Solve: $f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x).f(\frac{1}{y})$

Question

Here i have one functional equation: If $$f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x)\cdot f(\tfrac{1}{y})\text{ for all x},y\in\mathbb{R}-{0}$$ and $f(0) = \frac{1}{2}$ , then find the value of $f(4)$.

Answer 1

$f(x)=\frac{1}{2}\cos x$ satisfies the condition.

Answer 2

As noted in the comments, $f(x)$ is symmetric. Set $a=f(1), b=f(2), c=f(3), d=f(4)$. Then we have the following four equations:

$$d+f(0)=2b^2$$ $$d+b=2ac$$ $$c+a=2ab$$ $$b+f(0)=2a^2$$

This has two solutions according to alpha; in both solutions $d=-\frac{1}{2}$. Finding them by hand doesn't seem too difficult.

Answer 3

following Rahul's comment.

$f(x + z) + f(x-z) = 2f(x)f(z)$

take $x=z\neq 0$,

$f(2x) = 2f(x)^2 - 1/2$

$f(3x) + f(x) = 2f(x)f(2x) = 4f(x)^3 - f(x)$

thus

$f(3x) = 4f(x)^3 - 2f(x)$

and

$f(4x) + f(2x) = 2f(3x)f(x) = 2(4f(x)^3 - 2f(x))f(x) = 8 f(x)^4 - 4f(x)^2$

denote $f(x)$ as $f$ for short.

$f(4x) = 8f^4 - 4f^2 - (2f^2 - 1/2) = 8f^4 - 6f^2 + 1/2$

also

$f(4x) = 2f(2x)^2 - 1/2 = 2(2f^2 - 1/2)^2 - 1/2 = 2(4f^4 + 1/4 - 2f^2) - 1/2 = 8f^4 - 4f^2$

then $2f^2 = 1/2$, $f^2 = 1/4$, $f(x) = 1/2$ or $f(x) = -1/2$. for any $x\neq 0$.

which leads to $f(2x) = 0$ for any $x\neq 0$, contradiction. $f$ does not exist.