Nice "if and only if" function problem

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Nice "if and only if" function problem

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22 posts

#1 h May 23, 2025, 7:23 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 $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$ . )

Please do it at 9th grade level. Thank you!

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#2 h May 24, 2025, 9:35 AM

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Please someone

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wh0nix

#3 h May 24, 2025, 9:52 AM

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Hint

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#4 h May 24, 2025, 1:00 PM

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wh0nix wrote:

Hint

Thank you but I need a 9th grade level solution. I did an induction for the "<=" part, but for "=>" i don t have idea what to do

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#5 h May 24, 2025, 3:15 PM

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Please help

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wh0nix

#6 h May 24, 2025, 4:35 PM

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Ok, here it is a solution by induction:
${{f}_{2}}(x=f(f(x))=\frac{f(x)}{1+f(x)}=\frac{x}{1+2x}$ . Suppose that ${{f}_{n}}(x)=\frac{x}{1+nx}$ we have to prove that ${{f}_{n+1}}(x)=\frac{x}{1+(n+1)x}$ . Indeed ${{f}_{n+1}}(x)={{f}_{n}}(f(x))=\frac{f(x)}{1+nf(x)}=\frac{x}{1+(n+1)x}.$

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#7 h May 24, 2025, 5:26 PM

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wh0nix wrote:

Ok, here it is a solution by induction:
${{f}_{2}}(x=f(f(x))=\frac{f(x)}{1+f(x)}=\frac{x}{1+2x}$ . Suppose that ${{f}_{n}}(x)=\frac{x}{1+nx}$ we have to prove that ${{f}_{n+1}}(x)=\frac{x}{1+(n+1)x}$ . Indeed ${{f}_{n+1}}(x)={{f}_{n}}(f(x))=\frac{f(x)}{1+nf(x)}=\frac{x}{1+(n+1)x}.$

Yes, but this is only the reciprocal implication ! He needs help with the direct one! (the one where i know the value of f_n , and i need to show the value of f).

This post has been edited 1 time. Last edited by BBNoDollar, May 24, 2025, 5:26 PM

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#8 h May 24, 2025, 6:59 PM

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Please help me guys

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#9 h May 24, 2025, 10:06 PM

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PLEASE PLS

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#10 h May 24, 2025, 10:43 PM

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am i tweaking or does $n=2$ just work
since $f(x)=\frac{x}{1+x}$ should be the only solution to $f(f(x))=\frac{x}{1+2x}$
@below ohhhhhh

This post has been edited 1 time. Last edited by aidan0626, May 24, 2025, 10:52 PM

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#11 h May 24, 2025, 10:49 PM • 1 Y

Y by aidan0626

problem statment is confusing, it's $(\exists n \quad (\forall x \quad f_n(x) = \frac{x}{1+nx} )) \iff (\forall x \quad f(x) = \frac{x}{1+x})$ not $\exists n \quad ((\forall x \quad f_n(x) = \frac{x}{1+nx}) \iff (\forall x \quad f(x) = \frac{x}{1+x}))$

This post has been edited 2 times. Last edited by maromex, May 24, 2025, 10:52 PM

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27 posts

wh0nix

#12 h May 25, 2025, 6:26 AM

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The reciprocal proof is more difficult! Here is my attempt:
Let ${{f}_{k}}(x)=\frac{{{a}_{k}}x+{{b}_{k}}}{{{c}_{k}}x+{{d}_{k}}}\Rightarrow {{f}_{k+1}}(x)=\frac{{{a}_{k+1}}x+{{b}_{k+1}}}{{{c}_{k+1}}x+{{d}_{k+1}}}$ but ${{f}_{k+1}}(x)={{f}_{k}}(f(x))$ hence result ${{f}_{k+1}}(x)=\frac{({{a}_{k}}a+{{b}_{k}}c)x+{{b}_{k}}d+{{a}_{k}}b}{\left( {{c}_{k}}a+{{d}_{k}}c \right)x+{{d}_{k}}d+{{c}_{k}}b}=\frac{x}{nx+1},\forall x\ge 0$ result $A{{x}^{2}}+Bx+C=0,\forall x\ge 0\Leftrightarrow A=B=C=0$ and result
${{a}_{k}}b+{{b}_{k}}d=0$ , $a(n{{a}_{k}}-{{c}_{k}})=c({{d}_{k}}-n{{b}_{k}})$ , $a{{c}_{k}}+c{{d}_{k}}=b{{c}_{k}}+d{{d}_{k}}$ and we must solve this system in $a,b,c,d$

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#13 h May 25, 2025, 7:01 AM

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wh0nix wrote:

The reciprocal proof is more difficult! Here is my attempt:
Let ${{f}_{k}}(x)=\frac{{{a}_{k}}x+{{b}_{k}}}{{{c}_{k}}x+{{d}_{k}}}\Rightarrow {{f}_{k+1}}(x)=\frac{{{a}_{k+1}}x+{{b}_{k+1}}}{{{c}_{k+1}}x+{{d}_{k+1}}}$ but ${{f}_{k+1}}(x)={{f}_{k}}(f(x))$ hence result ${{f}_{k+1}}(x)=\frac{({{a}_{k}}a+{{b}_{k}}c)x+{{b}_{k}}d+{{a}_{k}}b}{\left( {{c}_{k}}a+{{d}_{k}}c \right)x+{{d}_{k}}d+{{c}_{k}}b}=\frac{x}{nx+1},\forall x\ge 0$ result $A{{x}^{2}}+Bx+C=0,\forall x\ge 0\Leftrightarrow A=B=C=0$ and result
${{a}_{k}}b+{{b}_{k}}d=0$ , $a(n{{a}_{k}}-{{c}_{k}})=c({{d}_{k}}-n{{b}_{k}})$ , $a{{c}_{k}}+c{{d}_{k}}=b{{c}_{k}}+d{{d}_{k}}$ and we must solve this system in $a,b,c,d$

This isn't ok, i dont know why you came up with ak, bk, ck ,dk. Please someone

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#14 h May 25, 2025, 7:07 AM

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Let $f : [0, \infty) \to [0, \infty)$ be defined by:
$f(x) = \frac{ax + b}{cx + d}, \quad \text{where } a, d > 0 \text{ and } b, c \geq 0.$ We need to prove:
$\text{There exists } n \in \mathbb{N}^* \text{ such that } f_n(x) = \frac{x}{1 + nx} \quad \text{if and only if} \quad f(x) = \frac{x}{1 + x}.$ First, check that composing $f(x) = \frac{x}{1 + x}$ with itself $n$ times gives $\frac{x}{1 + nx}$ .
For $n = 1$ : $f_1(x) = f(x) = \frac{x}{1 + x}.$ For $n = 2$ : $f_2(x) = f(f(x)) = \frac{\frac{x}{1 + x}}{1 + \frac{x}{1 + x}} = \frac{x}{(1 + x) + x} = \frac{x}{1 + 2x}.$ By induction: Assume $f_k(x) = \frac{x}{1 + kx}$ for some $k \geq 1$ . Then:
$f_{k+1}(x) = f(f_k(x)) = \frac{\frac{x}{1 + kx}}{1 + \frac{x}{1 + kx}} = \frac{x}{(1 + kx) + x} = \frac{x}{1 + (k+1)x}.$ Suppose $f(x) = \frac{ax + b}{cx + d}$ satisfies $f_n(x) = \frac{x}{1 + nx}$ for some $n$ . We show $f(x) = \frac{x}{1 + x}$ .
Behavior at $x = 0$ : $f(0) = \frac{b}{d} \quad \text{and} \quad f_n(0) = \frac{0}{1 + n \cdot 0} = 0.$ Since $f_n(0) = 0$ , repeated composition forces $f(0) = 0$ . Thus, $b = 0$ .
Simplify $f(x)$ : Now $f(x) = \frac{ax}{cx + d}$ . Let’s write it as:
$f(x) = \frac{kx}{mx + 1} \quad \text{where } k = \frac{a}{d}, \ m = \frac{c}{d}.$ Compose $f$ twice: Compute $f_2(x) = f(f(x))$ :
$f_2(x) = \frac{k \cdot \frac{kx}{mx + 1}}{m \cdot \frac{kx}{mx + 1} + 1} = \frac{k^2 x}{(mk + m)x + 1}.$ For $f_n(x) = \frac{x}{1 + nx}$ , we need:
$\frac{k^2 x}{(mk + m)x + 1} = \frac{x}{1 + 2x} \implies k^2 = 1 \text{ and } mk + m = 2.$ Solve the equations: From $k^2 = 1$ and $k > 0$ , we get $k = 1$ . Then:
$m(1) + m = 2 \implies 2m = 2 \implies m = 1.$ Thus, $f(x) = \frac{x}{x + 1}$ .
The only function satisfying $f_n(x) = \frac{x}{1 + nx}$ is:
$\boxed{f(x) = \frac{x}{1 + x}}$

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27 posts

wh0nix

#15 h May 25, 2025, 7:31 AM

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ICE_CNME_4 wrote:

wh0nix wrote:

The reciprocal proof is more difficult! Here is my attempt:
Let ${{f}_{k}}(x)=\frac{{{a}_{k}}x+{{b}_{k}}}{{{c}_{k}}x+{{d}_{k}}}\Rightarrow {{f}_{k+1}}(x)=\frac{{{a}_{k+1}}x+{{b}_{k+1}}}{{{c}_{k+1}}x+{{d}_{k+1}}}$ but ${{f}_{k+1}}(x)={{f}_{k}}(f(x))$ hence result ${{f}_{k+1}}(x)=\frac{({{a}_{k}}a+{{b}_{k}}c)x+{{b}_{k}}d+{{a}_{k}}b}{\left( {{c}_{k}}a+{{d}_{k}}c \right)x+{{d}_{k}}d+{{c}_{k}}b}=\frac{x}{nx+1},\forall x\ge 0$ result $A{{x}^{2}}+Bx+C=0,\forall x\ge 0\Leftrightarrow A=B=C=0$ and result
${{a}_{k}}b+{{b}_{k}}d=0$ , $a(n{{a}_{k}}-{{c}_{k}})=c({{d}_{k}}-n{{b}_{k}})$ , $a{{c}_{k}}+c{{d}_{k}}=b{{c}_{k}}+d{{d}_{k}}$ and we must solve this system in $a,b,c,d$

This isn't ok, i dont know why you came up with ak, bk, ck ,dk. Please someone

Just because you don't know why I used ak, bk, ck, dk doesn't mean it's not correct. You just don't understand my attempt, which is very correct, it just needs to be done. You should be more polite to those who are trying to help you, because if you do, they may not help you anymore!

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