Computing functions

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#1 h May 18, 2025, 5:25 PM

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Let $f : [0, \infty) \to [0, \infty)$ , $f(x) = \dfrac{ax + b}{cx + d}$ , with $a, d \in (0, \infty)$ , $b, c \in [0, \infty)$ . Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
$f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.$ (For $n \in \mathbb{N}^*$ and $x \geq 0$ , the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$ . )

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#2 h May 18, 2025, 7:41 PM

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Here's a sketch of a method which solves every single mobius tranform problem I saw.
Let $z_1,z_2$ be the two complex roots of $f(z)=z$ . Then, since a mobius transform is just a combination of shifting,scaling,rotating, and inversion, for any complex number $z$ we have:
$(z , \infty ; z_1,z_2) = (f(z) , \frac{a}{c} ; z_1,z_2)$ If you write this $n$ times you'd get:
$k^n .\frac{z-z_1}{z-z_2} = \frac{f_n - z_1}{f_n - z_2}$ Where $k = \frac{a/c - z_1}{a/c - z_2}$ .Now let $f_n(x) = \frac{x}{1 + nx}$ for some $n$ . One can easily get $k^n=1$ (by comparing the coefficient of $x$ in the respective polynomial identity) and so $x_1=x_2$ (comparing $x^2$ s).
Now, $x_1=x_2$ means we have a double root for $f(x)=x$ and delta=0 so $(d-a)^2+4bc=0$ . Combining with the fact that $x_1,x_2$ are fix points of every $f_k$ , we'll get $(n-1)^2+0=0$ and $n=1$

This post has been edited 2 times. Last edited by alinazarboland, May 18, 2025, 7:42 PM

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172 posts

#3 h May 18, 2025, 7:47 PM

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Here are two old problems one from $2012$ IMC and one from Iranian Olympiad which are trivial with this method
https://artofproblemsolving.com/community/c7h491145p2754513
https://artofproblemsolving.com/community/c6h368215p2026678

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#4 h May 18, 2025, 10:14 PM

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alinazarboland wrote:

Here's a sketch of a method which solves every single mobius tranform problem I saw.
Let $z_1,z_2$ be the two complex roots of $f(z)=z$ . Then, since a mobius transform is just a combination of shifting,scaling,rotating, and inversion, for any complex number $z$ we have:
$(z , \infty ; z_1,z_2) = (f(z) , \frac{a}{c} ; z_1,z_2)$ If you write this $n$ times you'd get:
$k^n .\frac{z-z_1}{z-z_2} = \frac{f_n - z_1}{f_n - z_2}$ Where $k = \frac{a/c - z_1}{a/c - z_2}$ .Now let $f_n(x) = \frac{x}{1 + nx}$ for some $n$ . One can easily get $k^n=1$ (by comparing the coefficient of $x$ in the respective polynomial identity) and so $x_1=x_2$ (comparing $x^2$ s).
Now, $x_1=x_2$ means we have a double root for $f(x)=x$ and delta=0 so $(d-a)^2+4bc=0$ . Combining with the fact that $x_1,x_2$ are fix points of every $f_k$ , we'll get $(n-1)^2+0=0$ and $n=1$

Thank you very much, i appreciate this solution ! I can understand it, but i need a 9th grade solution. I solved the ''reciprocal'' implication by induction, now i need to demonstrate the ''direct'' one. Can you or anyone help me ?

This post has been edited 1 time. Last edited by BBNoDollar, May 18, 2025, 10:15 PM

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#5 h May 19, 2025, 12:01 PM

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Bumping this

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#6 h May 19, 2025, 9:00 PM

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Bump. Bump

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#7 h May 20, 2025, 4:28 PM

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BUMPING for 9th grade solution

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#8 h May 23, 2025, 3:50 PM

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Someone for this?

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wh0nix

#9 h May 24, 2025, 10:01 AM

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Hint

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