How often will $a_{n+1}=|a_n|^psin a_n$ tend to $0$?
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Consider sequence $a_1=x$, $a_{n+1}=|a_n|^psin a_n$. Denote $S$ the set $Big{x Bigvert limlimits_{ntoinfty}a_n=0Big}$, whats the value of $$lim_{xtoinfty}frac1{2x}mu([-x,x]cap S)?$$
(where $mu(M)$ denotes the measure of $M$)
I tried starting with $p=2$. I plotted the graph of $ln |a_n|$ with many $x$. It seems that almost every $x$ will make the limit tends to $0$.
For $0le p< 1$, I can prove that the limit is $1$. All of the $x$ will make $limlimits_{ntoinfty}a_n$ tend to $0$.
Also, I can calculate that for $p=1$ the limit we want to find is also $1$.
sequences-and-series convergence
asked Nov 14 at 7:51
Kemono Chen
1,601330
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up vote
3
down vote
favorite
Consider sequence $a_1=x$, $a_{n+1}=|a_n|^psin a_n$. Denote $S$ the set $Big{x Bigvert limlimits_{ntoinfty}a_n=0Big}$, whats the value of $$lim_{xtoinfty}frac1{2x}mu([-x,x]cap S)?$$
(where $mu(M)$ denotes the measure of $M$)
I tried starting with $p=2$. I plotted the graph of $ln |a_n|$ with many $x$. It seems that almost every $x$ will make the limit tends to $0$.
For $0le p< 1$, I can prove that the limit is $1$. All of the $x$ will make $limlimits_{ntoinfty}a_n$ tend to $0$.
Also, I can calculate that for $p=1$ the limit we want to find is also $1$.
sequences-and-series convergence
asked Nov 14 at 7:51
Kemono Chen
1,601330
Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider sequence $a_1=x$, $a_{n+1}=|a_n|^psin a_n$. Denote $S$ the set $Big{x Bigvert limlimits_{ntoinfty}a_n=0Big}$, whats the value of $$lim_{xtoinfty}frac1{2x}mu([-x,x]cap S)?$$
(where $mu(M)$ denotes the measure of $M$)
I tried starting with $p=2$. I plotted the graph of $ln |a_n|$ with many $x$. It seems that almost every $x$ will make the limit tends to $0$.
For $0le p< 1$, I can prove that the limit is $1$. All of the $x$ will make $limlimits_{ntoinfty}a_n$ tend to $0$.
Also, I can calculate that for $p=1$ the limit we want to find is also $1$.
sequences-and-series convergence
asked Nov 14 at 7:51
Kemono Chen
1,601330
Consider sequence $a_1=x$, $a_{n+1}=|a_n|^psin a_n$. Denote $S$ the set $Big{x Bigvert limlimits_{ntoinfty}a_n=0Big}$, whats the value of $$lim_{xtoinfty}frac1{2x}mu([-x,x]cap S)?$$
(where $mu(M)$ denotes the measure of $M$)
I tried starting with $p=2$. I plotted the graph of $ln |a_n|$ with many $x$. It seems that almost every $x$ will make the limit tends to $0$.
For $0le p< 1$, I can prove that the limit is $1$. All of the $x$ will make $limlimits_{ntoinfty}a_n$ tend to $0$.
Also, I can calculate that for $p=1$ the limit we want to find is also $1$.
sequences-and-series convergence
sequences-and-series convergence
asked Nov 14 at 7:51
Kemono Chen
1,601330
asked Nov 14 at 7:51
Kemono Chen
1,601330
edited Nov 14 at 8:21
asked Nov 14 at 7:51
Kemono Chen
1,601330
asked Nov 14 at 7:51
Kemono Chen
1,601330
asked Nov 14 at 7:51
Kemono Chen
1,601330
1,601330
Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30
add a comment |
Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30
Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30
add a comment |
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Do you require $p>0$ ?
– Gabriel Romon
Nov 14 at 8:27
No. When $p<-1$ the limit I want to find can be easily found as $0$.
– Kemono Chen
Nov 14 at 8:53
When $p>1$, $a_n to 0$ iff $|a_n| < 1$ for some $n$. Therefore, it suffices to show $cup_{n ge 0} f^{-n}(E)$ has full measure, where $f(x) = x^2sin(x)$ and $E = {x in mathbb{R} : x^2|sin(x)| > 1}$. Note that $E$ looks like $cup_{k in mathbb{Z}} [2pi k - epsilon_k, 2pi k+epsilon_k]$ for $epsilon_k sim frac{1}{4pi^2 k^2}$.
– mathworker21
Nov 15 at 15:30