How to solve $f(x) = x^3 - f(2x)$

Question

I'm trying to solve this functional equation:

Find all continuous functions $f:\mathbb R \to \mathbb R$ fulfilling $$ f(x) = x^{3} - f(2x). $$

What I did: $$ \begin{align*} f(x) & = x^3 - f(2x) \\ f\left(\frac{x}{2}\right) & = \left(\frac{x}{2}\right)^{3} - f(x) \\ & \vdots \\ f\left(\frac{x}{2^{n}}\right) & = \left(\frac{x}{2^{n}}\right)^{3} - f\left(\frac{x}{2^{n-1}}\right) \end{align*} $$

Now if I substitute $f(x)$ in the second equation with it's value in the first I get: $$ f\left(\frac{x}{2}\right) = \left(\frac{x}{2}\right)^3 - \left(x^3 - f(2x)\right) $$

So I keep going like that and eventually I get this: $$ f\left(\frac{x}{2^{n}}\right) = \left(\frac{x}{2^{n}}\right)^{3} - \cdots + (-1)^{n + 1} \left(\frac{x}{2}\right)^3 + (-1)^n f(x) $$

Then if I take the limit I should be able to find $f(x)$. However, I struggle with determining the limit of the alternating series I get: $$ \sum_{n=1}^\infty (-1)^{n+1} 2^{-3n} $$

The left-hand side $f\left(\frac{x}{2^{n}}\right) \to 0$ when $n \to \infty$.

I also searched online and found another solution to a similar problem:

If I follow it, I get this answer:

Let $ f(x) = kx^{3}$, then $f(2x) = 8kx^{3}$ and their sum is $9kx^{3}$. So $k = \frac{1}{9}$ which means that $f(x) = \frac{1}{9}x^{3}$ and that does seem to satisfy the equation.

My question is this: is it possible to get the same with my approach?

Thanks in advance.

Answer 1

The sum you got is a trivial geometric progression of ratio $-\frac18$

$$(-1) \cdot \sum_{n=1}^{\infty} (-1)^n (\frac18)^n = (-1) \cdot \frac{-\frac18}{1 + \frac18} = \frac19$$

the bigger problem with your approach is the term $(-1)^n$ multiplying $f(x)$

$$0 = \frac {x^3}9 \pm f(x)$$ (assuming that $\lim_{n \to \infty} (-1)^n$ is either 1 or -1 which it isn't)

Your approach is not quite formal, perhaps if you jump from two to two instead of making a recurrence from $n \to n+1$ you may get rid of that $(-1)^n$ and get a formal proof.

Answer 2

Another way is like this: $$f(x)=x^3-f(2x)$$ Formally, suppose $f(x)=a_0+a_1x+a_2x^2+\cdots$ then $$a_0+a_1x+a_2x^2+\cdots=x^3-a_0-2a_1x-2^2a_2x^2-\cdots,$$ so $$2a_0+3a_1x+5a_2x^2+9a_3x^3+17a_4x^4+33a_5x^5+\cdots=x^3,$$ so $9a_3=1\implies a_3=1/9$ and $a_{i\neq 3}=0$. So $$f(x)=\frac{x^3}{9}.$$

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But as @hellofriends points out, to evaluate your expression you can use the fact that: $$\frac{x}{1+x}=x-x^2+x^3-x^4+\cdots,$$ (it's easy to convince yourself of this, just multiply up by $1+x$ and expand, or by geometric progression formula), and then plug in $1/8$.