# IMC 2023 problem 1

Find all functions $$f : \mathbb{R} \to \mathbb{R}$$ that have continuous second derivative and for which the equality $$f(7x+1)=49f(x)$$ holds for all $$x \in \mathbb{R}$$.

I found a solution online to the problem and I want to ask about a part of the solution.

The solution is the following:

Solution. Differentiating twice gives that $$f''(7x + 1) = f''(x)$$. Repeatedly applying this equation (in both directions), we get that $$f''(x) = f''(7^n (x + 1/6) - 1/6)$$ for all $$n \in \mathbb{Z}$$. Since $$f''$$ is continuous, letting $$n \rightarrow -\infty$$, we get $$f''(x) = f''(-1/6)$$, and thus $$f''(x)$$ is constant for all $$x \in \mathbb{R}$$. This immediately implies that $$f$$ is quadratic, and we can then directly substitute in to find that the set of solutions is $$f(x) = ax^2 + \frac{1}{3}a x + \frac{1}{36}a,$$for any $$a \in \mathbb{R}$$.

what I wanted to ask about is what is meant by this "repeated application of the equation"? I'm not very familiar with recursion, but I suspect that there is some recursive argument going on with this.

It is a bit weird that they leave this part out, it seems pretty non-obvious.

Let's look at the equation $$f''(7x+1)=f''(x)$$

Say we substitute $$\frac{x}{7}-\frac{1}{7}$$ for $$x$$. This will give us $$f''(7(\frac{x}{7}-\frac{1}{7})+1)=f''(\frac{x}{7}-\frac{1}{7})$$ $$f''(x)=f''(\frac{x}{7}-\frac{1}{7})$$

Let's do the same substitution again. We get $$f''(\frac{x}{7}-\frac{1}{7})=f''(\frac{x}{49}-\frac{8}{49})$$ But we know that $$f''(x)=f''(\frac{x}{7}-\frac{1}{7})$$ so we have $$f''(x)=f''(\frac{x}{49}-\frac{8}{49})$$

If we perform the same substitution over and over again to infinity(with the same reduction back to $$f''(x)=...$$), we will end up with $$f''(x)=f''(\frac{-1}{7} + \frac{1}{7}(\frac{-1}{7}+\frac{1}{7}(\frac{-1}{7}+\cdots)))$$ To evaluate this, we can exploit self-similarity. Let $$y=\frac{-1}{7} + \frac{1}{7}(\frac{-1}{7}+\frac{1}{7}(\frac{-1}{7}+\cdots))$$. Then, we can notice that $$y$$ appears again in the expression and substitute as follows: $$y=\frac{-1}{7}+\frac{1}{7} y$$ $$\frac{6}{7}y=\frac{-1}{7}$$ $$y=\frac{-1}{6}$$ and hence $$f''(x)=f''(\frac{-1}{6})$$ and the solution proceeds as in OP.

If you want to rigorously define "repeating substitution infinitely many times," then you can show that after $$n$$ substitutions the equation you get is $$f''(x)=f''(7^{-n}(x+1/6)-1/6)$$ and then you take the limit as $$n \to \infty$$. You will end up with $$f''(x)=f''(\frac{-1}{6})$$ and again the solution proceeds as in OP.

Note: here, I only substitute in one direction, but by doing the reverse substitution $$x \to 7x+1$$ instead of $$x \to \frac{x}{7}-\frac{1}{7}$$ you get a similar equation.

• I think they ended up with $7^n$ instead of $7^{-n}$. Is there some difference between what you did here compared to their solution? @daniel-geyfman Aug 4, 2023 at 20:13
• @Victor The $7^{n}$ in their solution results from the substition $x \to 7x+1$(at least I assume). In my solution, I use $x \to \frac{x}{7}-\frac{1}{7}$, which is the "inverse" of their substitution. So, $n$ substitutions of $x \to 7x+1$ is equivalent to "$-n$" substitutions of $x \to \frac{x}{7}-\frac{1}{7}$, which is where my $7^{-n}$ comes from. If you were to recreate my solution with their substitution, you would end up with a $7^{n}$ Aug 4, 2023 at 21:09

Repeated application means apply the formula multiple times. For example if you apply it $$2$$ times you will have:

$$f''(x) = f''(7x+1) = f''(7(7x+1) + 1) = f\left(7^2 x + 8\right) = f\left(7^2 \left(x +\frac16\right) - \frac{7^2}6 + 8\right) = f\left(7^2 \left(x +\frac16\right) - \frac16\right)$$