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Find all real numbers $a,b,c\in\mathbb{R}$ for which there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that: $$ f(f(x))=ax^2+bx+c $$ for all $x\in\mathbb{R}$.

The only thing I could deduce is: $$ f(ax^2+bx+c)=af(x)^2+bf(x)+c $$ Which doesn't help much. How to tackle the problem?

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  • $\begingroup$ Could you just calculate $f(f(0)), f(f(1)), f(f(-1))$ and then solve the resulting equations to find $a, b, c$ in terms of $f(f(0)), f(f(1)), f(f(-1))$? $\endgroup$
    – Mufasa
    Dec 31, 2015 at 14:33
  • $\begingroup$ You could.. but then? $\endgroup$ Dec 31, 2015 at 14:40
  • $\begingroup$ My question was - is that sufficient or does the OP want $a,b,c$ in terms not involving $f$ at all? $\endgroup$
    – Mufasa
    Dec 31, 2015 at 14:42
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    $\begingroup$ It is from me; I stumbled upon a special case (namely to show that there is no function with $f(f(x))=x^2-2$) and thought it would be interesting to generalize a bit. $\endgroup$ Dec 31, 2015 at 15:18
  • $\begingroup$ From your case you can extend the result to all second-degree polynomials $h(x)$ in right hand side for which $h(x)-x$ and $h(h(x))-x$ have distinct real roots.. $\endgroup$ Dec 31, 2015 at 16:22

4 Answers 4

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SHORT ANSWER: A general solution to this problem is not known in the closed form, but some special cases can be solved. Sorry.

LONG ANSWER: Notice that if a function $f$ satisfies $$f(x)=(g^{-1} \circ h \circ g)(x)$$ for some $g,h$, then $$f^n(x)=(g^{-1} \circ h^n \circ g)(x)$$ where the superscript denotes functional iteration rather than exponentiation.

We can solve this equation for a whole class of quadratics $$f^2(x)=q(x)=ax^2+bx+c$$ for which $$c=\frac{b^2-2b}{4a}$$ This is because we can rewrite $$q(x)=ax^2+bx+\frac{b^2-2b}{4a}$$ as $$q(x)=a\bigg(x+\frac{b}{2a}\bigg)^2-\frac{b}{2a}$$ and so, by letting $g(x)=x-\frac{b}{2a}$ and $h(x)=ax^2$, $$q(x)=(g^{-1} \circ h \circ g)(x)$$ and thus $$q^n(x)=(g^{-1} \circ h^n \circ g)(x)$$ and since the formula for $h^n$ is $$h^n(x)=a^{2^n-1}x^{2^n}$$ we have $$q^n(x)=a^{2^n-1}\bigg(x+\frac{b}{2a}\bigg)^{2^n}-\frac{b}{2a}$$ and, finally, $$f(x)=q^{1/2}(x)=a^{\sqrt 2-1}\bigg(x+\frac{b}{2a}\bigg)^{\sqrt 2}-\frac{b}{2a}$$ So there's the solution for that special case. There is another special case when $$c=\frac{b^2-2b-8}{4a}$$ but the solution to that case is much longer, and so I will omit it and leave it to you for independent research.

There is another special case involving trigonometric functions. Note that if we let $$g(x)=\arccos x$$ $$h(x)=2x$$ we have $$(g^{-1}\circ h\circ g)(x)=\cos(2\arccos x)$$ and, using the double-angle formula, $$(g^{-1}\circ h\circ g)(x)=\cos^2(\arccos x)-\sin^2(\arccos x)$$ $$(g^{-1}\circ h\circ g)(x)=x^2-(1-x^2)$$ $$(g^{-1}\circ h\circ g)(x)=2x^2-1$$ and so $g^{-1}\circ h\circ g$ is a quadratic. Now let $$q(x)=(g^{-1}\circ h\circ g)(x)$$ so that $$q^n(x)=(g^{-1}\circ h^n\circ g)(x)$$ Now, since $$h^n(x)=2^n x$$ we have $$q^n(x)=\cos(2^n\arccos x)$$ and $$f(x)=q^{1/2}(x)=\cos(\sqrt{2}\arccos x)$$ which solves yet another special case. However, notice that this is only solved on the domain $[-1,1]$, because that is where $\arccos x$ is defined.

I am sure there are other special cases involving trigonometric functions, but I will leave those for you to find.

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Here are some reflections for continuous and differentiable $f$. It turns out that there is a unique solution if some conditions are satisfied (you can look at the end for the final answer).


  • $$ f(f(x))=ax^2+bx+c\implies f(ax^2+bx+c)=af(x)^2+bf(x)+c $$
  • If $f(x)=f(y)$, $x\neq y$ then: $$ a(x+y)=b. $$ If $a(x+y)=b$ then in any case $ff(x)=ff(y)$ which means that either $f(x)=f(y)$ or $a(f(x)+f(y))=b$. In any case there are two non-equal numbers $x'$ and $y'$ such that $f(x')=f(y')$ and they lie on two different sides of $\frac{-b}{2a}$.
  • If $f(x)=x$ then: $f(x)=ax^2+bx+c$ hence: $$ ax^2+(b-1)x+c=0\implies (b-1)^2\geq 4ac; x=\frac{-b+1\pm\sqrt{(b-1)^2-4ac} }{2a}. $$
  • $f'(ax^2+bx+c)(2ax+b)=f'(x)(2af(x)+b)$ hence: $$ x=\frac{-b}{2a}\implies f'(\frac{-b}{2a})=0\text{ or } f(\frac{-b}{2a})=\frac{-b}{2a}$$ On the other hand: $$ f'(f(x)).f'(x)=2ax+b. $$ So:

    If $f'(x)=0$ then $x=\frac{-b}{2a}$.

Note that from large enough $x$ and $y$ satisfying $a(x+y)=b$ and $f(x)=f(y)$

  • If $f(\frac{-b}{2a})=\frac{-b}{2a}$, then $f\circ f(\frac{-b}{2a})=\frac{-b}{2a}$ which means: $$ c-\frac{b^2}{4a}=-\frac b{2a}\implies b^2-4ac=2b. $$

moreover: $$ f'(f(\frac{-b}{2a})).f'(\frac{-b}{2a})=\implies f'(\frac{-b}{2a})=0. $$ If $b^2-4ac\neq 2b$, $f'(\frac{-b}{2a})=0$; so in any case:

$$ f'(\frac{-b}{2a})=0. $$

This means that the function is strictly increasing on the one side of $\frac{-b}{2a}$ and strictly decreasing on the other side.

Moreover if $f'(f(x))=0$ then $x=\frac{-b}{2a}$; so if there is $x$ such that $f(x)=\frac{-b}{2a}$ then $f'(f(x))=0$ which means that $x=\frac{-b}{2a}$. So:

$$f(\frac{-b}{2a})=\frac{-b}{2a}\implies b^2-4ac=2b$$

Without this condition there will be no answer.

  • But if $a(x+y)=b$ and $f(x)\neq f(y)$, then $a(f(x)+f(y))=b$; but this means that both of them cannot be in the same time bigger (or smaller) than $\frac{-b}{2a}$; which is a contradiction since $f'(\frac{-b}{2a})=0$ and $\frac{-b}{2a}$ is an extremum point of $f$. Hence:

    $$ f(x)=f(y), x\neq y\iff a(x+y)=b$$

Therefore it is enough to find the function for $x>\frac{-b}{2a}$.

  • See that after using the information derived above: $$ f(ax^2+bx+\frac{b^2-2b}{4a})=af(x)^2+bf(x)+\frac{b^2-2b}{4a}\implies\\ f\left(a(x+\frac{b}{2a})^2-\frac{b}{2a}\right)=a\left(f(x)+\frac{b}{2a}\right)^2-\frac{b}{2a}. $$.
    Note that if $a>0$, then $x>-\frac{b}{2a}\implies f(x)>-\frac{b}{2a}$; w.l.o.g. we assume $a>0$ and $x>-\frac b{2a}$. $$ af\left(a(x+\frac{b}{2a})^2-\frac{b}{2a}\right)+\frac{b}{2}=\left(af(x)+\frac{b}{2}\right)^2 $$ See that defining $g(x)=af(x)+\frac b2$, we get: $$ g(a(x+\frac{b}{2a})^2-\frac{b}{2a})=g(x)^2 $$ Define the function: $$ h(x)=\frac{\log g(x)}{\log(ax+\frac b2)}. $$ Then we get: $$ h(a(x+\frac{b}{2a})^2-\frac{b}{2a})=h(x) $$ This function can be shown to be a constant by constructing a decreasing sequence $u_n$ from an arbitrary $u$ and showing that $h(u)$ is equal to the limit of $u_n$ which is a constant value independent of $u$. Hence: $$ \frac{\log g(x)}{\log(ax+\frac b2)}=C\implies g(x)=(ax+\frac b2)^C\\ \implies f(x)=\frac 1a\left((ax+\frac b2)^C-\frac b2\right). $$ By plugging in to the original question, it turns out that $C=\sqrt 2$. Extending it for all $x$'s, we get:

$$ f(x)=\frac 1a\left(|ax+\frac b2|^{\sqrt 2}-\frac b2\right) \text{ if } b^2-4ac=2b. $$

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This is only a partial answer, but it might be of some help.

If $f(x)=mx+d$, then $f(f(x))=m(mx+d)+d=m^2x+(m+1)d$, so any triple of the form $(0,b,c)$ with $b\ge0$ works, by taking $m=\sqrt b$ and $d=c/(1+\sqrt b)$.

Likewise, if $f(x)=m|x|^\sqrt2$, then $f(f(x))=m|(m|x|^\sqrt2)|^\sqrt2=m|m|^\sqrt2x^2$, so any triple of the form $(a,0,0)$ works, by taking $m=sgn(a)|a|^{1/(1+\sqrt2)}$.

So it looks to me like there are two natural questions: 1) Does $a\not=0$ force $b=c=0$? and 2) are there any triples with $b\lt0$?

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    $\begingroup$ Barry, see this AOPS thread from the start for the likely beginnings of the problem artofproblemsolving.com/community/c6h14151p101406 $\endgroup$
    – Will Jagy
    Dec 31, 2015 at 18:16
  • $\begingroup$ the proof of nonexistence on the whole real line, for $(b-1)^2 > (4ac+1),$ appears valid. Not that there is always a solution from the largest fixpoint out to infinity. $\endgroup$
    – Will Jagy
    Dec 31, 2015 at 18:34
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I will divide the Case to these $3$: $$ \textbf{Case 1. }\bf{(b-1)^2-4ac>0.} \\ \textbf{Case 2. }\bf{(b-1)^2-4ac=0.} \\ \textbf{Case 3. }\bf{(b-1)^2-4ac<0.} $$

This is my solution: \begin{align} & \textbf{Case 1. } \bf{(b-1)^2-4ac > 0:} \\ \ \\ &\text{Let's think about $t_2$ which satisfies } f^2(t_2)=t_2. \\ \Rightarrow \ t_2 = \ & a{t_2}^2+bt_2+c. \Rightarrow a{t_2}^2+(b-1)t_2+c=0. \\ \ \\ \therefore \ t_2 = \ & \frac{-b+1\pm\sqrt{b^2-2b+1+4ac}}{2a}. \\ \Rightarrow \ & \text{I will let $t_2$=$\alpha_1$ and $\beta_1$.} \ \\ &\text{Now, let's think about $t_4$ which satisfies $f^4(t_4)=t_4$.} \\ \Rightarrow \ t_4 \ & = a(a{t_4}^2+bt_4+c)^2+b(a{t_4}^2+bt_4+c)+c \\ &=a^3{t_4}^4 + 2a^2b{t_4}^3+(2a^2c+ab^2+ab){t_4}^2+(2abc+b^2){t_4}+bc+c. \\ \therefore \ t_4 \ & \text{has $4$ solutions, including $t_2$.} \\ \ \\ &\text{I will let $t_4=\alpha_1, \beta_1, \alpha_2, \beta_2$.} \\ \ \\ \Rightarrow &\text{I'll define $\alpha_1$chain, $\beta_1$chain, $\alpha_2$chain and $\beta_2$chain:} \\ \ \\ \alpha_1\text{chain}: & \begin{bmatrix} \alpha_1 & \rightarrow & f(\alpha_1) \\ \uparrow & \fbox{$2_\text{cyc}$} & \downarrow \\ f(\alpha_1) &\leftarrow & \alpha_1 \end{bmatrix} \\ \ \\ \beta_1\text{chain}: & \begin{bmatrix} \beta_1 & \rightarrow & f(\beta_1) \\ \uparrow & \fbox{$2_\text{cyc}$} & \downarrow \\ f(\beta_1) &\leftarrow & \beta_1 \end{bmatrix} \\ \ \\ \alpha_2\text{chain}: & \begin{bmatrix} \alpha_2 & \rightarrow & f(\alpha_2) \\ \uparrow & \fbox{$4_\text{cyc}$} & \downarrow \\ f^3(\alpha_2) &\leftarrow & f^2(\alpha_2) \end{bmatrix} \\ \ \\ \beta_2\text{chain}: & \begin{bmatrix} \beta_2 & \rightarrow & f(\beta_2) \\ \uparrow & \fbox{$4_\text{cyc}$} & \downarrow \\ f^3(\beta_2) &\leftarrow & f^2(\beta_2) \end{bmatrix} \\ \ \\ &\text{I will let set } T_4=\{t_4\}=\{\alpha_1, \beta_1, \alpha_2, \beta_2\}. \\ \Rightarrow \ &f^n(\alpha_1), f^n(\alpha_2), f^n(\beta_1), f^n(\beta_2) \in T_4. \\ \ \\ &\text{Looking at $\alpha_2$ and $\beta_2$, you can easily show the contradiction.} \\ &\text{(just try to assume the value of $f(\alpha_2)$, for example.)} \\ \ \\ \therefore & \not\exists f: \mathbb{R} \to \mathbb{R} \text{ s.t. } f(f(x))=ax^2+bx+c, (b-1)^2-4ac>0. \end{align}

For Case 2, you can just think about 3 variables with $t_2$ and $t_4$.($n(T_4)=3, \ \exists! \ t_2.$)

For Case 3, there is a solution for few $f$.

I will let you think about Case 2. I can't answer about Case $3$, because this answer just wanted to show that $f$ satisfying the OP doesn't exist, totally.

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