I want to find all polynomials that has the following property

$$P(P'(x)) = P'(P(x))$$

where P(x) is a polynomial.

Can you please tell me a way to solve this problem?

  • $\begingroup$ I don't think it can be monic. Look at the leading term of the composition. $\endgroup$ – open problem Jan 24 at 3:15
  • 1
    $\begingroup$ $P(x)=x$ works, as does $P(x)=0$ $\endgroup$ – Rhys Hughes Jan 24 at 3:17
  • $\begingroup$ Ah yes exceptions for 1st and 0th order. I should have been more clear. $\endgroup$ – open problem Jan 24 at 3:20

Let $ax^{n}$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving:

$a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$

So $a^{n+1}n^{n} = a^{n}n$

So $a= \frac{1}{n^{n-1}}$

So one such family is $\frac{x^{n}}{n^{n-1}}$

Now consider the constant coefficient K and the linear coefficient L:

$P'(K) = P(L)$

So if $P(x) = Lx+K$ is linear then, LK=K

So linear functions of the form

$P(x) = x+k$


$P(x) = ax^{2}+Lx+K$

Has $a=\frac{1}{2}$ and $\frac{L^{2}}{2}+L^{2} + K = K + L$

So K is free, and Either L=0 or $L = \frac{2}{3}$.

$L = \frac{2}{3}$ doesn't work when plugging back into $P'(P(x)) = P(P'(x))$

So we get $\frac{x^{2}}{2}+k$.

At third order and higher P and P' are both non-linear which forces K to be a solution of a polynomial of order higher than 2 and so K can only take on so many values. It is likely that the first family given will describe all of the remaining polynomials. We will demonstrate for the third order polynomials.

$\frac{x^{3}}{9} + bx^{2} + Lx + K$

Looking at 5th order terms:

$\frac{b}{27} = \frac{2b}{81}$

Which forces $b=0$.

Then looking at second order terms:

$\frac{L^{2}}{3} = 0$

which forces $L=0$.

$$\frac{K^{2}}{3} + 2bK + L = \frac{L^{3}}{9} + bL^{2} + L^{2} + K$$

So $\frac{K^{2}}{3} - K = 0$.

So K=3 or K=0.

$\frac{1}{3}(\frac{x^{3}}{9} + K )^2 = \frac{1}{9}(\frac{x^{2}}{3})^3 + K$

Looking at the 3rd order term forces K=0.

So $\frac{x^{3}}{9}$ is the only 3rd order solution.


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