Smallest integers $alpha$ and $beta$ such that $x+beta leq sqrt{x^2+4} leq x+alpha$ for every $x in...
$begingroup$
let $alpha$ and $beta$ integers and define $f(x) = sqrt{x^2+4}$ with $x in [0,infty[$ , I want the smallest integers $alpha$ and $beta$ for which :
$x+beta leq f(x)leq x+alpha$ , My attempt is this choice : $alpha =beta=0$ yield to $|f(x)|=x $ which is not true , any way ?
real-analysis inequality
edited Dec 28 '18 at 17:26
Did
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
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closed as off-topic by Did, Namaste, mrtaurho, Cesareo, Eevee Trainer Dec 29 '18 at 0:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Namaste, mrtaurho, Cesareo, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
let $alpha$ and $beta$ integers and define $f(x) = sqrt{x^2+4}$ with $x in [0,infty[$ , I want the smallest integers $alpha$ and $beta$ for which :
$x+beta leq f(x)leq x+alpha$ , My attempt is this choice : $alpha =beta=0$ yield to $|f(x)|=x $ which is not true , any way ?
real-analysis inequality
edited Dec 28 '18 at 17:26
Did
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
$endgroup$
closed as off-topic by Did, Namaste, mrtaurho, Cesareo, Eevee Trainer Dec 29 '18 at 0:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Namaste, mrtaurho, Cesareo, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
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– tilper
Dec 28 '18 at 16:36
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Have you tried graphing $y=sqrt{x^2+4}$?
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– Steve Kass
Dec 28 '18 at 16:37
1
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10
|
show 3 more comments
$begingroup$
let $alpha$ and $beta$ integers and define $f(x) = sqrt{x^2+4}$ with $x in [0,infty[$ , I want the smallest integers $alpha$ and $beta$ for which :
$x+beta leq f(x)leq x+alpha$ , My attempt is this choice : $alpha =beta=0$ yield to $|f(x)|=x $ which is not true , any way ?
real-analysis inequality
edited Dec 28 '18 at 17:26
Did
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
$endgroup$
let $alpha$ and $beta$ integers and define $f(x) = sqrt{x^2+4}$ with $x in [0,infty[$ , I want the smallest integers $alpha$ and $beta$ for which :
$x+beta leq f(x)leq x+alpha$ , My attempt is this choice : $alpha =beta=0$ yield to $|f(x)|=x $ which is not true , any way ?
real-analysis inequality
real-analysis inequality
edited Dec 28 '18 at 17:26
Did
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
edited Dec 28 '18 at 17:26
Did
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
edited Dec 28 '18 at 17:26
Did
248k23225464
edited Dec 28 '18 at 17:26
Did
248k23225464
edited Dec 28 '18 at 17:26
Did
248k23225464
248k23225464
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
asked Dec 28 '18 at 16:27
zeraoulia rafikzeraoulia rafik
2,43411032
2,43411032
closed as off-topic by Did, Namaste, mrtaurho, Cesareo, Eevee Trainer Dec 29 '18 at 0:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Namaste, mrtaurho, Cesareo, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Namaste, mrtaurho, Cesareo, Eevee Trainer Dec 29 '18 at 0:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Namaste, mrtaurho, Cesareo, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
$endgroup$
– tilper
Dec 28 '18 at 16:36
$begingroup$
Have you tried graphing $y=sqrt{x^2+4}$?
$endgroup$
– Steve Kass
Dec 28 '18 at 16:37
1
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10
|
show 3 more comments
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
$endgroup$
– tilper
Dec 28 '18 at 16:36
$begingroup$
Have you tried graphing $y=sqrt{x^2+4}$?
$endgroup$
– Steve Kass
Dec 28 '18 at 16:37
1
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
$endgroup$
– tilper
Dec 28 '18 at 16:36
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
$endgroup$
– tilper
Dec 28 '18 at 16:36
$begingroup$
Have you tried graphing $y=sqrt{x^2+4}$?
$endgroup$
– Steve Kass
Dec 28 '18 at 16:37
$begingroup$
Have you tried graphing $y=sqrt{x^2+4}$?
$endgroup$
– Steve Kass
Dec 28 '18 at 16:37
1
1
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10
|
show 3 more comments
2 Answers
2
active
oldest
votes
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If you want that to be true over the entire domain $[0,infty)$ then it is clear that $alphageq 2$. You can prove that $alpha=2$ works by squaring the right inequality. If you wish to force $beta$ to be the "smallest integer" (which I assume you really mean largest in the case) that works for the left inequality, then you will find by similar argumentation that $beta=0$.
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
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add a comment |
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$x+beta leq sqrt{x^2+4} leq x+alpha$ is equivalent to $$sqrt{x^2+4}-xin [beta, alpha]$$
Since $sqrt{x^2+4}-x=frac{4}{sqrt{x^2+4}+x}$ is strictly decreasing and $sqrt{x^2+4}-xxrightarrow{xto0^+} 2$, we conclude that $alpha = 2$.
On the other hand, we also see $0 le sqrt{x^2+4}-xxrightarrow{xtoinfty} 0$ so $beta = 0$.
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want that to be true over the entire domain $[0,infty)$ then it is clear that $alphageq 2$. You can prove that $alpha=2$ works by squaring the right inequality. If you wish to force $beta$ to be the "smallest integer" (which I assume you really mean largest in the case) that works for the left inequality, then you will find by similar argumentation that $beta=0$.
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
$endgroup$
add a comment |
$begingroup$
If you want that to be true over the entire domain $[0,infty)$ then it is clear that $alphageq 2$. You can prove that $alpha=2$ works by squaring the right inequality. If you wish to force $beta$ to be the "smallest integer" (which I assume you really mean largest in the case) that works for the left inequality, then you will find by similar argumentation that $beta=0$.
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
$endgroup$
add a comment |
$begingroup$
If you want that to be true over the entire domain $[0,infty)$ then it is clear that $alphageq 2$. You can prove that $alpha=2$ works by squaring the right inequality. If you wish to force $beta$ to be the "smallest integer" (which I assume you really mean largest in the case) that works for the left inequality, then you will find by similar argumentation that $beta=0$.
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
$endgroup$
If you want that to be true over the entire domain $[0,infty)$ then it is clear that $alphageq 2$. You can prove that $alpha=2$ works by squaring the right inequality. If you wish to force $beta$ to be the "smallest integer" (which I assume you really mean largest in the case) that works for the left inequality, then you will find by similar argumentation that $beta=0$.
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
answered Dec 28 '18 at 16:33
DudeManDudeMan
1113
1113
add a comment |
add a comment |
$begingroup$
$x+beta leq sqrt{x^2+4} leq x+alpha$ is equivalent to $$sqrt{x^2+4}-xin [beta, alpha]$$
Since $sqrt{x^2+4}-x=frac{4}{sqrt{x^2+4}+x}$ is strictly decreasing and $sqrt{x^2+4}-xxrightarrow{xto0^+} 2$, we conclude that $alpha = 2$.
On the other hand, we also see $0 le sqrt{x^2+4}-xxrightarrow{xtoinfty} 0$ so $beta = 0$.
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
$endgroup$
add a comment |
$begingroup$
$x+beta leq sqrt{x^2+4} leq x+alpha$ is equivalent to $$sqrt{x^2+4}-xin [beta, alpha]$$
Since $sqrt{x^2+4}-x=frac{4}{sqrt{x^2+4}+x}$ is strictly decreasing and $sqrt{x^2+4}-xxrightarrow{xto0^+} 2$, we conclude that $alpha = 2$.
On the other hand, we also see $0 le sqrt{x^2+4}-xxrightarrow{xtoinfty} 0$ so $beta = 0$.
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
$endgroup$
add a comment |
$begingroup$
$x+beta leq sqrt{x^2+4} leq x+alpha$ is equivalent to $$sqrt{x^2+4}-xin [beta, alpha]$$
Since $sqrt{x^2+4}-x=frac{4}{sqrt{x^2+4}+x}$ is strictly decreasing and $sqrt{x^2+4}-xxrightarrow{xto0^+} 2$, we conclude that $alpha = 2$.
On the other hand, we also see $0 le sqrt{x^2+4}-xxrightarrow{xtoinfty} 0$ so $beta = 0$.
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
$endgroup$
$x+beta leq sqrt{x^2+4} leq x+alpha$ is equivalent to $$sqrt{x^2+4}-xin [beta, alpha]$$
Since $sqrt{x^2+4}-x=frac{4}{sqrt{x^2+4}+x}$ is strictly decreasing and $sqrt{x^2+4}-xxrightarrow{xto0^+} 2$, we conclude that $alpha = 2$.
On the other hand, we also see $0 le sqrt{x^2+4}-xxrightarrow{xtoinfty} 0$ so $beta = 0$.
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
answered Dec 28 '18 at 17:01
mechanodroidmechanodroid
28.8k62548
28.8k62548
add a comment |
add a comment |
$begingroup$
@MohammadZuhairKhan, $-infty notin Bbb Z$
$endgroup$
– tilper
Dec 28 '18 at 16:36
$begingroup$
Have you tried graphing $y=sqrt{x^2+4}$?
$endgroup$
– Steve Kass
Dec 28 '18 at 16:37
1
$begingroup$
Do you want the largest value of $beta?$
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 16:44
$begingroup$
You probably want the maximal $beta$ and the minimal $alpha$.
$endgroup$
– anomaly
Dec 28 '18 at 16:44
$begingroup$
desmos.com/calculator/qo30nzfk3y
$endgroup$
– Mason
Dec 28 '18 at 17:10