The relation between the number of solution of $f(x)=x$ and $f(f(x))=x$, when f is a quartic polynomial

Question

I'm trying to solve the following problem :

"Let $f$ be a quartic polynomial with a positive leading coefficent, and all coefficients are real numbers. Let $m$ be the number of real solution of $f(x)=x$, and $n$ be the number of real solution of $f(f(x))=x$. If $m,n$ are natural numbers, then how many ordered pairs of $(m,n)$ are possible?"

When $m=1$, it follows that $f(x) \ge x$, which in turn implies $f(f(x)) \ge f(x) \ge x $ and equality holds iff $f(x)=x$, thus $n=1$.

But in other cases, when $m=2,3,4$ I can't handle it. The only guess I have is as follows. If $f(x)=x$, then $f(f(x))=x$, thus $n \ge m$. And if $f(f(x))=x$ and $f(x) \ne x$, then there exists distinct $a$ and $b$ such that $f(a)=b , f(b)=a$. Thus $n$ should be $m+(even number)$.

Is there any other hints can I get? I desperately need help now. I'm in trouble..

Answer 1

Consider a simplified problem.

$f(x)=ax^2+bx+c$

$f(x)=x\implies ax^2+(b-1)x+c=0$

$\implies x= \frac{(1-b)\pm \sqrt{(b-1)^2-4ac}}{2a}$

$\frac{(b-1)^2}{4a}\ge c$

By FTA, we know there are at most 2 real solutions. This occurs iff $(b-1)^2-4ac>0$. There might be only one solution if $(b-1)^2=4ac$, and no real solutions if $(b-1)^2-4ac<0$. The term under the radical is the discriminant.

Suppose $b=0$ and $a=1$. Then the discriminant is $\sqrt{1-4c}$ So there are no fixed points for $c>1/4$, one distinct solution for $c=1/4$ and two solutions for $c<1/4$.

$f(f(x))=(x^2+c)^2+c=x^4+2cx^2+c^2+c=x$. Since $f(x)=x$ gives solutions of this equation too,

$x^4+2cx^2-x+c^2+c=(Ax^2+Bx+C)(x^2-x+c)$

$A=1, C=c+1, B=A=1$

So $x=\frac{-1\pm \sqrt{-3-4c}}{2}$ gives potentially $2$ new solutions. $-3-4c\ge 0 \implies -3/4 \ge c$

If $f(x)=x$, solutions are called fixed points. If $f(f(x))=x$ and $f(x) \ne x$, solutions are called periodic points of prime period 2.

In this simplified case, there can be $2,1$ or $0$ distinct fixed points and $2,1$ or $0$ period 2 points, (ignoring solutions of $f(f(x))=x$ where $f(x)=x$ ).

If $f(x)$ is a 4th degree polynomial, the discriminant is more complicated: 4th degree polynomial discriminant

In the second order version the solution you seek is a subset of the Cartesian Product $\{2,1,0\} \times \{4, 3,2 ,1, 0\}$

$(2,0)$ is not a possible ordered pair since since 2 solutions for $f(x)=x$ implies at least two solutions for $f(f(x))=x$. Note iff there is only 1 solution of $f(x)=x$, where $f(x)=x^2+c$, there is only one solution for $f(f(x))=x$, i.e. the one distinct fixed point, and since the discriminant of $f(f(x))<0$ in that case, no solutions having period 2.

Also keep in mind you can find possible solutions graphically.

Fixed points are intersections of $y=x$ and $y=f(x)$.

Period two points are $y=x$ intersecting $y=f(f(x))$.

Hope this helps. For more look up algorithms for finding fixed points and periodic points of prime period 2.

Answer 2

Unfortunately, I can provide only partial results and more or less promisimg ideas.

I do not know restrictions on possible pairs $(m,n)$ distinct from those mentioned in the question. So I used Mathcad to look for the polynomials $f(x)$, yielding different pairs $(m,n)$. We found that the following pairs $(m,n)$ can be realized by the respective polynomials.

$(2,2): x^4$,

$(2,4): x^4-1$,

$(2,8): x^4-4x^2-1$,

$(3,3): x^4-x^2+x$,

$(3,5): x^4-3x^2+x$,

$(3,9): x^4-4x^2+x$,

$(4,4): 0.1x^4-x^2+1$,

$(4,6): 0.1x^4-0.2x^3-x^2+x+0.5$,

$(4,10): 0.05x^4-0.1x^3-x^2+0.5x+0.75$,

$(4,12): 0.01x^4-x^2+1$,

$(4,16): 0.01x^4-x^2+10$,

This list looks incomplete, but I constructed the found polynomials by hand. But the search range can be done much bigger with the respective soft, for instance, by trying polynomials with random coefficients.

Also maybe the found polynomials can be modified to realize new pairs $(m,n)$.

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I also tried to describe the solutions of the equation $f(f(x))=x$. Namely, let $f(t)=a_4t^4+a_3t^3+a_2t^2+a_1t+a_0$ and $x$ be a solution of the equation $f(f(x))=x$ but not a solution of the equation $f(x)=x$. Let $y=f(x)$. Then $x-y=f(y)-f(x)$. Dividing both parts of the quality by $y-x$ we obtain that $-1$ equals $$a_4(x^3+x^2y+xy^2+y^3)+a_3(x^2+xy+y^2)+a_2(x+y)+a_1=$$ $$a_4(\sigma^3-2\sigma\rho)+a_3(\sigma^2-\rho)+a_2\sigma+a_1,$$ where $\sigma=x+y$ and $\rho=xy$ are the standard symmetric polynomials. Then $$\rho=\frac{a_4\sigma^3+a_3\sigma^2+a_2\sigma+a_1+1}{2a_4\sigma+a_3}$$ or $2a_4\sigma+a_3=0$ and $a_4\sigma^3+a_3\sigma^2+a_2\sigma+a_1+1=0$. Moreover, $x+y=f(y)+f(x)$, that is $\sigma$ equals $$a_4(\sigma^4-2\rho(2\sigma^2-\rho))+a_3(\sigma^3-3\sigma\rho)+a_2(\sigma^2-2\rho)+a_1\sigma+2a_0=$$ $$(2a_4\sigma+a_3)\sigma\rho-\sigma-2a_4\rho(2\sigma^2-\rho)-3a_3\sigma\rho-2a_2\rho+a_1\sigma+2a_0=$$ $$-\sigma-2a_4\sigma^2\rho-2a_4\rho^2-2a_3\sigma\rho-2a_2\rho+a_1\sigma+2a_0.$$ Substituting $\rho$, we obtain an equation of the sixth degree for $\sigma$.

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Also there is an other way to represent $f(x)$. For instance, let $m=4$. Then there exist real numbers $x_1<x_2<x_3<x_4$ such that $f(x)=P(x)+x$, where $P(x)=a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$. Then $$f(f(x))-x=(P(x)+x-x_1)(P(x)+x-x_2)\times$$ $$(P(x)+x-x_3)(P(x)+x-x_4)+P(x)=$$ $$P(x)(a(x-x_2)(x-x_3)(x-x_4)+1)\times$$ $$(a(x-x_1)(x-x_3)(x-x_4)+1)\times$$ $$(a(x-x_1)(x-x_2)(x-x_4)+1)\times$$ $$(a(x-x_1)(x-x_2)(x-x_3)+1)+1).$$