Continuity- function Let $f : R \to R $ be a function which is continuous at $0$ and $f(0)=1$ . Also assume that f satisfies the following relation for all $x$ $$f(x)-f\left(\frac{x}{2}\right)=\frac{3x^2}{4}+x$$ Find $f(3)$
[I have tried it solve. I will be grateful indeed if someone can review the following solution and made comments wherever necessary ]
Ans: $f(0)=1=\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-} f(x).........................(1) $
$f(x)-f\left(\frac{x}{2}\right)=\frac{3x^2}{4}+x$..........................$(2)$
$f(x)=f\left(\frac{x}{2}\right)+\frac{3x^2}{4}+x=f\left(\frac{x}{4}\right)+\left(\frac{3}{4}\frac{x^2}{2^2}\right)+3\frac{x^2}{4} +x .......$.(Using (2) and putting $\frac{x}{2}$ in place of $x$)
$=f\left(\frac{x}{4}\right)+\frac{3}{4}\left(x^2+\frac{x^2}{2^2}\right)+ \left(x+\frac{x}{2}\right)=f\left(\frac{x}{8}\right)+\frac{3}{4}\left(x^2+\frac{x^2}{2^2}+\frac{x^3}{2^3}\right)+\left(x+\frac{x}{2}+\frac{x}{4}\right)$
Continuing this way, for an infinitely large positive no. $n$ we have
$f(x)=f\left(\frac{x}{2^n}\right)+\frac{3}{4}x^2\left(1+\frac{1}{2^2}+\frac{1}{2^4}+..........+\frac{1}{2^{2n-2}}\right)+x\left(1+\frac{1}{2}+\frac{1}{2^2}+...............+\frac{1}{2^{n-1}}\right)$
Now , putting $x=3$ in the last eqn. and considering $n\to \infty$ , $$f\left(\frac{x}{2^n}\right)\to f(0)=1$$Assuming the two sums in the R.H.S. of $(A)$ as infinite G.P. (Am I correct here ?)
$$f(3)=f(0)+\frac{3}{4}.3^2\left(1+\frac{1}{2^2}+\frac{1}{2^4}+.........\right)+3.\left(1+\frac{1}{2}+\frac{1}{2^2}+.............\right)=1+\frac{27}{4}.\frac{4}{3}+3.2=1+9+6=16$$

hint
$$f (3)-f (\frac {3}{2})=\frac {3^3}{2^2}+3$$
$$f (\frac {3}{2})-f (\frac {3}{2^2})=\frac {3^3}{2^4}+\frac {3}{2} $$
...
$$f (\frac {3}{2^n})-f (\frac {3}{2^{n+1}})=\frac {3^3}{2^{2n+2}}+\frac {3}{2^n} $$
sum and $n\to +\infty $.
You will find
$$f (3)=f (0)+\frac {27}{4}\frac {1}{1-\frac {1}{4}}+3\frac {1}{1-\frac {1}{2}} $$
$$=16$$
