# At most $n$ functions satisfy $f(x)=\bigl(x-f(a_1)\bigr)\bigl(x-f(a_2)\bigr)\dots\bigl(x-f(a_n)\bigr)$?

Some background: I was trying to solve the functional equation $$f\bigl(f(x)\bigr)=\sin x$$. I realized that $$f(\pi n)$$ is a root of $$f$$ for all integers $$n$$, because $$f\bigl(f(\pi n)\bigr)=\sin(\pi n)=0$$. Thus, we can write $$f$$ as $$f(x)=A\bigl(x-f(0)\bigr)\bigl(x-f(\pi)\bigr)\bigl(x-f(-\pi)\bigr)\dots$$. The problem now is to find all functions $$f$$ that solve this. I couldn't even begin on this, so I decided to try letting the constant $$A$$ equal $$1$$, and to try some simpler versions first. E.g. For $$f(x)=x-f(0)$$ we find the unique solution $$f(x)=x$$; for $$f(x)=\bigl(x-f(0)\bigr)\Bigl(x-f\left(\sqrt{2}\right)\Bigr)$$ we find two solutions, namely $$f(x)\in \left\{x^2-1,x^2-2\left(\sqrt{2}-1\right)x\right\}$$.

I solved many more examples, and then conjectured that:

Let $$a_i\in \mathbb{C}$$ for $$i=1,2,\dots,n$$. Prove that the functional equation $$f(x)=\bigl(x-f(a_1)\bigr)\bigl(x-f(a_2)\bigr)\dots\bigl(x-f(a_n)\bigr)$$ has exactly $$n$$ distinct solutions.

Finally, I realized that if this is proved, then it would "follow" (I don't know if this is rigorous but I hope someone knows how to make it rigorous) that our original functional equation $$f(x)=\bigl(x-f(0)\bigr)\bigl(x-f(\pi)\bigr)\bigl(x-f(-\pi)\bigr)\dots$$ (I put $$A=1$$) has infinitely many solutions.

• I don't understand. Does $f(x) = x - f(0)$ not have uncountably many solutions depending on $f(0)$? Commented Sep 27, 2014 at 14:33
• No, $f(x)=x+c$, then $x-f(0)=x-c$, so $c=0$ Commented Sep 27, 2014 at 14:38

The first random examples I tried showed that your conjecture was mostly false. The only thing that may still be possibly true is that there is always at least one solution.

Let us put $b_i=f(a_i)(1\leq i\leq n)$. The conditions on $f$ can then be restated as :

$$b_i=\prod_{k=1}^{n} (a_i-b_k) \ (1\leq i \leq n) \ (1)$$

The number of solutions can be LESS than $n$ : consider the case $n=2,a_1=0,a_2=1$. Then (1) yields $(i)b_1=b_1b_2$ and $(ii)b_2=(1-b_1)(1-b_2)$. By (i) we have $b_1=0$ or $b_2=1$, but $b_2=1$ is impossible because of (ii). So $b_1=0$ and hence $b_2=\frac{1}{2}$. In this case, there is exactly one solution.

The number of solutions can be MORE than $n$ : consider the case $n=3,a_1=0,a_2=1,a_3=7$. The polynomial

$$P=2304X^4 - 39168X^3 + 214208X^2 - 402384X + 223191$$

has four real roots $\alpha_1\approx 0.97, \alpha_2\approx 2.02, \alpha_3\approx 6.98,\alpha_4\approx 7.01$, and is easily seen to be irreducible over $\mathbb Q$.

Let $\alpha$ be any of the four roots of $P$ ; let

$$\left\lbrace\begin{array}{lcl} b_1 &=& \alpha, \\ b_2 &=& \frac{1152\alpha^3 - 12240\alpha^2 + 32224\alpha - 20964}{525} \\ b_3 &=& \frac{-1152\alpha^3 + 12240\alpha^2 - 32944\alpha + 26004}{645} \\ \end{array}\right.$$

Then routine calculations show that $(x-b_1)(x-b_2)(x-b_3)$ is indeed a solution. In this case, there are thus at least four solutions.