Mixing
Proving if a sequence is Cauchy and/or convergent. [duplicate]
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
Let $M=(0,infty)$ be supplied with the metric function $d(x,y)=|arctan(x)-arctan(y)|$ and let ${n}_{n=1}^infty$ be a sequence of positive integers.
a) Is the sequence a Cauchy sequence in $(M,d)$?
b) Is the sequence a convergent sequence in $(M,d)$?
I am pretty sure it is Cauchy but not convergent. What I have so far is:
$|arctan(n)-arctan(m)|<epsilon$ if $n,m$ are sufficiently large: $|arctan(n)-arctan(m)|≤|arctan(n)-0|+|arctan(m)-0|=arctan(n)+arctan(m)<cdots$
This is where I'm stuck.
real-analysis convergence cauchy-sequences
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
marked as duplicate by Lord Shark the Unknown, Brahadeesh, Vidyanshu Mishra, José Carlos Santos real-analysis
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Nov 29 '18 at 9:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 4 more comments
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
Let $M=(0,infty)$ be supplied with the metric function $d(x,y)=|arctan(x)-arctan(y)|$ and let ${n}_{n=1}^infty$ be a sequence of positive integers.
a) Is the sequence a Cauchy sequence in $(M,d)$?
b) Is the sequence a convergent sequence in $(M,d)$?
I am pretty sure it is Cauchy but not convergent. What I have so far is:
$|arctan(n)-arctan(m)|<epsilon$ if $n,m$ are sufficiently large: $|arctan(n)-arctan(m)|≤|arctan(n)-0|+|arctan(m)-0|=arctan(n)+arctan(m)<cdots$
This is where I'm stuck.
real-analysis convergence cauchy-sequences
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
marked as duplicate by Lord Shark the Unknown, Brahadeesh, Vidyanshu Mishra, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Nov 29 '18 at 9:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37
|
show 4 more comments
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
Let $M=(0,infty)$ be supplied with the metric function $d(x,y)=|arctan(x)-arctan(y)|$ and let ${n}_{n=1}^infty$ be a sequence of positive integers.
a) Is the sequence a Cauchy sequence in $(M,d)$?
b) Is the sequence a convergent sequence in $(M,d)$?
I am pretty sure it is Cauchy but not convergent. What I have so far is:
$|arctan(n)-arctan(m)|<epsilon$ if $n,m$ are sufficiently large: $|arctan(n)-arctan(m)|≤|arctan(n)-0|+|arctan(m)-0|=arctan(n)+arctan(m)<cdots$
This is where I'm stuck.
real-analysis convergence cauchy-sequences
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
Let $M=(0,infty)$ be supplied with the metric function $d(x,y)=|arctan(x)-arctan(y)|$ and let ${n}_{n=1}^infty$ be a sequence of positive integers.
a) Is the sequence a Cauchy sequence in $(M,d)$?
b) Is the sequence a convergent sequence in $(M,d)$?
I am pretty sure it is Cauchy but not convergent. What I have so far is:
$|arctan(n)-arctan(m)|<epsilon$ if $n,m$ are sufficiently large: $|arctan(n)-arctan(m)|≤|arctan(n)-0|+|arctan(m)-0|=arctan(n)+arctan(m)<cdots$
This is where I'm stuck.
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
real-analysis convergence cauchy-sequences
real-analysis convergence cauchy-sequences
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
edited Nov 21 '18 at 18:38
Yadati Kiran
1,692619
1,692619
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
asked Nov 21 '18 at 18:17
Amanda VarvakAmanda Varvak
13
13
marked as duplicate by Lord Shark the Unknown, Brahadeesh, Vidyanshu Mishra, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Nov 29 '18 at 9:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, Brahadeesh, Vidyanshu Mishra, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Nov 29 '18 at 9:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37
|
show 4 more comments
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37
|
show 4 more comments
1 Answer
1
active
oldest
votes
Well, since
$$
lim_{nto+infty} arctan n = frac{pi}{2}
$$
you get
$$
d(m,n)to leftlvertfrac{pi}{2}-frac{pi}{2}rightrvert=0
$$
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Well, since
$$
lim_{nto+infty} arctan n = frac{pi}{2}
$$
you get
$$
d(m,n)to leftlvertfrac{pi}{2}-frac{pi}{2}rightrvert=0
$$
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
add a comment |
Well, since
$$
lim_{nto+infty} arctan n = frac{pi}{2}
$$
you get
$$
d(m,n)to leftlvertfrac{pi}{2}-frac{pi}{2}rightrvert=0
$$
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
add a comment |
Well, since
$$
lim_{nto+infty} arctan n = frac{pi}{2}
$$
you get
$$
d(m,n)to leftlvertfrac{pi}{2}-frac{pi}{2}rightrvert=0
$$
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
Well, since
$$
lim_{nto+infty} arctan n = frac{pi}{2}
$$
you get
$$
d(m,n)to leftlvertfrac{pi}{2}-frac{pi}{2}rightrvert=0
$$
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
answered Nov 21 '18 at 18:36
francescop21francescop21
1,012115
1,012115
add a comment |
add a comment |
Is $tan^{-1}=1/tan$? Or $tan^{-1}=arctan$?
– francescop21
Nov 21 '18 at 18:27
$tan^-1 =arctan$
– Amanda Varvak
Nov 21 '18 at 18:29
@AmandaVarvak Do you know if the sequence $arctan(n)$ converges as $n$ goes to positive infinity?
– irchans
Nov 21 '18 at 18:33
@francescop21: Aren't both notations the same?
– Yadati Kiran
Nov 21 '18 at 18:36
@irchans I believe arctan(n) diverges, so no it does not converge.
– Amanda Varvak
Nov 21 '18 at 18:37