right) \end{align} or, more explicitly,
\begin{align} a_0 + 2a_1\,x + 2^2a_2\,x^2 + 2^3a_3\,x^3 + 2^4a_4\,x^4 + 2^5a_5\,x^5 + \ldots = \Big(a_0 + a_1\,x + a_2\,x^2 + a_3\,x^3 + a_4\,x^4 + a_5\,x^5 + \ldots\Big) \cdot \Big( 1 - \dfrac{1}{2!}x^2 + \dfrac{1}{4!}x^4 - \dfrac{1}{6!}x^6 + \dfrac{1}{8!}x^8 - \ldots\Big) \end{align}
Collecting coefficients in front of powers of $\,x\,$ we get
\begin{align} a_0 &= a_0 \cdot 1 &\implies&& a_0&=\alpha \;\text{ – free parameter}\\ 2a_1 &= a_1 \cdot 1 &\implies&& a_1 &= 0\\ 2^2a_2 &= a_2 \cdot 1 +a_0\cdot \dfrac{-1}{2!} &\implies && a_2 &= \dfrac{1}{2^2-1}\left(\dfrac{-1}{2!}\right)a_0\\ &&&&&=\dfrac{1}{3!}\,a_0\\ 2^3a_3 &= a_3 \cdot 1 +a_1\cdot \dfrac{-1}{2!} &\implies && a_3 &= \dfrac{1}{2^3-1}\cdot\dfrac{-1}{2!}\,a_1 =0\\ 2^4a_4 &= a_4 \cdot 1 +a_2\cdot \dfrac{-1}{2!} +a_0\cdot \dfrac{1}{4!} &\implies&&a_4&=\frac1{2^4-1}\left(\frac{-1}{2!}\,a_2+\frac1{4!}\,a_0\right)\\ % &&&&&=\frac{1}{2^4-1}\left(\dfrac{-1}{2!}\cdot\dfrac{1}{3!} + \dfrac{1}{4!}\right)a_0\\ % &&&&&=\dfrac{2^2}{\left(2^2-1\right)\left(2^4-1\right)}a_0\\ &&&&&=\dfrac{1}{5!}\,a_0\\ 2^5a_5 &= a_5 \cdot 1 +a_2\cdot \dfrac{-1}{2!} +a_0\cdot \dfrac{1}{4!} &\implies &&a_5 & = 0\\ 2^6a_6 &= a_6\cdot1 +a_4\cdot\dfrac{-1}{2!} +a_2\cdot\dfrac1{4!} +a_0\cdot\dfrac{-1}{6!} &\implies &&a_6 & = \dfrac{1}{2^6-1} \left( a_4\cdot\dfrac{-1}{2!} +a_2\cdot\dfrac1{4!} +a_0\cdot\dfrac{-1}{2!} \right)\\ &&&&&=\dfrac{-1}{7!}\,a_0\\ 2^7a_7 &= a_7 \cdot 1 +a_2\cdot \dfrac{-1}{2!} +a_0\cdot \dfrac{1}{4!} &\implies &&a_7 & = 0\\ &&\cdots \end{align} Generalizing formulas above we get \begin{align} a_n = \begin{cases} 0,& n = 2k+1,& k\in\mathbb R\\ \dfrac{\left(-1\right)^\frac{n}{2}}{\left(n+1\right)!},& n = 2k,& k\in\mathbb R \end{cases} \end{align} so that $$ f\left(x\right)=\sum_{k=0}^{\infty} \dfrac{\left(-1\right)^k}{\left(2k+1\right)!} = \dfrac{\sin x}{x} $$
We have $$f(x) = f\left(\dfrac{x}2\right)\cos\left(\dfrac{x}2\right) = f\left(\dfrac{x}4\right)\cos\left(\dfrac{x}4\right)\cos\left(\dfrac{x}2\right)$$ Hence, we have $$f(x) = f\left(\dfrac{x}{2^{n}}\right) \prod_{k=1}^n \cos\left(\dfrac{x}{2^k}\right) = \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right)\dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n} \right)}$$ Hence, we have that $$f(x) = \lim_{n \to \infty} \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right) \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right) \lim_{n \to \infty} \dfrac1{2^n} \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \dfrac{\sin(x)}{x} \lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right)$$ If we assume that $\lim_{x \to 0} f(x)$ exists at the origin, we then have that $$\lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right) = c$$ from which we obtain that $$f(x) = c\cdot \dfrac{\sin(x)}x$$