If $p(x)$ is polynomial such that $$p(x^2+1)=(p(x))^2 +1$$ and $p(0)=0$ then $p'(0)$ is equal to _____.


While looking on it and putting $x=0$ I got $p(1)=1$ and also the equation given had also the similar form (in comparison) of $$y^2+1=(y^2)+1$$. So, I put this polynomial in test and found that it satisfies this as $p(x^2+1)=x^2+1$ and and $p(x)=x$ and hence it satisfied for every real x and so $p'(0)=1$. But I was not able to progress in the direction of correct solving.

Also, the given related problem is there on this site but, I have not been taught the coefficient determination and hence that answer there is out of my scope for now. But,I think there is this derivative method which I am currently learning and can understand about it.So, I would like to know how to approach this in derivative method and also if there is any other easier way without proving $p(x)$ to be the unique solution.

Also, it may happen that if any other exists,then many persons will get to know about the new method. It will be more helpful to them and me inlcuded.

So, is there any hints or suggestions on this problem?

Thanks for the help!

  • $\begingroup$ Apparently $p(x)=x$ is a solution, and then $p'(0)=1$. The phrasing of the question implies that $p'(0)$ is uniquely determined by the given functional equation, and so $p'(0)=1$ is the answer. $\endgroup$ – Servaes Feb 20 at 14:01
  • $\begingroup$ @daw Then it will be $p'(x^2+1)2x=2p(x)p'(x)$ $\endgroup$ – jayant98 Feb 20 at 14:01
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    $\begingroup$ @daw It's not immediately obvious how that helps, as inserting $x = 0$ into that relation just tells us that $0 =0$. So why don't you flesh it out a bit and make an actual answer out of it? $\endgroup$ – Arthur Feb 20 at 14:02
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    $\begingroup$ Related: Determine all real polynomials with $P(0)=0$ and $P(x^2+1)=(P(x))^2+1$ $\endgroup$ – Martin R Feb 20 at 14:02
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    $\begingroup$ @Servaes: Related (and perhaps useful), but not necessarily a duplicate. There might be an easier proof of $p`(0)=1$, without showing first that $p(x)=x$ is the only solution. $\endgroup$ – Martin R Feb 20 at 14:09

The value of $p'(0)$ cannot determined uniquely unless we utilize the information that $p$ is polynomial critically at some point. Indeed,

Claim. For any $c \in [0, \infty)$, there exists a smooth function $p : \mathbb{R} \to \mathbb{R}$ such that

$$p(x^2 + 1) = p(x)^2 + 1, \qquad p(0) = 0, \qquad p'(0) = c. $$

Because the proof is rather technical and I do not want to digress, let me skip the full proof. But the idea is that, starting with any function $p_0 : [0, 1) \to [0, 1)$ with $p_0(0) = 0$, we can define $p$ on $[0, \infty)$ by

$$ p(x) = \big(f^{\circ n} \circ p_0 \circ f^{\circ(-n)} \big)(x), \qquad \text{if} \quad x \in \big[ f^{\circ n}(0), f^{\circ(n+1)}(0) \big)$$

and then extend to all of $\mathbb{R}$ by the relation $p(-x) = -p(x)$.

That said, one must take advantage of the fact that $p$ is polynomial. But since the derivative method does not distinguish polynomials from other smooth functions, I see no easy way of utilizing this method to solve OP's problem other than showing $p(x) = x$ first.

In this regard, let me introduce a solution which relies on the following simple fact:

Lemma. If $q$ is a polynomial which has infinitely many zeros, then $q(x) = 0$ for all $x$.

Now define $x_0 = 0$ and $x_{n+1} = x_n^2 + 1$. Then we can inductively prove that $p(x_n) = x_n$. Indeed, this is true for $n = 0$ by the assumption. Moreover, if $p(x_n) = x_n$ is true for $n$, then

$$ p(x_{n+1}) = p(x_n^2 + 1) = p(x_n)^2 + 1 = x_n^2 + 1 = x_{n+1}. $$

Then applying Lemma to the polynomial $p(x) - x$, which has infinitely many zeros $x_0, x_1, \cdots$, we conclude that $p(x) = x$ and hence $p'(0) = 1$. $\square$.

  • $\begingroup$ Nice.... shows that it's unlikely there is an elementary solution that avoids showing $p(x)=x$, but more importantly, shows that proving $p(x)=x$ is itself simple and short and only needs knowing that a (non-$0$) real polynomial has only finitely many roots (no calculus nor coefficient calculations are needed) $\endgroup$ – Ned Feb 22 at 16:32
  • $\begingroup$ @Ned, That's how I believe so. Indeed, the derivative method is not quite suited for distinguishing polynomials from other smooth functions, so I can hardly think of any easy solution in this direction. $\endgroup$ – Sangchul Lee Feb 22 at 16:35

Of course the link proposed as a duplicate solves this.

a more advanced perspective
There are very limited possibilities where two polynomials commute under composition (in characteristic zero).

Note that $S(x):=x^2+1$ is not conjugate to the degree $2$ Chebyshev polynomial. Therefore, the only polynomials that commute with $S$ under composition are the compositional powers $S^{[n]}$, $n=0,1,2,\cdots$. The only compositional power $p$ with $S^{[n]}(0)=0$ is $S^{[0]}$ (the constant terms increase with $n$). So $p(x) = x$.


The derivative of $p$ satisfies the equation $$ p'(x^2+1)\cdot 2x = 2p(x)p'(x). $$ Dividing by $2x$, and passing to the limit $x\to 0$ we get $$ p'(1) = p'(0)^2, $$ because $p(x)/x \to p'(0)$ for $x\to0$.

I do not see how to obtain $p'(0)$ without proving $p(x)=x$. The relation $p'(1) = p'(0)^2$ can also be obtained by comparing the polynomial coefficients in the original function equation.


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