# Solve the functional equation $f (2x)=f (x)\cos x$

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 $$f (2^n x) = f (x) \cos (x) \cos (2x) ... \cos (2^{n-1} x)$$

• By inspection $f(x) = \frac{\sin x}{x}$ for $x \neq 0$ and $f(0) = 1$ is one solution. Dec 21, 2015 at 5:07
• There are some rather serious problems with this post, so you should consider using the edit to improve it. The tags aren't relevant, and you should include your own thoughts and efforts in this problem: What context did you find this in? What techniques are you aware of? What have you tried? You may also find it helpful to include an English translation of the question.
– user296602
Dec 21, 2015 at 5:11
• If English is not your first language, I would suggest using the translation-request tag. Hopefully, one of our users proficient in both languages will be able to provide an adequate translation. Dec 21, 2015 at 5:32
• I did fix the issues, I am just reading this from original book in Arabic, and then I found a similar question in French. I am just practicing some problems in Analysis.
– user65304
Dec 21, 2015 at 5:38

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 $$f(x) = c\cdot \dfrac{\sin(x)}x$$

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 $$f\left(x\right)=\sum_{k=0}^{\infty} \dfrac{\left(-1\right)^k}{\left(2k+1\right)!} = \dfrac{\sin x}{x}$$