Solving a Functional Equation Problem: Finding f(3)-f(0)

[Saitama]

Problem:
Let $f:R\rightarrow R$ be a continuous function such that
$$f(x)-2f\left(\frac{x}{2}\right)+f \left( \frac{x}{4} \right)=x^2$$
Find $f(3)-f(0)$.

Attempt:
I really don't know how should I approach this problem. I could only deduce that $f(0)=0$. Then I tried putting a few values for $x$ i.e $x=1,2,3$ but that doesn't seem to help. I honestly don't know what should I do to find the value of $f(3)$.

Any help is appreciated. Thanks!

(I am not sure if this should belong to the Algebra section)

 

[Opalg]

Problem:
Let $f:R\rightarrow R$ be a continuous function such that
$$f(x)-2f\left(\frac{x}{2}\right)+f \left( \frac{x}{4} \right)=x^2$$
Find $f(3)-f(0)$.

Attempt:
I really don't know how should I approach this problem. I could only deduce that $f(0)=0$. Then I tried putting a few values for $x$ i.e $x=1,2,3$ but that doesn't seem to help. I honestly don't know what should I do to find the value of $f(3)$.

To start with, I do not see why $f(0)$ should be $0$. If you put $x=0$ in the above equation then you just get $0=0$.

Next, suppose that you repeatedly apply the definition to fractions of $x$, like this: $$\begin{aligned} f(x) &= x^2 + 2f\bigl(\tfrac x2\bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + 2\Bigl(\bigl(\tfrac x2\bigr)^2 + 2f\bigl(\tfrac x4\bigr) - f\bigl(\tfrac x8\bigr)\Bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3f\bigl(\tfrac x4\bigr) - 2f\bigl(\tfrac x8\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3\Bigl(\bigl(\tfrac x4\bigr)^2 + 2f\bigl(\tfrac x8\bigr) - f\bigl(\tfrac x{16}\bigr)\Bigr) - f\bigl(\tfrac x8\bigr) \\ &= \ldots .\end{aligned}$$ Continuing in that way, you should find a series for $f(x)$ that converges to a limit, giving a expression for $f(x) - f(0).$

 

[Saitama]

Hi Opalg! :)

To start with, I do not see why $f(0)$ should be $0$. If you put $x=0$ in the above equation then you just get $0=0$.

Sorry about that.

Next, suppose that you repeatedly apply the definition to fractions of $x$, like this: $$\begin{aligned} f(x) &= x^2 + 2f\bigl(\tfrac x2\bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + 2\Bigl(\bigl(\tfrac x2\bigr)^2 + 2f\bigl(\tfrac x4\bigr) - f\bigl(\tfrac x8\bigr)\Bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3f\bigl(\tfrac x4\bigr) - 2f\bigl(\tfrac x8\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3\Bigl(\bigl(\tfrac x4\bigr)^2 + 2f\bigl(\tfrac x8\bigr) - f\bigl(\tfrac x{16}\bigr)\Bigr) - f\bigl(\tfrac x8\bigr) \\ &= \ldots .\end{aligned}$$ Continuing in that way, you should find a series for $f(x)$ that converges to a limit, giving a expression for $f(x) - f(0).$

I don't get this. If I continue from where you left,
$$f(x)=\frac{3x^2}{2}+\frac{3x^2}{4^2}+5f\left( \frac{x}{8}\right)-3\left(\frac{x}{16}\right)$$
$$=\frac{3x^2}{2}+\frac{3x^2}{4^2}+5\left(\frac{x^2}{4^3}+2f\left(\frac{x}{16}\right)-f\left(\frac{x}{32}\right)\right)-3f\left(\frac{x}{16}\right)$$
$$=\frac{3x^2}{2}+\frac{3x^2}{4^2}+\frac{5x^2}{4^3}+7\left(\frac{x}{16}\right)-5f\left(\frac{x}{32}\right)$$
I don't see what I have to do with the above, I can keep on expanding the above and I will end up with a series plus (something)*f(x/something').

 

[Opalg]

To start with, I do not see why $f(0)$ should be $0$. If you put $x=0$ in the above equation then you just get $0=0$.

Next, suppose that you repeatedly apply the definition to fractions of $x$, like this: $$\begin{aligned} f(x) &= x^2 + 2f\bigl(\tfrac x2\bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + 2\Bigl(\bigl(\tfrac x2\bigr)^2 + 2f\bigl(\tfrac x4\bigr) - f\bigl(\tfrac x8\bigr)\Bigr) - f\bigl(\tfrac x4\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3f\bigl(\tfrac x4\bigr) - 2f\bigl(\tfrac x8\bigr) \\ &= x^2 + \tfrac{2x^2}4 + 3\Bigl(\bigl(\tfrac x4\bigr)^2 + 2f\bigl(\tfrac x8\bigr) - f\bigl(\tfrac x{16}\bigr)\Bigr) - \color{red}{2}f\bigl(\tfrac x8\bigr)\quad \color{red}{\text{I left out a 2 from the previous line!}} \\ &= \ldots .\end{aligned}$$ Continuing in that way, you should find a series for $f(x)$ that converges to a limit, giving a expression for $f(x) - f(0).$

I don't get this. If I continue from where you left,
$$f(x)=\frac{3x^2}{2}+\frac{3x^2}{4^2}+5f\left( \frac{x}{8}\right)-3\left(\frac{x}{16}\right)$$
$$=\frac{3x^2}{2}+\frac{3x^2}{4^2}+5\left(\frac{x^2}{4^3}+2f\left(\frac{x}{16}\right)-f\left(\frac{x}{32}\right)\right)-3f\left(\frac{x}{16}\right)$$
$$=\frac{3x^2}{2}+\frac{3x^2}{4^2}+\frac{5x^2}{4^3}+7\left(\frac{x}{16}\right)-5f\left(\frac{x}{32}\right)$$
I don't see what I have to do with the above, I can keep on expanding the above and I will end up with a series plus (something)*f(x/something').

Putting in the $2$ that I left out, you should end up with a series looking like $$\sum_{k=0}^n \frac{(k+1)x^2}{4^k} + (n+2)\,f\bigl(\tfrac x{2^{n+1}}\Bigr) - (n+1)\,f\bigl(\tfrac x{2^{n+2}}\Bigr).$$ As $n\to\infty$, the continuity of $f$ at $x=0$ tells you that $$f(x) = \sum_{k=0}^\infty \frac{(k+1)x^2}{4^k} + f(0).$$ Then you just have to sum that series.

 

[Saitama]

Putting in the $2$ that I left out, you should end up with a series looking like $$\sum_{k=0}^n \frac{(k+1)x^2}{4^k} + (n+2)\,f\bigl(\tfrac x{2^{n+1}}\Bigr) - (n+1)\,f\bigl(\tfrac x{2^{n+2}}\Bigr).$$ As $n\to\infty$, the continuity of $f$ at $x=0$ tells you that $$f(x) = \sum_{k=0}^\infty \frac{(k+1)x^2}{4^k} + f(0).$$ Then you just have to sum that series.

That is great! Thanks a lot Opalg! :)

I am sorry for being careless.

(I really miss the thanks button.)