Matematické Fórum

stuart clark · 05. 03. 2015 05:46

If $f(x) = x - \lfloor x \rfloor .$ Then no. of real solution of $f(f(x)) = 0$ and $f(f(f(x))) = 0$ and $f(f(f(f(x)))) = 0$

Where $0 \leq x<1$ and $\lfloor x \rfloor$ is floor function of $x$

byk7 · 05. 03. 2015 10:17

Is that correct? Because $0 \leq x<1\Rightarrow \lfloor x \rfloor =0$ so $f(x) = x .$

Freedy · 08. 03. 2015 02:27

If $f(x)=x-\lfloor x\rfloor$ , then
$f(f(x))=x-\lfloor x\rfloor-\underbrace{\lfloor x-\lfloor x\rfloor\rfloor}_{0}=x-\lfloor x\rfloor=f(x)$ so I think I don't understand the question as well.

↑ byk7:
If $\{x\}=x-\lfloor x\rfloor$ , then $\forall x\in \mathbb{R}$ is $0\le \{x\}<1$

byk7 · 09. 03. 2015 22:53

↑ Freedy:

Since $0 \leq x<1$ we have $\lfloor x \rfloor =0$ and thus $f(x)=x-\lfloor x\rfloor=x-0=x$ .
So we get $f(x)=f^2(x)=f^3(x)=f^4(x)=x$ where the upper index means no. of iterations.
And every equation $f^k(x)=0,k\in\{2,3,4\}$ has the only root $x=0$ .

This seems too easy, and that's why I am asking about corectness of the problem.

stuart clark · 10. 03. 2015 04:09

Sorry ↑ byk7: . Now i have edited my question.

If $f(x) = \frac{3x}{2} - \lfloor \frac{3x}{2} \rfloor .$ Then no. of real solution of $f(f(x)) = 0$ and $f(f(f(x))) = 0$ and $f(f(f(f(x)))) = 0$

Where $0 \leq x<1$ and $\lfloor x \rfloor$ is floor function of $x$

Freedy · 10. 03. 2015 07:28 — Editoval Freedy (14. 03. 2015 13:15)

Hi,

for $x\in \langle0;\frac{2}{3})$ we have
$f(x)=\frac{3x}{2}-0=\frac{3x}{2}$
$f(f(x))=\frac{3(\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor)}{2}-\frac{3\lfloor \frac{3x}{2}-\lfloor\frac{3x}{2} \rfloor \rfloor}{2}=\frac{9x}{4}$
$f(f(f(x)))=\frac{9(\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor)}{4}=\frac{27x}{8}$ .
All of these functions (and others) have only one real solution x = 0.

for
$x\in \langle\frac{2}{3};1)$ we have:
$f(x)=\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor=\frac{3x}{2}-1=\frac{3x-2}{2}$ one solution x = 2/3 but it's not in the interval
$f(f(x))=\frac{3(\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor)}{2}-\lfloor \frac{3(\frac{3x}{2}-\lfloor \frac{3x}{2} \rfloor)}{2} \rfloor=\frac{3(\frac{3x}{2}-1)}{2}=\frac{9x-6}{4}$
$f(f(f(x)))=\frac{27x-18}{8}$ ...
We can find for $f(x),f(f(x)),f(f(f(x)))$ real solution $x=\frac{2}{3}$ and this solution is from the interval $\langle\frac{2}{3};1)$ , therefore there are 2 real solutions for any $f(x),f(f(x)),f(f(f(x)))...$

Freedy

Kdosi · 11. 03. 2015 15:38 — Editoval Kdosi (11. 03. 2015 15:39)

Hi, after your respons ↑ Freedy: I'm not sure of my solution, so if U'll find mistake, let me know.
First: $f^{n}(x)$ is function $f(x)$ after $n$ itteration.
If $f(x) = \frac{3x}{2} - \lfloor \frac{3x}{2} \rfloor$ and $0 \leq x<1$ . $f(x)=0 \Leftrightarrow x=0 \vee x=\frac{2}{3}$ . It's easy to show there are 4 solutions for $f^{2}(x)=0$ .
I think there is $2^n$ solutions for $f^{n}(x)=0$ . I'll prove it by induction.
For n=1 it's correct.
Then, for $n=l+1$ : $f^{l+1}(x)=0 \Leftrightarrow f^{l}(x)=0 \vee f^{l}(x)=\frac{2}{3}$ . So number of solutions is eaqual to $2^l+2^l=2\cdot 2^l=2^{l+1}$ .
I know, I should show there is x in R, such, that $f^{l}(x)=0 \vee f^{l}(x)=\frac{2}{3}$ for all $l$ . But I realized it just when I was writting this and I dont have proved it yet.
Now it's really easy to find answer on↑ stuart clark:'s question.

My apologies, if I'm wrong.