bounding a certain quantity
$begingroup$
Is there a way to bound the following quantity
$$
|sqrt{a_n - b_n} - sqrt{a-b}|
$$
with an expression composed of the expressions $a_n-a$ and $b_n -b$?
inequality
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
$endgroup$
add a comment |
$begingroup$
Is there a way to bound the following quantity
$$
|sqrt{a_n - b_n} - sqrt{a-b}|
$$
with an expression composed of the expressions $a_n-a$ and $b_n -b$?
inequality
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
$endgroup$
$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40
add a comment |
$begingroup$
Is there a way to bound the following quantity
$$
|sqrt{a_n - b_n} - sqrt{a-b}|
$$
with an expression composed of the expressions $a_n-a$ and $b_n -b$?
inequality
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
$endgroup$
Is there a way to bound the following quantity
$$
|sqrt{a_n - b_n} - sqrt{a-b}|
$$
with an expression composed of the expressions $a_n-a$ and $b_n -b$?
inequality
inequality
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
asked Dec 29 '18 at 3:54
Julienne FranzJulienne Franz
1955
1955
$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40
add a comment |
$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40
$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is bounded by $frac {|a_n-a|+|b_b-a|} {sqrt {|a-b|}}$. You get this multiplying numerator and denominator by $sqrt{a_n-b_n}+sqrt {a-b}$.
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
You can use
$$ |sqrt{x} -sqrt{y}|^2 leqslant |sqrt{x}-sqrt{y}|, |sqrt{x} + sqrt{y}| = |x-y|$$
which implies
$$|sqrt{x}-sqrt{y}| leqslant sqrt{|x-y|}$$
where $x = a_n - b_n$ and $y = a-b$.
Thus,
$$|sqrt{a_n-b_n}-sqrt{a-b}| leqslant sqrt{|(a_n-a)-(b_n-b)|}$$
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
$endgroup$
add a comment |
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
It is bounded by $frac {|a_n-a|+|b_b-a|} {sqrt {|a-b|}}$. You get this multiplying numerator and denominator by $sqrt{a_n-b_n}+sqrt {a-b}$.
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
It is bounded by $frac {|a_n-a|+|b_b-a|} {sqrt {|a-b|}}$. You get this multiplying numerator and denominator by $sqrt{a_n-b_n}+sqrt {a-b}$.
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
It is bounded by $frac {|a_n-a|+|b_b-a|} {sqrt {|a-b|}}$. You get this multiplying numerator and denominator by $sqrt{a_n-b_n}+sqrt {a-b}$.
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
It is bounded by $frac {|a_n-a|+|b_b-a|} {sqrt {|a-b|}}$. You get this multiplying numerator and denominator by $sqrt{a_n-b_n}+sqrt {a-b}$.
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 29 '18 at 4:58
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
61.8k42262
add a comment |
add a comment |
$begingroup$
You can use
$$ |sqrt{x} -sqrt{y}|^2 leqslant |sqrt{x}-sqrt{y}|, |sqrt{x} + sqrt{y}| = |x-y|$$
which implies
$$|sqrt{x}-sqrt{y}| leqslant sqrt{|x-y|}$$
where $x = a_n - b_n$ and $y = a-b$.
Thus,
$$|sqrt{a_n-b_n}-sqrt{a-b}| leqslant sqrt{|(a_n-a)-(b_n-b)|}$$
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
$endgroup$
add a comment |
$begingroup$
You can use
$$ |sqrt{x} -sqrt{y}|^2 leqslant |sqrt{x}-sqrt{y}|, |sqrt{x} + sqrt{y}| = |x-y|$$
which implies
$$|sqrt{x}-sqrt{y}| leqslant sqrt{|x-y|}$$
where $x = a_n - b_n$ and $y = a-b$.
Thus,
$$|sqrt{a_n-b_n}-sqrt{a-b}| leqslant sqrt{|(a_n-a)-(b_n-b)|}$$
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
$endgroup$
add a comment |
$begingroup$
You can use
$$ |sqrt{x} -sqrt{y}|^2 leqslant |sqrt{x}-sqrt{y}|, |sqrt{x} + sqrt{y}| = |x-y|$$
which implies
$$|sqrt{x}-sqrt{y}| leqslant sqrt{|x-y|}$$
where $x = a_n - b_n$ and $y = a-b$.
Thus,
$$|sqrt{a_n-b_n}-sqrt{a-b}| leqslant sqrt{|(a_n-a)-(b_n-b)|}$$
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
$endgroup$
You can use
$$ |sqrt{x} -sqrt{y}|^2 leqslant |sqrt{x}-sqrt{y}|, |sqrt{x} + sqrt{y}| = |x-y|$$
which implies
$$|sqrt{x}-sqrt{y}| leqslant sqrt{|x-y|}$$
where $x = a_n - b_n$ and $y = a-b$.
Thus,
$$|sqrt{a_n-b_n}-sqrt{a-b}| leqslant sqrt{|(a_n-a)-(b_n-b)|}$$
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
edited Dec 29 '18 at 5:34
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
answered Dec 29 '18 at 5:19
RRLRRL
51.7k42573
51.7k42573
add a comment |
add a comment |
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$begingroup$
Is using $a_n + a$ and $b_n + b$ ok? Because that's possible by squaring and using AM-GM inequality.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:15
$begingroup$
I actually need the differences $a_n - a$ and $b_n - b$ on the right side of the resulting inequality.
$endgroup$
– Julienne Franz
Dec 29 '18 at 4:31
$begingroup$
If this is about sequences then try multiplying top and bottom by the "conjugate" $|sqrt{a_n - b_n} + sqrt{a-b}|$ and use the convergence of $a_n$ and $b_n$ to bound the denominator for sufficiently large $n$.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:40