Prove that permutation of a sequence does not affect convergence.
$begingroup$
Given a sequence ${y_k}$ is obtained by permuting the sequence ${x_n}$ and:
$$
{forall nin mathbb N exists k_n in mathbb N:x_n = y_{k_n}, n_1 ne n_2 implies k_{n_1} ne k_{n_2}}
$$
and:
$$
{forall kin mathbb N exists n_k in mathbb N:y_k = x_{n_k}, k_1 ne k_2 implies n_{k_1} ne n_{k_2}}
$$
Prove that:
$$
lim_{nto infty}x_n = a implies lim_{kto infty}y_k=a
$$
I've been thinking of the following. Let there exist a bijection $P:mathbb Nto mathbb N$ which converts $k$ to $n_k$. So basically we have that if $x_n$ is convergent to some $a$ then:
$$
|x_n - a| < varepsilon
$$
Thus :
$$
forall varepsilon>0 exists N in mathbb N:n ge N implies |x_n - a| < varepsilon
$$
This is only valid after some $N$. The only terms violating that condition are $x_1, x_2, dots, x_N$. My guess was that we could take $max{P(n)}$ and that would imply that $|y_{P(n)} - a| < varepsilon$, so:
$$
forall varepsilon>0 exists M = max{n: 1 le P(n) le N}: n>M implies |y_n-a| < varepsilon
$$
I decided to test this idea graphically and here is what I got.
$$
x_n = frac{10}{n}
$$
So from the graph for $varepsilon = 4$ it follows that for $N ge 3 $ all terms of $x_n$ are satisfying the condition of convergence. Now if we take the maximum value for $P(n)$ where $n le 3$ we get $16$. But after $y_{16}$ there exists a value $y_{19}$ which falls out of the neighborhood of $0$.
Where did I get it wrong and how to prove the statement in the problem section?
calculus sequences-and-series limits convergence epsilon-delta
asked Nov 29 '18 at 18:52
romanroman
2,18521224
$endgroup$
add a comment |
$begingroup$
Given a sequence ${y_k}$ is obtained by permuting the sequence ${x_n}$ and:
$$
{forall nin mathbb N exists k_n in mathbb N:x_n = y_{k_n}, n_1 ne n_2 implies k_{n_1} ne k_{n_2}}
$$
and:
$$
{forall kin mathbb N exists n_k in mathbb N:y_k = x_{n_k}, k_1 ne k_2 implies n_{k_1} ne n_{k_2}}
$$
Prove that:
$$
lim_{nto infty}x_n = a implies lim_{kto infty}y_k=a
$$
I've been thinking of the following. Let there exist a bijection $P:mathbb Nto mathbb N$ which converts $k$ to $n_k$. So basically we have that if $x_n$ is convergent to some $a$ then:
$$
|x_n - a| < varepsilon
$$
Thus :
$$
forall varepsilon>0 exists N in mathbb N:n ge N implies |x_n - a| < varepsilon
$$
This is only valid after some $N$. The only terms violating that condition are $x_1, x_2, dots, x_N$. My guess was that we could take $max{P(n)}$ and that would imply that $|y_{P(n)} - a| < varepsilon$, so:
$$
forall varepsilon>0 exists M = max{n: 1 le P(n) le N}: n>M implies |y_n-a| < varepsilon
$$
I decided to test this idea graphically and here is what I got.
$$
x_n = frac{10}{n}
$$
So from the graph for $varepsilon = 4$ it follows that for $N ge 3 $ all terms of $x_n$ are satisfying the condition of convergence. Now if we take the maximum value for $P(n)$ where $n le 3$ we get $16$. But after $y_{16}$ there exists a value $y_{19}$ which falls out of the neighborhood of $0$.
Where did I get it wrong and how to prove the statement in the problem section?
calculus sequences-and-series limits convergence epsilon-delta
asked Nov 29 '18 at 18:52
romanroman
2,18521224
$endgroup$
add a comment |
$begingroup$
Given a sequence ${y_k}$ is obtained by permuting the sequence ${x_n}$ and:
$$
{forall nin mathbb N exists k_n in mathbb N:x_n = y_{k_n}, n_1 ne n_2 implies k_{n_1} ne k_{n_2}}
$$
and:
$$
{forall kin mathbb N exists n_k in mathbb N:y_k = x_{n_k}, k_1 ne k_2 implies n_{k_1} ne n_{k_2}}
$$
Prove that:
$$
lim_{nto infty}x_n = a implies lim_{kto infty}y_k=a
$$
I've been thinking of the following. Let there exist a bijection $P:mathbb Nto mathbb N$ which converts $k$ to $n_k$. So basically we have that if $x_n$ is convergent to some $a$ then:
$$
|x_n - a| < varepsilon
$$
Thus :
$$
forall varepsilon>0 exists N in mathbb N:n ge N implies |x_n - a| < varepsilon
$$
This is only valid after some $N$. The only terms violating that condition are $x_1, x_2, dots, x_N$. My guess was that we could take $max{P(n)}$ and that would imply that $|y_{P(n)} - a| < varepsilon$, so:
$$
forall varepsilon>0 exists M = max{n: 1 le P(n) le N}: n>M implies |y_n-a| < varepsilon
$$
I decided to test this idea graphically and here is what I got.
$$
x_n = frac{10}{n}
$$
So from the graph for $varepsilon = 4$ it follows that for $N ge 3 $ all terms of $x_n$ are satisfying the condition of convergence. Now if we take the maximum value for $P(n)$ where $n le 3$ we get $16$. But after $y_{16}$ there exists a value $y_{19}$ which falls out of the neighborhood of $0$.
Where did I get it wrong and how to prove the statement in the problem section?
calculus sequences-and-series limits convergence epsilon-delta
asked Nov 29 '18 at 18:52
romanroman
2,18521224
$endgroup$
Given a sequence ${y_k}$ is obtained by permuting the sequence ${x_n}$ and:
$$
{forall nin mathbb N exists k_n in mathbb N:x_n = y_{k_n}, n_1 ne n_2 implies k_{n_1} ne k_{n_2}}
$$
and:
$$
{forall kin mathbb N exists n_k in mathbb N:y_k = x_{n_k}, k_1 ne k_2 implies n_{k_1} ne n_{k_2}}
$$
Prove that:
$$
lim_{nto infty}x_n = a implies lim_{kto infty}y_k=a
$$
I've been thinking of the following. Let there exist a bijection $P:mathbb Nto mathbb N$ which converts $k$ to $n_k$. So basically we have that if $x_n$ is convergent to some $a$ then:
$$
|x_n - a| < varepsilon
$$
Thus :
$$
forall varepsilon>0 exists N in mathbb N:n ge N implies |x_n - a| < varepsilon
$$
This is only valid after some $N$. The only terms violating that condition are $x_1, x_2, dots, x_N$. My guess was that we could take $max{P(n)}$ and that would imply that $|y_{P(n)} - a| < varepsilon$, so:
$$
forall varepsilon>0 exists M = max{n: 1 le P(n) le N}: n>M implies |y_n-a| < varepsilon
$$
I decided to test this idea graphically and here is what I got.
$$
x_n = frac{10}{n}
$$
So from the graph for $varepsilon = 4$ it follows that for $N ge 3 $ all terms of $x_n$ are satisfying the condition of convergence. Now if we take the maximum value for $P(n)$ where $n le 3$ we get $16$. But after $y_{16}$ there exists a value $y_{19}$ which falls out of the neighborhood of $0$.
Where did I get it wrong and how to prove the statement in the problem section?
calculus sequences-and-series limits convergence epsilon-delta
calculus sequences-and-series limits convergence epsilon-delta
asked Nov 29 '18 at 18:52
romanroman
2,18521224
asked Nov 29 '18 at 18:52
romanroman
2,18521224
asked Nov 29 '18 at 18:52
romanroman
2,18521224
asked Nov 29 '18 at 18:52
romanroman
2,18521224
asked Nov 29 '18 at 18:52
romanroman
2,18521224
2,18521224
add a comment |
add a comment |
1 Answer
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$begingroup$
if ${a}$ converges then for any $epsilon > 0$ there is an $N>0$ such than $n>N implies |a_n - a| < epsilon$
This means that there are only finitely many $a_n$ such that $|a_n - a| > epsilon$
When we permute ${a}$ to construct ${b}$ then there are only finitely many $b_m$ such that $|b_m - a| > epsilon$
And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < epsilon$
and $m>M implies |b_m - a| < epsilon$
What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
if ${a}$ converges then for any $epsilon > 0$ there is an $N>0$ such than $n>N implies |a_n - a| < epsilon$
This means that there are only finitely many $a_n$ such that $|a_n - a| > epsilon$
When we permute ${a}$ to construct ${b}$ then there are only finitely many $b_m$ such that $|b_m - a| > epsilon$
And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < epsilon$
and $m>M implies |b_m - a| < epsilon$
What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
if ${a}$ converges then for any $epsilon > 0$ there is an $N>0$ such than $n>N implies |a_n - a| < epsilon$
This means that there are only finitely many $a_n$ such that $|a_n - a| > epsilon$
When we permute ${a}$ to construct ${b}$ then there are only finitely many $b_m$ such that $|b_m - a| > epsilon$
And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < epsilon$
and $m>M implies |b_m - a| < epsilon$
What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
$endgroup$
add a comment |
$begingroup$
if ${a}$ converges then for any $epsilon > 0$ there is an $N>0$ such than $n>N implies |a_n - a| < epsilon$
This means that there are only finitely many $a_n$ such that $|a_n - a| > epsilon$
When we permute ${a}$ to construct ${b}$ then there are only finitely many $b_m$ such that $|b_m - a| > epsilon$
And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < epsilon$
and $m>M implies |b_m - a| < epsilon$
What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
$endgroup$
if ${a}$ converges then for any $epsilon > 0$ there is an $N>0$ such than $n>N implies |a_n - a| < epsilon$
This means that there are only finitely many $a_n$ such that $|a_n - a| > epsilon$
When we permute ${a}$ to construct ${b}$ then there are only finitely many $b_m$ such that $|b_m - a| > epsilon$
And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < epsilon$
and $m>M implies |b_m - a| < epsilon$
What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
answered Nov 29 '18 at 18:59
Doug MDoug M
44.8k31854
44.8k31854
add a comment |
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