<s>[INST] $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$ 
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.

 [/INST] OK, since the hint seems not to be enough, here is the solution.
First,
Setting $g(x)=f(x)-f(\frac{x}{2})$ we have 
$$g(x)-g(\frac{x}{2})=x^2 \\
f(x)-f(\frac{x}{2})=g(x) \tag{*}$$
We solve the first equation.
By induction 
$$g(x)-g(\frac{x}{2^{n}})=x^2+\frac{x^2}{4}+\frac{x^2}{16}+...+\frac{x^2}{4^{n-1}}=x^2 \frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}$$
Since $f$ is continuous at $x=0$, so is $g$. Taking the limit by $n$ you get
$$g(x)-g(0)=\frac{4}{3} x^2$$
Moreover, $f(x)-f(\frac{x}{2})=g(x)$ implies $g(0)=0$.
Now, we need to solve 
$$f(x)-f(\frac{x}{2})=\frac{4}{3}x^2$$
which is exactly the equation above, multiplies by $\frac{4}{3}$. Therefore, solving exactly as above we get
$$f(x)-f(0)=\frac{4}{3}\frac{4}{3}x^2=\frac{16}{9}x^2$$
This shows that 
$$f(x)=\frac{16}{9}x^2 +c$$
where $c=f(0)$ 
are all the solutions.
P.S. To make this more clear, the equation can be reduced via (*) to two equations of the type
$$h(x)-h(\frac{x}{2})=r(x)$$
with $h, r$ continuous. This equation can be solved as above: By induction
$$h(x)-h(\frac{x}{2^n})=\sum_{k=0}^{n-1}r(\frac{x}{2^n})$$
Using the continuity of $h$ at $x=0$ (we don't even need $h$ to be continuous at other points) we get
$$h(x)-h(0)=\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
(Note here that, if the series is divergent it implies that there is no solution which is defined at that $x$).
So, by calculating the series, you get the solution
$$h(x)=c+\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
where $c=h(0)$.
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