An interesting functional equation

We have:

$\displaystyle \left( f(x)-f\left( \frac{x}{2} \right) \right)-\left( f\left( \frac{x}{2} \right)-f\left( \frac{x}{4} \right) \right)=x^{2}$

Replacing $x$ by $\frac{x}{2}$ :

$\displaystyle \left( f\left( \frac{x}{2} \right)-f\left( \frac{x}{4} \right) \right)-\left( f\left( \frac{x}{4} \right)-f\left( \frac{x}{8} \right) \right)=\frac{x^{2}}{4}$

Now again replace $x$ by $\frac{x}{4}$ and write the next term. Continuing this process uptil $n+1$ terms:

$\displaystyle \left( f\left( \frac{x}{2^{n}} \right)-f\left( \frac{x}{2^{n+1}} \right) \right)-\left( f\left( \frac{x}{2^{n+1}} \right)-f\left( \frac{x}{2^{n+2}} \right) \right)=\frac{x^{2}}{2^{2n}}$

Now adding everything and taking $n\rightarrow \infty$ we obtain:

$\displaystyle f(x)-f\left( \frac{x}{2} \right)=\frac{4x^{2}}{3}$

Note that right hand side was obtained by adding infinite terms of a G.P. Further note that as $n\rightarrow \infty$ we have:

$\displaystyle f\left( \frac{x}{2^{n+1}} \right)=f\left( \frac{x}{2^{n+2}} \right)$

Now we shall again repeat this process. Replacing $x$ by $\frac{x}{2}$ , $\frac{x}{4}$ and so on and then adding all the terms we obtain:

$\displaystyle f(x)-f(0)=\frac{16{x}^{2}}{9}$

Again note that as $n\rightarrow \infty$ we have:

$\displaystyle f\left( \frac{x}{2^{n+1}} \right)=f(0)$

Now just substitute $x=3$ to obtain the answer ! :)

This is my first solution, so please be nice in the comments as a lot of effort has gone into typing it :)

By hit and error let f(x) be kx²+c so From Given equation get that k=16/9 Hence easily find the required difference.

I kinda assumed that f is a polynomial which seemed reasonable as we have a polynomial on one side and f is continuous. Let one such term of f be a*x^n. If n=2, we want the coefficient of the LHS (which is a polynomial) to be 1 and the rest of the terms to drop out. If you just look at the x^n term, you see that (coefficient) x^n (2^2n -2^n+1 +1)/2^2n=0 whenever n not equal to 2. However, this is zero only when coefficient=0, x=0, or n=0 (constant term). But we want this to always equal to 0: so the coefficent is 0. Thus the only terms present are a quadratic and a constant term. WLOG you can set the constant term to 0 since whatever it is, it cancels out on the LHS. Then you say f=k *x^2, and you can find that k=16/9. Thus f(3)-f(0)=16-0=16.

The problem becomes a lot easier if you assume f is also three-times differentiable. Application of the chain rule shows f'(0) = 0, f''(0) = 32/9, and all higher-order derivatives are identically 0. Therefore, Taylor's Theorem yields f(3) = f(0) + 0 + (32/9) (1/2) (3-0)^2 + 0, which implies f(3) - f(0) = 16

You never know with functional equations, and I'm really wobbly with them. For no good reason, other than the presence of the ${ x }^{ 2}$ on the right, I assumed a cubic polynomial.

• define a cubic with arbitrary coefficients
• rewrite the given functional equation in terms of the cubic, expand and simplify
• compare like terms in the indeterminate
• check that the resulting cubic satisfies the functional equation
• calculate required result