How does squaring give you a monotonic transformation?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
324111
add a comment |
add a comment |
2 Answers
2
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.
$endgroup$
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 $[0,+infty)$.
$endgroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3114891%2fhow-does-squaring-give-you-a-monotonic-transformation%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
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 $[0,+infty)$.
$endgroup$
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 $[0,+infty)$.
$endgroup$
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 $[0,+infty)$.
$endgroup$
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 $[0,+infty)$.
edited Feb 16 at 10:35
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3114891%2fhow-does-squaring-give-you-a-monotonic-transformation%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown