Mixing
Can any diverging increasing real sequence be arbitrarily approached by an arithmetic progression?
$begingroup$
Let $(x_n)_{n in mathbb{N}}, x_nin mathbb{R^+}$ be a increasing diverging sequence.
Let $epsilon(n) : mathbb{N} to mathbb{R^+_*}$ be a decreasing strictly positive function.
I also add the condition that the $x_n$ are "spaced out".
$x_{n+1} geq x_n+epsilon(n)$
$(1)$ Does there always exist an arithmetic progression of the form
$p_n = nalpha$ with $alpha in mathbb{R}$
such as for an infinite number of $n in mathbb{N}$, $|x_n-p_n| leq epsilon(n)$.
For example if I take $x_n = n + frac{1}{n}$, and $epsilon(n) = frac{1}{2^n}$. I don't see why the progression $p_n = pi n$ could not go arbitrarily close to $x_n$ an infinite amount of times.
This stems from an exercice I was given to prove which is related to the question.
The exercice states
Let $f:mathbb{R^+}tomathbb{R^+}$ be a continuous function.
Show that
$(2)$ $(forall alpha in mathbb{R},space alpha > 0 land lim limits_{n in mathbb{N} space n to +infty} f(alpha n) to 0) implies lim limits_{x in mathbb{R} space x to +infty} f(x) to 0$
It is easy to show that this is true with uniform continuity, however for a simple continuous function I have tried to find a counterexample to give myself some intuition about why that should be true.
Let $g_{t, w}(x):mathbb{R^+}tomathbb{R^+}$ be a triangle of width $w$ and height $1$ centered around $t$.
$g_{t,w}(x) = max(0,1-frac{|x-t|}{w})$
Let $G:mathbb{R^+}tomathbb{R^+}$ $G(x) = sum limits_{n=0} limits^{+infty} g_{x_n,epsilon(n)}(x)$
Since the $x_n$ are spaced out G is continuous as a sum of continuous functions with disjoint support.
If $(1)$ is false then I can find some $G$ of this form which will satisfy the first part of the implication of $(2)$ but clearly does not converges to 0.
Therefore
$$ space space space space space lnot (1) implies lnot (2) \
Leftrightarrow (2) implies (1)
$$
But since I was asked to prove $(2)$ it should be true, therefore (1) should be true too but it looks weird since $epsilon(n)$ can get very small very fast.
real-analysis sequences-and-series continuity
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
$endgroup$
add a comment |
$begingroup$
Let $(x_n)_{n in mathbb{N}}, x_nin mathbb{R^+}$ be a increasing diverging sequence.
Let $epsilon(n) : mathbb{N} to mathbb{R^+_*}$ be a decreasing strictly positive function.
I also add the condition that the $x_n$ are "spaced out".
$x_{n+1} geq x_n+epsilon(n)$
$(1)$ Does there always exist an arithmetic progression of the form
$p_n = nalpha$ with $alpha in mathbb{R}$
such as for an infinite number of $n in mathbb{N}$, $|x_n-p_n| leq epsilon(n)$.
For example if I take $x_n = n + frac{1}{n}$, and $epsilon(n) = frac{1}{2^n}$. I don't see why the progression $p_n = pi n$ could not go arbitrarily close to $x_n$ an infinite amount of times.
This stems from an exercice I was given to prove which is related to the question.
The exercice states
Let $f:mathbb{R^+}tomathbb{R^+}$ be a continuous function.
Show that
$(2)$ $(forall alpha in mathbb{R},space alpha > 0 land lim limits_{n in mathbb{N} space n to +infty} f(alpha n) to 0) implies lim limits_{x in mathbb{R} space x to +infty} f(x) to 0$
It is easy to show that this is true with uniform continuity, however for a simple continuous function I have tried to find a counterexample to give myself some intuition about why that should be true.
Let $g_{t, w}(x):mathbb{R^+}tomathbb{R^+}$ be a triangle of width $w$ and height $1$ centered around $t$.
$g_{t,w}(x) = max(0,1-frac{|x-t|}{w})$
Let $G:mathbb{R^+}tomathbb{R^+}$ $G(x) = sum limits_{n=0} limits^{+infty} g_{x_n,epsilon(n)}(x)$
Since the $x_n$ are spaced out G is continuous as a sum of continuous functions with disjoint support.
If $(1)$ is false then I can find some $G$ of this form which will satisfy the first part of the implication of $(2)$ but clearly does not converges to 0.
Therefore
$$ space space space space space lnot (1) implies lnot (2) \
Leftrightarrow (2) implies (1)
$$
But since I was asked to prove $(2)$ it should be true, therefore (1) should be true too but it looks weird since $epsilon(n)$ can get very small very fast.
real-analysis sequences-and-series continuity
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
$endgroup$
$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26
add a comment |
$begingroup$
Let $(x_n)_{n in mathbb{N}}, x_nin mathbb{R^+}$ be a increasing diverging sequence.
Let $epsilon(n) : mathbb{N} to mathbb{R^+_*}$ be a decreasing strictly positive function.
I also add the condition that the $x_n$ are "spaced out".
$x_{n+1} geq x_n+epsilon(n)$
$(1)$ Does there always exist an arithmetic progression of the form
$p_n = nalpha$ with $alpha in mathbb{R}$
such as for an infinite number of $n in mathbb{N}$, $|x_n-p_n| leq epsilon(n)$.
For example if I take $x_n = n + frac{1}{n}$, and $epsilon(n) = frac{1}{2^n}$. I don't see why the progression $p_n = pi n$ could not go arbitrarily close to $x_n$ an infinite amount of times.
This stems from an exercice I was given to prove which is related to the question.
The exercice states
Let $f:mathbb{R^+}tomathbb{R^+}$ be a continuous function.
Show that
$(2)$ $(forall alpha in mathbb{R},space alpha > 0 land lim limits_{n in mathbb{N} space n to +infty} f(alpha n) to 0) implies lim limits_{x in mathbb{R} space x to +infty} f(x) to 0$
It is easy to show that this is true with uniform continuity, however for a simple continuous function I have tried to find a counterexample to give myself some intuition about why that should be true.
Let $g_{t, w}(x):mathbb{R^+}tomathbb{R^+}$ be a triangle of width $w$ and height $1$ centered around $t$.
$g_{t,w}(x) = max(0,1-frac{|x-t|}{w})$
Let $G:mathbb{R^+}tomathbb{R^+}$ $G(x) = sum limits_{n=0} limits^{+infty} g_{x_n,epsilon(n)}(x)$
Since the $x_n$ are spaced out G is continuous as a sum of continuous functions with disjoint support.
If $(1)$ is false then I can find some $G$ of this form which will satisfy the first part of the implication of $(2)$ but clearly does not converges to 0.
Therefore
$$ space space space space space lnot (1) implies lnot (2) \
Leftrightarrow (2) implies (1)
$$
But since I was asked to prove $(2)$ it should be true, therefore (1) should be true too but it looks weird since $epsilon(n)$ can get very small very fast.
real-analysis sequences-and-series continuity
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
$endgroup$
Let $(x_n)_{n in mathbb{N}}, x_nin mathbb{R^+}$ be a increasing diverging sequence.
Let $epsilon(n) : mathbb{N} to mathbb{R^+_*}$ be a decreasing strictly positive function.
I also add the condition that the $x_n$ are "spaced out".
$x_{n+1} geq x_n+epsilon(n)$
$(1)$ Does there always exist an arithmetic progression of the form
$p_n = nalpha$ with $alpha in mathbb{R}$
such as for an infinite number of $n in mathbb{N}$, $|x_n-p_n| leq epsilon(n)$.
For example if I take $x_n = n + frac{1}{n}$, and $epsilon(n) = frac{1}{2^n}$. I don't see why the progression $p_n = pi n$ could not go arbitrarily close to $x_n$ an infinite amount of times.
This stems from an exercice I was given to prove which is related to the question.
The exercice states
Let $f:mathbb{R^+}tomathbb{R^+}$ be a continuous function.
Show that
$(2)$ $(forall alpha in mathbb{R},space alpha > 0 land lim limits_{n in mathbb{N} space n to +infty} f(alpha n) to 0) implies lim limits_{x in mathbb{R} space x to +infty} f(x) to 0$
It is easy to show that this is true with uniform continuity, however for a simple continuous function I have tried to find a counterexample to give myself some intuition about why that should be true.
Let $g_{t, w}(x):mathbb{R^+}tomathbb{R^+}$ be a triangle of width $w$ and height $1$ centered around $t$.
$g_{t,w}(x) = max(0,1-frac{|x-t|}{w})$
Let $G:mathbb{R^+}tomathbb{R^+}$ $G(x) = sum limits_{n=0} limits^{+infty} g_{x_n,epsilon(n)}(x)$
Since the $x_n$ are spaced out G is continuous as a sum of continuous functions with disjoint support.
If $(1)$ is false then I can find some $G$ of this form which will satisfy the first part of the implication of $(2)$ but clearly does not converges to 0.
Therefore
$$ space space space space space lnot (1) implies lnot (2) \
Leftrightarrow (2) implies (1)
$$
But since I was asked to prove $(2)$ it should be true, therefore (1) should be true too but it looks weird since $epsilon(n)$ can get very small very fast.
real-analysis sequences-and-series continuity
real-analysis sequences-and-series continuity
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
asked Jan 31 at 16:21
Pâris DouadyPâris Douady
11
11
$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26
add a comment |
$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26
$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26
add a comment |
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$begingroup$
How about the sequence of factorials?
$endgroup$
– kimchi lover
Jan 31 at 16:55
$begingroup$
For your exam question, take a look at this: math.stackexchange.com/questions/3065572/… .
$endgroup$
– Mindlack
Jan 31 at 17:05
$begingroup$
@kimchi lover The progression p(n) = 1n covers infinite number of terms of x_n = n! Which you are proposing.
$endgroup$
– Pâris Douady
Jan 31 at 17:23
$begingroup$
@Mindlack thank you for your link, it's indeed the same problem.
$endgroup$
– Pâris Douady
Jan 31 at 17:26