Convergent sequence rigorous definition
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I know what a convergent sequence is and how it works and everything but whenever I look at the definition my mind cant suddenly make the link with my intuition part of the brain so I realise that I am actually going over the same thing in my brain and still get no good answer for my questions about the definition. A sequence $x_n$ of real numbers converge to a point $L$ if given $epsilon>0$, $exists$ $n_0=n_0(epsilon)$ such that $$ |x_n-L| < epsilon quad for everyquad n > n_0. $$
My first question is why do we use this definition? My second question is what does each term exactly mean and in general how did you come to digest this thing in your brain. I do seem to have general idea of it but i want to be precise and would like some directions and explanations. THANKS.
calculus
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show 1 more comment
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I know what a convergent sequence is and how it works and everything but whenever I look at the definition my mind cant suddenly make the link with my intuition part of the brain so I realise that I am actually going over the same thing in my brain and still get no good answer for my questions about the definition. A sequence $x_n$ of real numbers converge to a point $L$ if given $epsilon>0$, $exists$ $n_0=n_0(epsilon)$ such that $$ |x_n-L| < epsilon quad for everyquad n > n_0. $$
My first question is why do we use this definition? My second question is what does each term exactly mean and in general how did you come to digest this thing in your brain. I do seem to have general idea of it but i want to be precise and would like some directions and explanations. THANKS.
calculus
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$begingroup$
You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
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– Henry
Jul 17 '15 at 20:12
1
$begingroup$
The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
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– Ennar
Jul 17 '15 at 20:22
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Already corrected!
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– Chuks
Jul 17 '15 at 20:23
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Cant $n_0$ be any real number?
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– Elliti123
Jul 17 '15 at 20:24
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@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
$endgroup$
– Chuks
Jul 17 '15 at 21:38
|
show 1 more comment
$begingroup$
I know what a convergent sequence is and how it works and everything but whenever I look at the definition my mind cant suddenly make the link with my intuition part of the brain so I realise that I am actually going over the same thing in my brain and still get no good answer for my questions about the definition. A sequence $x_n$ of real numbers converge to a point $L$ if given $epsilon>0$, $exists$ $n_0=n_0(epsilon)$ such that $$ |x_n-L| < epsilon quad for everyquad n > n_0. $$
My first question is why do we use this definition? My second question is what does each term exactly mean and in general how did you come to digest this thing in your brain. I do seem to have general idea of it but i want to be precise and would like some directions and explanations. THANKS.
calculus
$endgroup$
I know what a convergent sequence is and how it works and everything but whenever I look at the definition my mind cant suddenly make the link with my intuition part of the brain so I realise that I am actually going over the same thing in my brain and still get no good answer for my questions about the definition. A sequence $x_n$ of real numbers converge to a point $L$ if given $epsilon>0$, $exists$ $n_0=n_0(epsilon)$ such that $$ |x_n-L| < epsilon quad for everyquad n > n_0. $$
My first question is why do we use this definition? My second question is what does each term exactly mean and in general how did you come to digest this thing in your brain. I do seem to have general idea of it but i want to be precise and would like some directions and explanations. THANKS.
calculus
calculus
edited Jul 17 '15 at 20:20
Chuks
835417
835417
asked Jul 17 '15 at 20:06
Elliti123Elliti123
497
497
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You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
$endgroup$
– Henry
Jul 17 '15 at 20:12
1
$begingroup$
The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
$endgroup$
– Ennar
Jul 17 '15 at 20:22
$begingroup$
Already corrected!
$endgroup$
– Chuks
Jul 17 '15 at 20:23
$begingroup$
Cant $n_0$ be any real number?
$endgroup$
– Elliti123
Jul 17 '15 at 20:24
$begingroup$
@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
$endgroup$
– Chuks
Jul 17 '15 at 21:38
|
show 1 more comment
$begingroup$
You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
$endgroup$
– Henry
Jul 17 '15 at 20:12
1
$begingroup$
The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
$endgroup$
– Ennar
Jul 17 '15 at 20:22
$begingroup$
Already corrected!
$endgroup$
– Chuks
Jul 17 '15 at 20:23
$begingroup$
Cant $n_0$ be any real number?
$endgroup$
– Elliti123
Jul 17 '15 at 20:24
$begingroup$
@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
$endgroup$
– Chuks
Jul 17 '15 at 21:38
$begingroup$
You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
$endgroup$
– Henry
Jul 17 '15 at 20:12
$begingroup$
You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
$endgroup$
– Henry
Jul 17 '15 at 20:12
1
1
$begingroup$
The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
$endgroup$
– Ennar
Jul 17 '15 at 20:22
$begingroup$
The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
$endgroup$
– Ennar
Jul 17 '15 at 20:22
$begingroup$
Already corrected!
$endgroup$
– Chuks
Jul 17 '15 at 20:23
$begingroup$
Already corrected!
$endgroup$
– Chuks
Jul 17 '15 at 20:23
$begingroup$
Cant $n_0$ be any real number?
$endgroup$
– Elliti123
Jul 17 '15 at 20:24
$begingroup$
Cant $n_0$ be any real number?
$endgroup$
– Elliti123
Jul 17 '15 at 20:24
$begingroup$
@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
$endgroup$
– Chuks
Jul 17 '15 at 21:38
$begingroup$
@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
$endgroup$
– Chuks
Jul 17 '15 at 21:38
|
show 1 more comment
5 Answers
5
active
oldest
votes
$begingroup$
M: Here is a good sequence $(x_n)_{ngeq1}$ for you. It converges to $pi$. For $10$ dollars it's yours.
E: I looked at your $x_5$. It's $=4$. That's far away from $pi$.
M: What precision do you have in mind?
E: The error should be at most $0.01$, say.
M: Look here at this $x_{53}$ for example. It's $=3.14367$.
E: Maybe I should lower my tolerance to $0.0001$.
M: No problem. Let me check. Here it is: $x_{1538}=3.141512$.
E: I'm not convinced. These could be numerical coincidences.
M: There is a warranty booklet coming with the sequence. It contains a proof that the error is less than $0.0001$ for all $n>1600$.
E: How about even higher precision?
M: For only five more dollars you get the guarantee extended to $0.000001$. If I remember correctly the corresponding paperback contains a proof that the error is less than $10^{-6}$ for all $n>600,000$.
And on, and on.
$endgroup$
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
add a comment |
$begingroup$
I think about the $epsilon$ as sort of an "allowable error" term. You specify how much error you are willing to allow: I want all the points to be within $epsilon$ of $L$. Then we say the limit of the sequence is $L$ if you can find a point after which all the terms are "acceptable" in the sense that they are not too far away. The important thing is that you have to be able to find the point at which they are all acceptable regardless of what the "allowable error" is, as long as you allow some wiggle room (i.e. $epsilon$ cannot be $0$).
$endgroup$
add a comment |
$begingroup$
As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool.
As for the meaning, you can visualize this somewhat simply. Plot a point on a number line and then start plotting a sequence of points that converges to the original point.
Now put an interval centered at the original point. The definition says that from some term onward, every point you plot lies in the interval.
That is, from some term ($n_0$) onward ($n ge n_0$), every point ($x_n$) lies in the interval ($|x_n - L| < epsilon$).
$endgroup$
$begingroup$
Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
$endgroup$
– Elliti123
Jul 17 '15 at 20:17
add a comment |
$begingroup$
The OP asks
My second question is what does each term exactly mean and in general
how did you come to digest this thing in your brain.
Each term on its own means nothing. The sequence $(x_n)_{,n ge 0}$ is 'carving out a story' and the value of any term $x_k$, for a fixed $k$, says nothing. It is how the sequence 'continues, and keeps acting' at each subsequent term that imparts the 'message'.
If $w$ is any real number and the sequence $(x_n)_{,n ge 0}$ 'never stops showing up' with values $x_m$ that can get arbitrarily close $w$, then $w$ is said to be a cluster point. If the sequence is bounded it must have one cluster point, and if there is only one 'it keep approaching', that number is called the limit point.
I offered the above intuitive verbiage to describe how one might be able to 'digest this thing' into the brain. But to do it formally, you need to be able to understand logical quantifiers and be comfortable with real numbers.
Of course the reason investigating sequences is so rewarding is that the real line has no gaps (it is a 'complete' system of numbers).
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+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
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– Paramanand Singh
Nov 27 '18 at 5:17
add a comment |
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The notion of a limit a sequence $(x_n)$, in simple terms, is a point $L$ such that all of the later terms $x_n$ in the sequence are arbitrarily close to $L$. Geometrically, this means that, given any small $varepsilon>0$, we can find a large integer $N$ such that, whenever $ngeq N$, we have $x_nin[L-varepsilon,L+varepsilon]$.
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
M: Here is a good sequence $(x_n)_{ngeq1}$ for you. It converges to $pi$. For $10$ dollars it's yours.
E: I looked at your $x_5$. It's $=4$. That's far away from $pi$.
M: What precision do you have in mind?
E: The error should be at most $0.01$, say.
M: Look here at this $x_{53}$ for example. It's $=3.14367$.
E: Maybe I should lower my tolerance to $0.0001$.
M: No problem. Let me check. Here it is: $x_{1538}=3.141512$.
E: I'm not convinced. These could be numerical coincidences.
M: There is a warranty booklet coming with the sequence. It contains a proof that the error is less than $0.0001$ for all $n>1600$.
E: How about even higher precision?
M: For only five more dollars you get the guarantee extended to $0.000001$. If I remember correctly the corresponding paperback contains a proof that the error is less than $10^{-6}$ for all $n>600,000$.
And on, and on.
$endgroup$
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
add a comment |
$begingroup$
M: Here is a good sequence $(x_n)_{ngeq1}$ for you. It converges to $pi$. For $10$ dollars it's yours.
E: I looked at your $x_5$. It's $=4$. That's far away from $pi$.
M: What precision do you have in mind?
E: The error should be at most $0.01$, say.
M: Look here at this $x_{53}$ for example. It's $=3.14367$.
E: Maybe I should lower my tolerance to $0.0001$.
M: No problem. Let me check. Here it is: $x_{1538}=3.141512$.
E: I'm not convinced. These could be numerical coincidences.
M: There is a warranty booklet coming with the sequence. It contains a proof that the error is less than $0.0001$ for all $n>1600$.
E: How about even higher precision?
M: For only five more dollars you get the guarantee extended to $0.000001$. If I remember correctly the corresponding paperback contains a proof that the error is less than $10^{-6}$ for all $n>600,000$.
And on, and on.
$endgroup$
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
add a comment |
$begingroup$
M: Here is a good sequence $(x_n)_{ngeq1}$ for you. It converges to $pi$. For $10$ dollars it's yours.
E: I looked at your $x_5$. It's $=4$. That's far away from $pi$.
M: What precision do you have in mind?
E: The error should be at most $0.01$, say.
M: Look here at this $x_{53}$ for example. It's $=3.14367$.
E: Maybe I should lower my tolerance to $0.0001$.
M: No problem. Let me check. Here it is: $x_{1538}=3.141512$.
E: I'm not convinced. These could be numerical coincidences.
M: There is a warranty booklet coming with the sequence. It contains a proof that the error is less than $0.0001$ for all $n>1600$.
E: How about even higher precision?
M: For only five more dollars you get the guarantee extended to $0.000001$. If I remember correctly the corresponding paperback contains a proof that the error is less than $10^{-6}$ for all $n>600,000$.
And on, and on.
$endgroup$
M: Here is a good sequence $(x_n)_{ngeq1}$ for you. It converges to $pi$. For $10$ dollars it's yours.
E: I looked at your $x_5$. It's $=4$. That's far away from $pi$.
M: What precision do you have in mind?
E: The error should be at most $0.01$, say.
M: Look here at this $x_{53}$ for example. It's $=3.14367$.
E: Maybe I should lower my tolerance to $0.0001$.
M: No problem. Let me check. Here it is: $x_{1538}=3.141512$.
E: I'm not convinced. These could be numerical coincidences.
M: There is a warranty booklet coming with the sequence. It contains a proof that the error is less than $0.0001$ for all $n>1600$.
E: How about even higher precision?
M: For only five more dollars you get the guarantee extended to $0.000001$. If I remember correctly the corresponding paperback contains a proof that the error is less than $10^{-6}$ for all $n>600,000$.
And on, and on.
answered Jul 18 '15 at 10:05
Christian BlatterChristian Blatter
173k7113326
173k7113326
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
add a comment |
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
1
1
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
$begingroup$
I thought you were going to give a wrong analogy until you mentioned "for all $n > 1600$. But off-topic, your sequence is hardly worth ten cents, as it converges so slowly. What a scam! =P
$endgroup$
– user21820
Jul 23 '15 at 12:11
add a comment |
$begingroup$
I think about the $epsilon$ as sort of an "allowable error" term. You specify how much error you are willing to allow: I want all the points to be within $epsilon$ of $L$. Then we say the limit of the sequence is $L$ if you can find a point after which all the terms are "acceptable" in the sense that they are not too far away. The important thing is that you have to be able to find the point at which they are all acceptable regardless of what the "allowable error" is, as long as you allow some wiggle room (i.e. $epsilon$ cannot be $0$).
$endgroup$
add a comment |
$begingroup$
I think about the $epsilon$ as sort of an "allowable error" term. You specify how much error you are willing to allow: I want all the points to be within $epsilon$ of $L$. Then we say the limit of the sequence is $L$ if you can find a point after which all the terms are "acceptable" in the sense that they are not too far away. The important thing is that you have to be able to find the point at which they are all acceptable regardless of what the "allowable error" is, as long as you allow some wiggle room (i.e. $epsilon$ cannot be $0$).
$endgroup$
add a comment |
$begingroup$
I think about the $epsilon$ as sort of an "allowable error" term. You specify how much error you are willing to allow: I want all the points to be within $epsilon$ of $L$. Then we say the limit of the sequence is $L$ if you can find a point after which all the terms are "acceptable" in the sense that they are not too far away. The important thing is that you have to be able to find the point at which they are all acceptable regardless of what the "allowable error" is, as long as you allow some wiggle room (i.e. $epsilon$ cannot be $0$).
$endgroup$
I think about the $epsilon$ as sort of an "allowable error" term. You specify how much error you are willing to allow: I want all the points to be within $epsilon$ of $L$. Then we say the limit of the sequence is $L$ if you can find a point after which all the terms are "acceptable" in the sense that they are not too far away. The important thing is that you have to be able to find the point at which they are all acceptable regardless of what the "allowable error" is, as long as you allow some wiggle room (i.e. $epsilon$ cannot be $0$).
answered Jul 17 '15 at 20:51
TravisJTravisJ
6,36831730
6,36831730
add a comment |
add a comment |
$begingroup$
As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool.
As for the meaning, you can visualize this somewhat simply. Plot a point on a number line and then start plotting a sequence of points that converges to the original point.
Now put an interval centered at the original point. The definition says that from some term onward, every point you plot lies in the interval.
That is, from some term ($n_0$) onward ($n ge n_0$), every point ($x_n$) lies in the interval ($|x_n - L| < epsilon$).
$endgroup$
$begingroup$
Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
$endgroup$
– Elliti123
Jul 17 '15 at 20:17
add a comment |
$begingroup$
As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool.
As for the meaning, you can visualize this somewhat simply. Plot a point on a number line and then start plotting a sequence of points that converges to the original point.
Now put an interval centered at the original point. The definition says that from some term onward, every point you plot lies in the interval.
That is, from some term ($n_0$) onward ($n ge n_0$), every point ($x_n$) lies in the interval ($|x_n - L| < epsilon$).
$endgroup$
$begingroup$
Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
$endgroup$
– Elliti123
Jul 17 '15 at 20:17
add a comment |
$begingroup$
As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool.
As for the meaning, you can visualize this somewhat simply. Plot a point on a number line and then start plotting a sequence of points that converges to the original point.
Now put an interval centered at the original point. The definition says that from some term onward, every point you plot lies in the interval.
That is, from some term ($n_0$) onward ($n ge n_0$), every point ($x_n$) lies in the interval ($|x_n - L| < epsilon$).
$endgroup$
As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool.
As for the meaning, you can visualize this somewhat simply. Plot a point on a number line and then start plotting a sequence of points that converges to the original point.
Now put an interval centered at the original point. The definition says that from some term onward, every point you plot lies in the interval.
That is, from some term ($n_0$) onward ($n ge n_0$), every point ($x_n$) lies in the interval ($|x_n - L| < epsilon$).
answered Jul 17 '15 at 20:13
Umberto P.Umberto P.
38.9k13064
38.9k13064
$begingroup$
Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
$endgroup$
– Elliti123
Jul 17 '15 at 20:17
add a comment |
$begingroup$
Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
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– Elliti123
Jul 17 '15 at 20:17
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Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
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– Elliti123
Jul 17 '15 at 20:17
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Oh ok so i would like to ask if what i made from the definition is right or no. If a sequence is convergent then it is possible to make the distance from the point of convergence less than any positive real number we like after taking n large enough this would happen since if it is a converging to a point then the distance has to become smaller.So that's why we have to prove that the distance can be made less than any other number if n is large enough.
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– Elliti123
Jul 17 '15 at 20:17
add a comment |
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The OP asks
My second question is what does each term exactly mean and in general
how did you come to digest this thing in your brain.
Each term on its own means nothing. The sequence $(x_n)_{,n ge 0}$ is 'carving out a story' and the value of any term $x_k$, for a fixed $k$, says nothing. It is how the sequence 'continues, and keeps acting' at each subsequent term that imparts the 'message'.
If $w$ is any real number and the sequence $(x_n)_{,n ge 0}$ 'never stops showing up' with values $x_m$ that can get arbitrarily close $w$, then $w$ is said to be a cluster point. If the sequence is bounded it must have one cluster point, and if there is only one 'it keep approaching', that number is called the limit point.
I offered the above intuitive verbiage to describe how one might be able to 'digest this thing' into the brain. But to do it formally, you need to be able to understand logical quantifiers and be comfortable with real numbers.
Of course the reason investigating sequences is so rewarding is that the real line has no gaps (it is a 'complete' system of numbers).
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+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
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– Paramanand Singh
Nov 27 '18 at 5:17
add a comment |
$begingroup$
The OP asks
My second question is what does each term exactly mean and in general
how did you come to digest this thing in your brain.
Each term on its own means nothing. The sequence $(x_n)_{,n ge 0}$ is 'carving out a story' and the value of any term $x_k$, for a fixed $k$, says nothing. It is how the sequence 'continues, and keeps acting' at each subsequent term that imparts the 'message'.
If $w$ is any real number and the sequence $(x_n)_{,n ge 0}$ 'never stops showing up' with values $x_m$ that can get arbitrarily close $w$, then $w$ is said to be a cluster point. If the sequence is bounded it must have one cluster point, and if there is only one 'it keep approaching', that number is called the limit point.
I offered the above intuitive verbiage to describe how one might be able to 'digest this thing' into the brain. But to do it formally, you need to be able to understand logical quantifiers and be comfortable with real numbers.
Of course the reason investigating sequences is so rewarding is that the real line has no gaps (it is a 'complete' system of numbers).
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$begingroup$
+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
$endgroup$
– Paramanand Singh
Nov 27 '18 at 5:17
add a comment |
$begingroup$
The OP asks
My second question is what does each term exactly mean and in general
how did you come to digest this thing in your brain.
Each term on its own means nothing. The sequence $(x_n)_{,n ge 0}$ is 'carving out a story' and the value of any term $x_k$, for a fixed $k$, says nothing. It is how the sequence 'continues, and keeps acting' at each subsequent term that imparts the 'message'.
If $w$ is any real number and the sequence $(x_n)_{,n ge 0}$ 'never stops showing up' with values $x_m$ that can get arbitrarily close $w$, then $w$ is said to be a cluster point. If the sequence is bounded it must have one cluster point, and if there is only one 'it keep approaching', that number is called the limit point.
I offered the above intuitive verbiage to describe how one might be able to 'digest this thing' into the brain. But to do it formally, you need to be able to understand logical quantifiers and be comfortable with real numbers.
Of course the reason investigating sequences is so rewarding is that the real line has no gaps (it is a 'complete' system of numbers).
$endgroup$
The OP asks
My second question is what does each term exactly mean and in general
how did you come to digest this thing in your brain.
Each term on its own means nothing. The sequence $(x_n)_{,n ge 0}$ is 'carving out a story' and the value of any term $x_k$, for a fixed $k$, says nothing. It is how the sequence 'continues, and keeps acting' at each subsequent term that imparts the 'message'.
If $w$ is any real number and the sequence $(x_n)_{,n ge 0}$ 'never stops showing up' with values $x_m$ that can get arbitrarily close $w$, then $w$ is said to be a cluster point. If the sequence is bounded it must have one cluster point, and if there is only one 'it keep approaching', that number is called the limit point.
I offered the above intuitive verbiage to describe how one might be able to 'digest this thing' into the brain. But to do it formally, you need to be able to understand logical quantifiers and be comfortable with real numbers.
Of course the reason investigating sequences is so rewarding is that the real line has no gaps (it is a 'complete' system of numbers).
answered Nov 26 '18 at 19:55
CopyPasteItCopyPasteIt
4,1531628
4,1531628
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+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
$endgroup$
– Paramanand Singh
Nov 27 '18 at 5:17
add a comment |
$begingroup$
+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
$endgroup$
– Paramanand Singh
Nov 27 '18 at 5:17
$begingroup$
+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
$endgroup$
– Paramanand Singh
Nov 27 '18 at 5:17
$begingroup$
+1 for the special mention of completeness and the fact it is rewarding to study such stuff.
$endgroup$
– Paramanand Singh
Nov 27 '18 at 5:17
add a comment |
$begingroup$
The notion of a limit a sequence $(x_n)$, in simple terms, is a point $L$ such that all of the later terms $x_n$ in the sequence are arbitrarily close to $L$. Geometrically, this means that, given any small $varepsilon>0$, we can find a large integer $N$ such that, whenever $ngeq N$, we have $x_nin[L-varepsilon,L+varepsilon]$.
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add a comment |
$begingroup$
The notion of a limit a sequence $(x_n)$, in simple terms, is a point $L$ such that all of the later terms $x_n$ in the sequence are arbitrarily close to $L$. Geometrically, this means that, given any small $varepsilon>0$, we can find a large integer $N$ such that, whenever $ngeq N$, we have $x_nin[L-varepsilon,L+varepsilon]$.
$endgroup$
add a comment |
$begingroup$
The notion of a limit a sequence $(x_n)$, in simple terms, is a point $L$ such that all of the later terms $x_n$ in the sequence are arbitrarily close to $L$. Geometrically, this means that, given any small $varepsilon>0$, we can find a large integer $N$ such that, whenever $ngeq N$, we have $x_nin[L-varepsilon,L+varepsilon]$.
$endgroup$
The notion of a limit a sequence $(x_n)$, in simple terms, is a point $L$ such that all of the later terms $x_n$ in the sequence are arbitrarily close to $L$. Geometrically, this means that, given any small $varepsilon>0$, we can find a large integer $N$ such that, whenever $ngeq N$, we have $x_nin[L-varepsilon,L+varepsilon]$.
answered Jul 17 '15 at 20:27
SamMSamM
1,308510
1,308510
add a comment |
add a comment |
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You want the difference between all values in the tail of the sequence and the limit $L$ to be smaller than any given positive real number, i.e. the tail of the sequence should be arbitrarily close to the limit.
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– Henry
Jul 17 '15 at 20:12
1
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The definition you wrote is almost correct except that this need to be true for any $epsilon > 0$. That is: $$(forallepsilon > 0)(exists n_0inmathbb N) ngeq n_0implies | x_n - L | < epsilon $$ point being that $n_0$ depends on $epsilon$ and can change when $epsilon$ changes.
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– Ennar
Jul 17 '15 at 20:22
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Already corrected!
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– Chuks
Jul 17 '15 at 20:23
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Cant $n_0$ be any real number?
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– Elliti123
Jul 17 '15 at 20:24
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@ Elliti123, no, it can't (by definition). Since it depends on $epsilon$, if $epsilon$ changes, then you can change your $n_0$ to the nearest natural number that will still make the stated condition to be satisfied.
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– Chuks
Jul 17 '15 at 21:38