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Start with $f(x)=x^2+bx+c$.

Then, attempt to solve for $x$ in $f(x)=x$.

It is easily found that $x=\frac{1-b\pm\sqrt{(b-1)^2-4c}}{2}$.

Then, start again with $f(x)=x$ and apply the function $f$ to both sides:

$$f(f(x))=f(x)$$

We see that the right side is equivalent to the original equation's left side, hence we have

$$f(f(x))=f(x)=x$$

Which we will remove the middle to get:

$$f(f(x))=x$$

Now, solving this problem is not equivalent to the original problem because we have now 'picked up' new solutions when we applied the function $f$.

However, we have not lost solutions.

Also, this, with manipulation, becomes

$$f(f(x))-x=0$$

Given that there are two solutions found through

$$f(x)-x=0$$

Because we have not lost any of the original solutions.

Since $f(x)-x$ contains solutions to the problem $f(f(x))-x$, is a polynomial, and does not contain any other solutions than the ones for $f(f(x))-x$, it is safe to assume that $f(x)-x$ is a factor of $f(f(x))-x$, which allows us to find the other two solutions that we 'picked up.'

And since it is a factor, we may divide to find the other factors:

$$\frac{f(f(x))-x}{f(x)-x}=P_1$$

Now, $P_1$ is a special polynomial that I have found to be:

$$P_1=x^2+(b+1)x+c+b+1$$

This is quite indeed an interesting quadratic in my opinion, and helpful as well, for it is the second quadratic factor of $f(f(x))-x$.

Furthermore, $f(f(f(x)))-x$ has factors $f(f(x))-x$ and $f(x)-x$ by a sort of recursive definition.

And if we proceed to find the unknown new roots we picked up, we will use division.

$$\frac{f(f(f(x)))-x}{[f(f(x))-x]\cdot[f(x)-x]}=P_2$$

Interestingly, $P_2$ is also a quadratic.

And if we proceed in this manner, we will find $P_n$ to be defined as:

$$P_n=\frac{f^n(x)-x}{\Pi_{i=1}^{n-1}\left(f^i(x)-x\right)}$$

Where we have $f^n(x)=f(f(f(\dots f(x)\dots)))$ where we have $n$ iterations of $f$.

Is there anything you can find that is special about $P_n$? Is there any easier method by which I can find what $P_n$ is?

I note that the definition is somewhat recursive because we have:

$$\frac{f(f(f(x)))-x}{[f(f(x))-x]\cdot[f(x)-x]}=\frac{f(f(f(x)))-x}{[(f(x)-x)(P_1)]\cdot[f(x)-x]}=\frac{f(f(f(x)))-x}{P_1[f(x)-x]^2}=P_2$$

A similar definition can be made for $P_n$, but it is very difficult for me to find it.

So three questions:

Is there anything you can find that is special about $P_n$? Is there any easier method by which I can find what $P_n$ is? Can you simplify the recursive formula for $P_n$?

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  • $\begingroup$ $\frac{f(f(x))-x}{f(x)-x}=P_1$ I thought $f(x) - x$ was zero? $\endgroup$ – Neil Feb 12 '16 at 13:18
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    $\begingroup$ @Neil for some $x$. $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 13:19
  • $\begingroup$ Oh, so $x$ is a constant? $\endgroup$ – Neil Feb 12 '16 at 13:29
  • $\begingroup$ @Neil It is a variable and $f(x)-x$ is zero only for some $x$. If it is a value $x=?$ where we have $f(x)-x=0$, then we also have $f(f(x))-x=0$. Other than that situation, division will result in a special quadratic. $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 21:27
  • $\begingroup$ The coefficient of $x$ in $P_1$ can be simplified to $1 + b +\frac{b^2}{b+c}$. Not much of an improvement, but slightly better. $\endgroup$ – Paul Sinclair Feb 13 '16 at 4:21
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Let $f(x) = x^2$, then $f^n(x) = x^{2^n}$.

$$f^n(x) - x = x^{2^n}-x = x\left(x^{2^n-1} - 1\right)$$ Thus the roots of $f^n(x) - x$ consist of $0$ and the $(2^n - 1)$th roots of unity. In particular, non-zero roots of

  • $f^1(x) - x$ are just $1$.
  • $f^2(x) - x$ are the $3$rd roots of unity $1, e^{i2\pi/3}, e^{i4\pi/3}$
  • $f^3(x) - x$ are the $7$th roots of unity $1, e^{i2\pi/7}, e^{i4\pi/7}, e^{i6\pi/7}, e^{i8\pi/7}, e^{i10\pi/7}, e^{i12\pi/7}$.

By your claim

Furthermore, $f(f(f(x)))−x$ has factors $f(f(x))−x$ and $f(x)−x$ by a sort of recursive definition.

the roots of $f^2(x) - x$ should be included among the roots of $f^3(x) - x$, but this is evidently not the case.

For general $f$, letting $x_0$ be a root of $f^n(x) - x$, we see that $$f^{n+1}(x_0) = f(f^n(x_0)) = f(x_0).$$ Thus $x_0$ is only a root of $f^{n+1}(x) - x$ if it is also a root of $f(x) - x$.

I think there is still an interesting recursion here, but it is not the one shown. Instead, ask what happens if $P_1$ takes the place of $f$. That is:

$Q_0(x)$ is any quadratic, and for $n \in \Bbb N$, $$Q_{n+1}(x) = \frac{Q_n(Q_n(x)) - x}{Q_n(x) - x}$$

Addendum since it turns out $Q_{n+1}(x) = Q_n(x) + x + b + 1$ for all $n$, this version isn't all that interesting after all. Oh well.

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  • $\begingroup$ Ah I see what you mean. Will think about this. $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 20:13
  • $\begingroup$ Isn't $P_n$ a subset of $f$? $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 20:18
  • $\begingroup$ I'm going to think about this. $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 20:23
  • $\begingroup$ @SimpleArt - "Isn't $P_n$ a subset of $f$?" - Since $P_n$ and $f$ are functions, not sets, I have no clue what you mean by this. $\endgroup$ – Paul Sinclair Feb 13 '16 at 23:00
  • $\begingroup$ I just meant that $f$ should be $x^2+bx+c$ and $P_n$, some quadratic, then $P_n$ is a version (subset?) of $f$? Also, thanks for answering my questions and more. :D $\endgroup$ – Simply Beautiful Art Feb 14 '16 at 12:23
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Considering a quadratic map: $z_{n+1}=z_{n}^{2}+c$ with $z_{0}=c$. For $c\in \mathbb{C}$ such that $z_{n+p}=z_{n}$, we say $c$ is a critical point with period $p$. In particular $n=0$, $c$ is attractive. Otherwise, it's called Misiurewicz point. Of course $\{c:z_{n+p}=z_{n}\} \subset \{c:z_{nk+pm}=z_{nk}\}$. For example,

$z_{1}=z_{0} \implies c^{2}+c=c \implies c^{2}=0$

$z_{2}=z_{1} \implies (c^{2}+c)^{2}+c=c^{2}+c \implies c^{3}(c+2)=0$

$z_{2}=z_{0} \implies (c^{2}+c)^{2}+c=c \implies c^{2}(c+1)^{2}=0$

$z_{3}=z_{2} \implies ((c^{2}+c)^{2}+c)^{2}+c=(c^{2}+c)^{2}+c \implies c^{4}(c+2)(c^{3}+2c^{2}+2c+2)=0$

$z_{3}=z_{1} \implies ((c^{2}+c)^{2}+c)^{2}+c=c^{2}+c \implies c^{3}(c+1)^{2}(c+2)(c^{2}+1)=0$

$z_{3}=z_{0} \implies ((c^{2}+c)^{2}+c)^{2}+c=c \implies c^{2}(c^{3}+2c^{2}+c+1)=0$

In fact, $\displaystyle M=\{c\in \mathbb{C}: \sup_{n\in \mathbb{N}} z_{n} < \infty \}$ is the famous Mandelbrot Set which is a fractal.

enter image description here

For the period doubling properties, see https://en.wikipedia.org/wiki/Feigenbaum_constants

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  • $\begingroup$ I think this is both similar and interesting. However, I think my problem is concerning $z_{n+1}=z_n^2+bz_n+c$? And because of my $f^n(x)-x$, each next polynomial in my sequence is factorable by all the polynomials before it. $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 13:51
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Trivial solution of $f(f(x)) \equiv x$ is $f(x) \equiv \pm x$ and there's no other polynomial solution otherwise the power will be raised. However, there's bilinear solution: $\displaystyle f(x) \equiv \frac{ax+b}{cx \pm a}$ where $bc \neq -a^{2}$. In particular $a=0$ gives $\displaystyle f(x) \equiv \frac{b}{cx}$ where $b, c, x\neq 0$.

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  • $\begingroup$ That is false. Suppose we have $f(x)=x^2-1$, then we also have $f(f(x))=(x^2-1)^2-1=x^4-2x^2$. Now subtract $x$ from each:$$f(x)-x=x^2-x-1$$ and $$f(f(x))-x=x^4-2x^2-x$$. Any solution to the former is a solution to the latter because it is a factor of the latter. However, if we have$$\frac{f(f(x))-x}{f(x)-x}$$then it equals$$=x^2-x$$The resulting polynomial solution you say doesn't exist. $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 21:31
  • $\begingroup$ Seem I've misinterpreted to functional equation. See another answer for your further interest. $\endgroup$ – Ng Chung Tak Feb 13 '16 at 7:04

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