Trivial Functional Equation

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Trivial Functional Equation

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#1 h May 10, 2021, 8:05 PM • 2 Y

Y by samrocksnature, HWenslawski

For all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ , solve the functional equation $f(f(n))+f(n)=2$ .

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#2 h May 10, 2021, 8:18 PM • 7 Y

Y by samrocksnature, HWenslawski, bubbletea070821, megarnie, Mango247, Mango247, Mango247

some solutions

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#3 h May 10, 2021, 8:22 PM • 4 Y

Y by samrocksnature, Jc426, mathleticguyyy, Mango247

1 soln.

glory to god
-tony BARKER

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#4 h May 10, 2021, 8:29 PM • 1 Y

Y by samrocksnature

Why can't you just set $f(n)=x$ and get $f(x)+x=2 \implies f(x)=2-x$ ?

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#5 h May 10, 2021, 8:31 PM • 4 Y

Y by samrocksnature, mira74, Mango247, Mango247

OlympusHero wrote:

Why can't you just set $f(n)=x$ and get $f(x)+x=2 \implies f(x)=2-x$ ?

We don't know the range of $f(n)$ , so we can't say $f(x)=2-x$ for all $x \in \mathbb{R}$ . (Might be wrong though.)

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#6 h May 10, 2021, 8:32 PM • 1 Y

Y by samrocksnature

OlympusHero wrote:

Why can't you just set $f(n)=x$ and get $f(x)+x=2 \implies f(x)=2-x$ ?

I don't think that works because not every value of $x$ is in the range of $f$ .

EDIT: sniped, but the previous user had the same idea, so it's probably correct

This post has been edited 1 time. Last edited by Doudou_Chen, May 10, 2021, 8:34 PM

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#8 h May 10, 2021, 9:06 PM • 1 Y

Y by tenebrine

The only solution over $\mathbb N$ (since the problem you got this from is over $\mathbb N$ ) is just $f\equiv1$ since $f\ge1$ .

If $A\subseteq\mathbb R$ , then $f(x)=\begin{cases}2-x&\text{if }x\in A\cup\{2-a|a\in A\}\\1&\text{otherwise}\end{cases}$ works.

This post has been edited 1 time. Last edited by jasperE3, May 10, 2021, 10:01 PM

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#9 h Jun 23, 2022, 12:13 PM

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OlympusHero wrote:

For all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ , solve the functional equation $f(f(n))+f(n)=2$ .

Either "find all" or "construct one"

OlympusHero wrote:

Why can't you just set $f(n)=x$ and get $f(x)+x=2 \implies f(x)=2-x$ ?

Prove that it's surjective

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#10 h Mar 31, 2025, 6:21 AM

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OlympusHero wrote:

For all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ , solve the functional equation $f(f(n))+f(n)=2\forall n\in\mathbb R$ .

General form should be this I think
Choose $A$ to be an arbitrary subset of $\mathbb R$ and define $B=A\cup\{2-a\mid a\in A\}$ . Then choose $g:\mathbb R\setminus B\to B$ arbitrarily, we have:
$f(x)=\begin{cases}2-x&\text{if }x\in B\\g(x)&\text{if }x\notin B\end{cases}.$

This post has been edited 1 time. Last edited by jasperE3, Mar 31, 2025, 6:21 AM

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