sup and inf of $A=left{frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n inBbb Nright}$
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$$A=left{frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n inBbb Nright}$$
$sup A$ doesn't exist, because $A$ is not bounded above:
$$begin{align}n&=1 \
frac{m^4+2}{m^2+1} &le M \
frac{m^4+2}{m^2} &le M \
m^2+frac{2}{m^2} &le M \
m^2&le Mend{align}$$
But I can choose $m=lfloor M rfloor+1$ from Archimedean property.
I think $inf A=3/2$. Am I right?
UPDATE: for $m=3$ and $n=4$ we get $-113/2$
UPDATE2: I think $inf A$ doesn't exist too.
I have to prove that e.g:
$$forall _m exists _n:m^2n>n^2+2m^2$$
real-analysis supremum-and-infimum
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
add a comment |
up vote
0
down vote
favorite
$$A=left{frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n inBbb Nright}$$
$sup A$ doesn't exist, because $A$ is not bounded above:
$$begin{align}n&=1 \
frac{m^4+2}{m^2+1} &le M \
frac{m^4+2}{m^2} &le M \
m^2+frac{2}{m^2} &le M \
m^2&le Mend{align}$$
But I can choose $m=lfloor M rfloor+1$ from Archimedean property.
I think $inf A=3/2$. Am I right?
UPDATE: for $m=3$ and $n=4$ we get $-113/2$
UPDATE2: I think $inf A$ doesn't exist too.
I have to prove that e.g:
$$forall _m exists _n:m^2n>n^2+2m^2$$
real-analysis supremum-and-infimum
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
2
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$A=left{frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n inBbb Nright}$$
$sup A$ doesn't exist, because $A$ is not bounded above:
$$begin{align}n&=1 \
frac{m^4+2}{m^2+1} &le M \
frac{m^4+2}{m^2} &le M \
m^2+frac{2}{m^2} &le M \
m^2&le Mend{align}$$
But I can choose $m=lfloor M rfloor+1$ from Archimedean property.
I think $inf A=3/2$. Am I right?
UPDATE: for $m=3$ and $n=4$ we get $-113/2$
UPDATE2: I think $inf A$ doesn't exist too.
I have to prove that e.g:
$$forall _m exists _n:m^2n>n^2+2m^2$$
real-analysis supremum-and-infimum
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
$$A=left{frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n inBbb Nright}$$
$sup A$ doesn't exist, because $A$ is not bounded above:
$$begin{align}n&=1 \
frac{m^4+2}{m^2+1} &le M \
frac{m^4+2}{m^2} &le M \
m^2+frac{2}{m^2} &le M \
m^2&le Mend{align}$$
But I can choose $m=lfloor M rfloor+1$ from Archimedean property.
I think $inf A=3/2$. Am I right?
UPDATE: for $m=3$ and $n=4$ we get $-113/2$
UPDATE2: I think $inf A$ doesn't exist too.
I have to prove that e.g:
$$forall _m exists _n:m^2n>n^2+2m^2$$
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
edited Nov 18 at 21:31
Lorenzo B.
1,6622519
1,6622519
asked Nov 17 at 21:44
matematiccc
1145
asked Nov 17 at 21:44
matematiccc
1145
asked Nov 17 at 21:44
matematiccc
1145
1145
2
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29
add a comment |
2
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29
2
2
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
For $n=3$ we have $frac{m^4+2n^2}{2m^2-m^2n+n^2}to-infty$ as $mto+infty$, hence $inf A=-infty$.
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For $n=3$ we have $frac{m^4+2n^2}{2m^2-m^2n+n^2}to-infty$ as $mto+infty$, hence $inf A=-infty$.
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
add a comment |
up vote
3
down vote
accepted
For $n=3$ we have $frac{m^4+2n^2}{2m^2-m^2n+n^2}to-infty$ as $mto+infty$, hence $inf A=-infty$.
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For $n=3$ we have $frac{m^4+2n^2}{2m^2-m^2n+n^2}to-infty$ as $mto+infty$, hence $inf A=-infty$.
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
For $n=3$ we have $frac{m^4+2n^2}{2m^2-m^2n+n^2}to-infty$ as $mto+infty$, hence $inf A=-infty$.
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
answered Nov 18 at 21:12
Fabio Lucchini
7,79311326
7,79311326
add a comment |
add a comment |
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2
About your first comment: then $;sup A=infty;$ ...isn't it? But you say "it doesn't exist"...and your question ends with $;sup A=frac32;$ . This is pretty confusing...and contradictory.
– DonAntonio
Nov 17 at 21:47
Sorry, It was only a typo.
– matematiccc
Nov 17 at 22:02
Do you have any idea how find infimum?
– matematiccc
Nov 17 at 23:29