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How to prove a sequence is bounded above or below
How do i prove a sequence is bounded above or below?
For a sequence like $frac{x}{x^2+1}$ the limit will be 0 as x --> infinity but that doesnt tell me about the bounds?
calculus sequences-and-series limits
asked Nov 4 '17 at 22:37
J.Doe
125
|
show 4 more comments
How do i prove a sequence is bounded above or below?
For a sequence like $frac{x}{x^2+1}$ the limit will be 0 as x --> infinity but that doesnt tell me about the bounds?
calculus sequences-and-series limits
asked Nov 4 '17 at 22:37
J.Doe
125
Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58
|
show 4 more comments
How do i prove a sequence is bounded above or below?
For a sequence like $frac{x}{x^2+1}$ the limit will be 0 as x --> infinity but that doesnt tell me about the bounds?
calculus sequences-and-series limits
asked Nov 4 '17 at 22:37
J.Doe
125
How do i prove a sequence is bounded above or below?
For a sequence like $frac{x}{x^2+1}$ the limit will be 0 as x --> infinity but that doesnt tell me about the bounds?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Nov 4 '17 at 22:37
J.Doe
125
asked Nov 4 '17 at 22:37
J.Doe
125
edited Nov 4 '17 at 22:58
asked Nov 4 '17 at 22:37
J.Doe
125
asked Nov 4 '17 at 22:37
J.Doe
125
asked Nov 4 '17 at 22:37
J.Doe
125
125
Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58
|
show 4 more comments
Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58
Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58
|
show 4 more comments
2 Answers
2
active
oldest
votes
First of all, is it a sequence defined on $mathbb{N}$ or a continuous function defined on $[0,infty)$?
Let $f(n) = frac{n}{n^2+1}$. You can show
$f(n+1) leq f(n)$
hence it is decreasing. Therefore $f(1)$ is the upper bound.
(Lower bound is 0)
If $f(x) = frac{x}{x^2+1}$, then find the solution of $f'(x)=0$, and verify that it is the maximum.
answered Nov 4 '17 at 23:04
Atbey
937411
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
add a comment |
$dfrac{x}{x^2+1}underset{xto +infty}{to}0iff forall varepsilon>0,exists A>0, s.t.quad x>Aimplies |f(x)|<varepsilon$
That means f is bounded on $(A,+infty)$
As $f$ is continuous on $[0,A]$ according the Extrem Value Theorem $f$ is bounded on $[0,A]$, $|f|<M$
Now we take $M'=max(varepsilon, M)$ then $|f|<M'$ on $[0,+infty)$
answered Nov 4 '17 at 23:30
Stu
1,1741413
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all, is it a sequence defined on $mathbb{N}$ or a continuous function defined on $[0,infty)$?
Let $f(n) = frac{n}{n^2+1}$. You can show
$f(n+1) leq f(n)$
hence it is decreasing. Therefore $f(1)$ is the upper bound.
(Lower bound is 0)
If $f(x) = frac{x}{x^2+1}$, then find the solution of $f'(x)=0$, and verify that it is the maximum.
answered Nov 4 '17 at 23:04
Atbey
937411
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
add a comment |
First of all, is it a sequence defined on $mathbb{N}$ or a continuous function defined on $[0,infty)$?
Let $f(n) = frac{n}{n^2+1}$. You can show
$f(n+1) leq f(n)$
hence it is decreasing. Therefore $f(1)$ is the upper bound.
(Lower bound is 0)
If $f(x) = frac{x}{x^2+1}$, then find the solution of $f'(x)=0$, and verify that it is the maximum.
answered Nov 4 '17 at 23:04
Atbey
937411
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
add a comment |
First of all, is it a sequence defined on $mathbb{N}$ or a continuous function defined on $[0,infty)$?
Let $f(n) = frac{n}{n^2+1}$. You can show
$f(n+1) leq f(n)$
hence it is decreasing. Therefore $f(1)$ is the upper bound.
(Lower bound is 0)
If $f(x) = frac{x}{x^2+1}$, then find the solution of $f'(x)=0$, and verify that it is the maximum.
answered Nov 4 '17 at 23:04
Atbey
937411
First of all, is it a sequence defined on $mathbb{N}$ or a continuous function defined on $[0,infty)$?
Let $f(n) = frac{n}{n^2+1}$. You can show
$f(n+1) leq f(n)$
hence it is decreasing. Therefore $f(1)$ is the upper bound.
(Lower bound is 0)
If $f(x) = frac{x}{x^2+1}$, then find the solution of $f'(x)=0$, and verify that it is the maximum.
answered Nov 4 '17 at 23:04
Atbey
937411
answered Nov 4 '17 at 23:04
Atbey
937411
answered Nov 4 '17 at 23:04
Atbey
937411
answered Nov 4 '17 at 23:04
Atbey
937411
937411
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
add a comment |
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
The sequence is defined for all real values of x. Also, i plotted the graph of that sequence and it has a maximum at 0.5 and minimum at -0.5 so how do i show this without plotting a graph because then its not strictly decreasing or increasing
– J.Doe
Nov 4 '17 at 23:15
1
1
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
Then it is a function, when you say sequence people tend to think a function defined from $mathbb{N}tomathbb{R}$. You can still find solutions of $f'(x)=0$, which will give you the upper and lower bounds as $1/2$ and $-1/2$.
– Atbey
Nov 4 '17 at 23:25
add a comment |
$dfrac{x}{x^2+1}underset{xto +infty}{to}0iff forall varepsilon>0,exists A>0, s.t.quad x>Aimplies |f(x)|<varepsilon$
That means f is bounded on $(A,+infty)$
As $f$ is continuous on $[0,A]$ according the Extrem Value Theorem $f$ is bounded on $[0,A]$, $|f|<M$
Now we take $M'=max(varepsilon, M)$ then $|f|<M'$ on $[0,+infty)$
answered Nov 4 '17 at 23:30
Stu
1,1741413
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
|
show 2 more comments
$dfrac{x}{x^2+1}underset{xto +infty}{to}0iff forall varepsilon>0,exists A>0, s.t.quad x>Aimplies |f(x)|<varepsilon$
That means f is bounded on $(A,+infty)$
As $f$ is continuous on $[0,A]$ according the Extrem Value Theorem $f$ is bounded on $[0,A]$, $|f|<M$
Now we take $M'=max(varepsilon, M)$ then $|f|<M'$ on $[0,+infty)$
answered Nov 4 '17 at 23:30
Stu
1,1741413
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
|
show 2 more comments
$dfrac{x}{x^2+1}underset{xto +infty}{to}0iff forall varepsilon>0,exists A>0, s.t.quad x>Aimplies |f(x)|<varepsilon$
That means f is bounded on $(A,+infty)$
As $f$ is continuous on $[0,A]$ according the Extrem Value Theorem $f$ is bounded on $[0,A]$, $|f|<M$
Now we take $M'=max(varepsilon, M)$ then $|f|<M'$ on $[0,+infty)$
answered Nov 4 '17 at 23:30
Stu
1,1741413
$dfrac{x}{x^2+1}underset{xto +infty}{to}0iff forall varepsilon>0,exists A>0, s.t.quad x>Aimplies |f(x)|<varepsilon$
That means f is bounded on $(A,+infty)$
As $f$ is continuous on $[0,A]$ according the Extrem Value Theorem $f$ is bounded on $[0,A]$, $|f|<M$
Now we take $M'=max(varepsilon, M)$ then $|f|<M'$ on $[0,+infty)$
answered Nov 4 '17 at 23:30
Stu
1,1741413
edited Nov 4 '17 at 23:36
answered Nov 4 '17 at 23:30
Stu
1,1741413
answered Nov 4 '17 at 23:30
Stu
1,1741413
answered Nov 4 '17 at 23:30
Stu
1,1741413
1,1741413
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
|
show 2 more comments
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
ok i think i know what that means according to the link given above :D but how do i show its bounded below too
– J.Doe
Nov 4 '17 at 23:41
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
$|f|<M$ means $-M<f<M$
– Stu
Nov 4 '17 at 23:42
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
thancs for the explanation
– J.Doe
Nov 4 '17 at 23:45
you are welcome
– Stu
Nov 4 '17 at 23:46
you are welcome
– Stu
Nov 4 '17 at 23:46
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
hey i got another question i understood the first line but i dont get the second line, we havent done extreme value theorem yet so i dont know whether i should include the second and third line in my answer i dont really know what the third line means either
– J.Doe
Nov 5 '17 at 0:08
|
show 2 more comments
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Have you tried anything?
– Gerard L.
Nov 4 '17 at 22:42
Every convergent sequence is bounded .
– S.H.W
Nov 4 '17 at 22:50
How do i tell if it is bounded above, below or both?
– J.Doe
Nov 4 '17 at 22:53
When sequence is bounded it means that sequence is bounded above and bounded below .
– S.H.W
Nov 4 '17 at 22:55
I should've phrased the question better but the question on my sheet says is it bounded or not and if so is it above, below or both and prove it.
– J.Doe
Nov 4 '17 at 22:58