Solve: $f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x).f(\frac{1}{y})$ 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 $f(4)$.

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.
