I had a test the day before yesterday and in that I got this question
$Let \; \psi : R \; \rightarrow \; R$ be a continuous function satisfying the relation $$\psi(x)-2\psi(\frac{x}{2})+ \psi (\frac{x}{4}) = x^2$$ and $\psi(0)=1$. Then graph of $y= \psi(x)$ is
The answer key says it's a parabola but I literally have no clue on how to get that.
I know it's not fair to ask questions here without showing my efforts but what should I show. I couldn't make any progress other than just rewriting the same equation multiple times and replacing $x$ with $\frac{x}{2}, \frac{x}{4}$...etc (but that didn't help me).
So can someone explain how to get the curve ?
Let $\phi(x)=\psi(x)-\psi\left(\frac{x}{2}\right)$. The initial equation becomes
$$\phi(x)-\phi\left(\frac{x}{2}\right)=x^2.$$ $\phi $ is continuous as $\psi$ is. And by induction for $n \ge 1$ $$\phi(x)-\phi\left(\frac{x}{2^n}\right)=x^2\frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}.$$ If you let $n \to \infty$, you get as $\phi$ is continuous and $\phi(0)=0$
$$\phi(x)=\frac{4}{3}x^2.$$ Applying similar techniques to $\psi$ which satisfies the equation $$\psi(x)-\psi\left(\frac{x}{2}\right)=\frac{4}{3}x^2.$$ we get
$$\psi(x)=\frac{16}{9}x^2+1.$$