$f: \mathbb{R} \to \mathbb{R},\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space$find $f(3)$ in terms of $f(0)$. 
$f: \mathbb{R} \to \mathbb{R},\space\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space\space\space$
  Find $f(3)$ in terms of $f(0)$.

My approach:
$$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$$
$$\Rightarrow f(\frac{x}{2})-2f(\frac{x}{4})+f(\frac{x}{8})=\frac{x^2}{4}$$
$$\Rightarrow f(\frac{x}{4})-2f(\frac{x}{8})+f(\frac{x}{16})=\frac{x^2}{16}$$
$$\vdots$$
$$\Rightarrow f(\frac{x}{2^{n-1}})-2f(\frac{x}{2^n})+f(\frac{x}{2^{n+1}})=\frac{x^2}{4^{n-1}}$$
Summing up these when n $\rightarrow$ $\infty$
$$f(x)-f(\frac{x}{2})=x^2\cdot\frac{4}{3}$$
I am unable to proceed further.
 A: I'll go ahead and assume that $f$ is continuous at $0$. 
For any $x \in \mathbb{R}$, let $P(x)$ denote the property that $f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2$. Then, for any $x$, we have the following: 
\begin{align}
P(x)&: \quad f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2
\\
P(\tfrac{x}{2})&: \quad f(\tfrac{x}{2}) - 2f(\tfrac{x}{4}) + f(\tfrac{x}{8}) = \tfrac{1}{4}x^2
\\
& \vdots
\\
P(\tfrac{x}{2^{n-1}})&: \quad f(\tfrac{x}{2^{n-1}}) - 2f(\tfrac{x}{2^n}) + f(\tfrac{x}{2^{n+1}}) = \tfrac{1}{4^{n-1}}x^2
\end{align}
Adding these up gives us $$f(x)-f(\tfrac{x}{2})-f(\tfrac{x}{2^n})+f(\tfrac{x}{2^{n+1}}) = \tfrac{4}{3}(1-\tfrac{1}{4^n})x^2$$
If we take the limit as $n \to \infty$ and use the assumption that $f$ is continuous at $0$, we get $$f(x)-f(\tfrac{x}{2})-f(0)+f(0) = \tfrac{4}{3}(1-0)x^2$$ i.e. $$f(x)-f(\tfrac{x}{2}) = \tfrac{4}{3}x^2.$$
Now, we can repeat the same trick. For any $x \in \mathbb{R}$, let $Q(x)$ denote the property that $f(x)-f(\tfrac{x}{2}) = \tfrac{4}{3}x^2$. Then, for any $x$, we have the following: 
\begin{align}
Q(x)&: \quad f(x) - f(\tfrac{x}{2}) = \tfrac{4}{3}x^2
\\
Q(\tfrac{x}{2})&: \quad f(\tfrac{x}{2}) - f(\tfrac{x}{4}) = \tfrac{1}{3}x^2
\\
& \vdots
\\
Q(\tfrac{x}{2^{n-1}})&: \quad f(\tfrac{x}{2^{n-1}}) - f(\tfrac{x}{2^n}) = \tfrac{1}{3 \cdot 4^{n-2}}x^2
\end{align}
Adding these up gives us $$f(x)-f(\tfrac{x}{2^n}) = \tfrac{16}{9}(1-\tfrac{1}{4^n})x^2$$
If we take the limit as $n \to \infty$ and use the assumption that $f$ is continuous at $0$, we get $$f(x)-f(0) = \tfrac{16}{9}x^2.$$ Hence, $f(x) = \tfrac{16}{9}x^2+f(0)$. It is easy to check that any function in the form $f(x) = \tfrac{16}{9}x^2+C$ for some constant $C$ satisfies the given functional equation. 
A: Making $y=2^x$ we have
$$
F(y)-2F(y-1)+F(y-2)=4^y
$$
which is a linear difference functional equation with solution
$$
F(y) = y \Phi_1(y)+\Phi_2(y)+\frac{4^{y+2}}{9}
$$
Here $\Phi_1(y),\Phi_2(y)$ are generic periodic functions with period $1$ Assuming $\Phi_1(y) = C_1,\Phi_2(y)= C_2$ we have
$$
f(x) = C_1\log_2 x + C_2 + \frac{16}{9}x^2
$$
then for the feasibility of $f(0)$ we have $C_1 = 0$ and
$$
f(0) = C_2
$$
and
$$
f(3) = f(0)+16
$$
