How does squaring give you a monotonic transformation?
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Consider the function: $$ f(x,y) = sqrt xy $$ Is the function $$ f_1(x,y) = x^2 y^2 $$ a monotonic transformation of $ f $?
I remember studying earlier that squaring does not give you a monotonic transformation since the order will not be preserved for negative values of, say, $x $. But my textbook says that $ f_1 $ is a monotonic transformation.
I understand that we're actually squaring twice here -- but that still won't preserve the order for negative values, right?
transformation
asked Feb 16 at 10:30
WorldGovWorldGov
324111
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add a comment |
$begingroup$
Consider the function: $$ f(x,y) = sqrt xy $$ Is the function $$ f_1(x,y) = x^2 y^2 $$ a monotonic transformation of $ f $?
I remember studying earlier that squaring does not give you a monotonic transformation since the order will not be preserved for negative values of, say, $x $. But my textbook says that $ f_1 $ is a monotonic transformation.
I understand that we're actually squaring twice here -- but that still won't preserve the order for negative values, right?
transformation
asked Feb 16 at 10:30
WorldGovWorldGov
324111
$endgroup$
add a comment |
$begingroup$
Consider the function: $$ f(x,y) = sqrt xy $$ Is the function $$ f_1(x,y) = x^2 y^2 $$ a monotonic transformation of $ f $?
I remember studying earlier that squaring does not give you a monotonic transformation since the order will not be preserved for negative values of, say, $x $. But my textbook says that $ f_1 $ is a monotonic transformation.
I understand that we're actually squaring twice here -- but that still won't preserve the order for negative values, right?
transformation
asked Feb 16 at 10:30
WorldGovWorldGov
324111
$endgroup$
Consider the function: $$ f(x,y) = sqrt xy $$ Is the function $$ f_1(x,y) = x^2 y^2 $$ a monotonic transformation of $ f $?
I remember studying earlier that squaring does not give you a monotonic transformation since the order will not be preserved for negative values of, say, $x $. But my textbook says that $ f_1 $ is a monotonic transformation.
I understand that we're actually squaring twice here -- but that still won't preserve the order for negative values, right?
transformation
transformation
asked Feb 16 at 10:30
WorldGovWorldGov
324111
asked Feb 16 at 10:30
WorldGovWorldGov
324111
asked Feb 16 at 10:30
WorldGovWorldGov
324111
asked Feb 16 at 10:30
WorldGovWorldGov
324111
asked Feb 16 at 10:30
WorldGovWorldGov
324111
324111
add a comment |
add a comment |
2 Answers
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There are no negative values. Under the usual interpretation of the $sqrt $ symbol, $sqrt xy $ is zero or the positive square root of $xy $. And on $[0,infty) $, the map $tlongmapsto t^2$ is monotone.
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
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add a comment |
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There are no negative values in the first place since the object you are squaring is the square root of something. The symbol $sqrt .$ is always assumed to output a non-negative number.
Note that the map $$xmapsto x^2$$ is monotonic if the domain is defined to be $[0,+infty)$.
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
$endgroup$
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are no negative values. Under the usual interpretation of the $sqrt $ symbol, $sqrt xy $ is zero or the positive square root of $xy $. And on $[0,infty) $, the map $tlongmapsto t^2$ is monotone.
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
There are no negative values. Under the usual interpretation of the $sqrt $ symbol, $sqrt xy $ is zero or the positive square root of $xy $. And on $[0,infty) $, the map $tlongmapsto t^2$ is monotone.
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
There are no negative values. Under the usual interpretation of the $sqrt $ symbol, $sqrt xy $ is zero or the positive square root of $xy $. And on $[0,infty) $, the map $tlongmapsto t^2$ is monotone.
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
There are no negative values. Under the usual interpretation of the $sqrt $ symbol, $sqrt xy $ is zero or the positive square root of $xy $. And on $[0,infty) $, the map $tlongmapsto t^2$ is monotone.
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
answered Feb 16 at 10:37
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
add a comment |
add a comment |
$begingroup$
There are no negative values in the first place since the object you are squaring is the square root of something. The symbol $sqrt .$ is always assumed to output a non-negative number.
Note that the map $$xmapsto x^2$$ is monotonic if the domain is defined to be $[0,+infty)$.
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
$endgroup$
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
add a comment |
$begingroup$
There are no negative values in the first place since the object you are squaring is the square root of something. The symbol $sqrt .$ is always assumed to output a non-negative number.
Note that the map $$xmapsto x^2$$ is monotonic if the domain is defined to be $[0,+infty)$.
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
$endgroup$
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
add a comment |
$begingroup$
There are no negative values in the first place since the object you are squaring is the square root of something. The symbol $sqrt .$ is always assumed to output a non-negative number.
Note that the map $$xmapsto x^2$$ is monotonic if the domain is defined to be $[0,+infty)$.
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
$endgroup$
There are no negative values in the first place since the object you are squaring is the square root of something. The symbol $sqrt .$ is always assumed to output a non-negative number.
Note that the map $$xmapsto x^2$$ is monotonic if the domain is defined to be $[0,+infty)$.
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
edited Feb 16 at 10:35
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
answered Feb 16 at 10:33
Arnaud MortierArnaud Mortier
20.2k22262
20.2k22262
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
add a comment |
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
1
1
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
... non-negaitve
$endgroup$
– Hagen von Eitzen
Feb 16 at 10:34
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
$begingroup$
@HagenvonEitzen No I assure you, it's really non-negative - check it out: en.wiktionary.org/wiki/non-negative
$endgroup$
– Arnaud Mortier
Feb 16 at 11:18
add a comment |
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