One limes

[SN54]

Let f be a function which satisfied following conditions:

1) f is of class C^1(ℝ), and ∣f'(x)∣< 1, ∀ x ∊ ℝ
2) f(x+1)=f(x), ∀ x ∊ ℝ (f is periodical with basic period ω=1)

Let p be a function defined with, p(x)=x+f(x), ∀ x ∊ ℝ

Prove that
$\displaystyle \lim_{n\to\infty}{\frac {x+p(x)+p(p(x))+...+p(p(...p(x)...))}{n^2} }$ (composition in last term is taken n-1 times)
exists, and doesn't depend on x. ...............................Latex edited

P.S. This is my first post here, I don't know how to tape this quotient. I ask admin, please, to type for me this correctly, just this time. (And delee this P.S.)
Thanks in advance

 

[AvgStudent]

Let $f$ be a function which satisfied the following conditions:

1) $f$ is of class $C^1(\R)$, and $|f'(x)| <1, \forall x \in \R$
2) $f(x+1)=f(x), \forall x \in \R$ ($f$ is periodical with basic period $\omega=1$)

Prove that:
$$\lim_{n\to\infty}{\frac {x+p(x)+p(p(x))+...+p(p(...p(x)...))}{n^2} }$$(Composition in last term is taken $n-1$ times) exists, and doesn't depend on $x$.

PS: I'm only here for the typesetting. You can hit "reply" to see the LaTeX code. Good day

 

[topsquark]

\lim_{n\to\infty}{\frac {x+p(x)+p(p(x))+...+p(p(...p(x)...))}{/n^2} } (composition in last term is taken n-1 times)

The code look good. Wrap the code in [ math ] tags.

-Dan

 

[blamocur]

I haven't yet solved it myself, but my intuition tells me that $|f(x)|$ must be limited.

 

[blamocur]

I believe I have proven it. My intuition about $|f(x)|$ is correct and relevant. There are also some properties of $p(x)$ which are easy to prove and which are relevant to the solution. Can you work with these hints?

 

[SN54]

Let $f$ be a function which satisfied the following conditions:

1) $f$ is of class $C^1(\R)$, and $|f'(x)| <1, \forall x \in \R$
2) $f(x+1)=f(x), \forall x \in \R$ ($f$ is periodical with basic period $\omega=1$)

Prove that:
$$\lim_{n\to\infty}{\frac {x+p(x)+p(p(x))+...+p(p(...p(x)...))}{n^2} }$$(Composition in last term is taken $n-1$ times) exists, and doesn't depend on $x$.

PS: I'm only here for the typesetting. You can hit "reply" to see the LaTeX code. Good day

 

[SN54]

This problem trouble me long time. I just proved, if $lim$ exist , he is independent on $x$.
I supose, result should to be $\frac{1}{2}f(0)$, but I don't know how to attack problem.
My thanks to AvgStudent.

​

 

[blamocur]

This problem trouble me long time. I just proved, if $lim$ exist , he is independent on $x$.
I supose, result should to be $\frac{1}{2}f(0)$, but I don't know how to attack problem.
My thanks to AvgStudent.

​

I've also proven that the limit is independent of $x$. I do not believe it is equal to $\frac{f(0)}{2}$, but I can prove that it lies in the $\left[\frac{f(0)}{2}-\frac{1}{4}, \frac{f(0)}{2}+\frac{1}{4} \right]$ interval.
And contrary to my earlier post I haven't actually proven yet that the limit exists

 

[SN54]

I think it should use $L (o) <L(x) <L(1)$ and $L (o) = L (1)$, and it''s obvious if [imath]f(0)/imath])\guad is \quad an \guad integer \guad that [imath]L (o)=\frac{1}{2}f(0)[/imath].

 

[blamocur]

Since I haven't been able to prove the convergence I'm posting what I have been able to prove. There is a small (yeah, right) chance that I've made a mistake somewhere which kept me from completing the problem. My apologies for a long post, and feel free to criticize and ridicule:

Lemma 1. $|f(x)-f(0)| \leq \frac{1}{2}$. Indeed, if this is not true for some
$x_1$ we can safely assume (because $f$ is periodic) that $0\leq x_1 < 1$.
But then $|x_1-b| \leq 1/2$ where $b=0$ or $b=1$. Note that $f(b) = f(0)$.

By taking absolute values of both sides in the Mean Value Theorem we conclude that there
must be a point $x_2$ between $x_1$ and $b$
such that
$$|f^\prime(x_2)| = \frac{|f(x_1)-f(b))|}{|x_1-b|}$$But in the above fraction the numerator is assumed to be larger than 1/2 and the
denominator is smaller than 1/2, which means that $f^\prime(x_2) > 1$, which contradicts
the problem's statement. $\blacksquare$

Using $c=f(0)-\frac{1}{2}$ we can write for all $x$:
$$c \leq f(x) \leq c+1$$and from the definition $p(x)=x+f(x)$ we now know that:
$$x + c \leq p(x) \leq x + c + 1$$Using $p_0(x) = x$, $p_1(x) = p(x)$, $p_2(x) = p(p(x)), ... ,p_n(x) = p (p_{n-1}(x)$ we have:
$$x+2c \leq p_2(x) \leq x+2c+2$$and by induction
$$x + kc \leq p_k(x) \leq x+k(c + 1)$$If $S_n(x) = \sum_{k=0}^n p_k(x)$ then
$$nx + \frac{n^2+n}{2}c \leq S_n(x)\leq nx + \frac{n^2+n}{2}(c+1)$$and switching from $n$ to $n-1$ we get
$$(n-1)x + \frac{n^2-n}{2}c \leq S_{n-1}(x)\leq (n-1)x + \frac{n^2-n}{2}(c+1)$$$$x\frac{n-1}{n^2} + \frac{n^2-n}{2n^2}c \leq \frac{S_{n-1}(x)}{n^2} \leq x\frac{n-1}{n^2} + \frac{n^2-n}{2n^2}(c+1)$$Both the left and the right expressions converge which means that if the limit of the expression in the middle exists then
$$\frac{2f(0)-1}{4} \leq \lim_{n\rightarrow\infty} \frac{\sum_{k=0}^{n-1} p_k(x)}{n^2} \leq \frac{2f(0)+1}{4}$$

 

[SN54]

If $L_n(x)=\frac{x+p(x)+\cdots+p(p(\cdots p(x)\cdots)}{n^2}$
your estimation prove that if $L_n\leq L_{n+1}, \ \text{for all}\ n_0\leq n$, all is proven. I was maybe wrong in attempt to establish exact value of $\lim_{n\to\infty}$.

 

[blamocur]

If $L_n(x)=\frac{x+p(x)+\cdots+p(p(\cdots p(x)\cdots)}{n^2}$
your estimation prove that if $L_n\leq L_{n+1}, \ \text{for all}\ n_0\leq n$, all is proven. I was maybe wrong in attempt to establish exact value of $\lim_{n\to\infty}$.

I don't see why $L_n\leq L_{n+1}$. In fact, my crude simulations did not show consistent monotonicity in either direction

 

[SN54]

My first thinking was, if $\lim_{n\to\infty}L_n(x)$ exists, and does not depend on $x$, then that must be some concrete value of $f(x)$, and seem to be it's logical to be $\frac12 f(0)$, while this is true when $f(0)\in Z$. If $f(0)\notin Z\ \text{then}\ f(0)=m+\alpha\text{\ for some}\ \alpha\in (0,1)\ \text{and}\ m\in Z$, and here became comlicated. Now, since $-x\le f(x)\le x$, this means $m=0,$, so $f(0)=\alpha$.

 

[SN54]

One exemple of function $f$ which satisfies both conditions 1) and 2) can be
$f(x)=\alpha+\beta \sin2\pi x$
where, $\alpha\in R, 0<\beta <\frac{1}{2\pi}$ are arbitrary.

 

[SN54]

I have another sugestion, to use Stolz–Cesàro theorem.
"It can be solved by applying the Stolz theorem, which reduces the problem to the definition of the Poincaré rotation number. If you search for that term on the Internet, you will find many sources in which your task is actually solved in that formulation, which it boils down to after applying Stolz's theorem".

Stolz–Cesàro theorem | Brilliant Math & Science Wiki

Stolz–Cesàro theorem is a powerful tool for evaluating limits of sequences, and it is a discrete version of L'Hôpital's rule. Reveal the answer ...

brilliant.org

Rotation number - Wikipedia

en.wikipedia.org