Applying square root inside a function lowers its order when it seemingly should not
$begingroup$
I have these two functions:
$f(x) = xsqrt{8-x^2}$
$g(x) = sqrt{8x^2 - x^4}$
If I factor out $x^2$ out of the root in $g(x)$, I will get:
$g(x) = sqrt{x^2(8-x^2)}$
If I attempt to apply the square root on $x^2$, I will be left with $g(x) = pm xsqrt{8-x^2}$, which is either identical to $f(x)$, or its opposite. Both of these results contradict the original function graph ($f(x)$ is red while $g(x)$ is blue. I offset them by a small value to make them distinct, as they were previously superimposed):
I know that if you "externally" simplify equations, you lower their order, and therefore alter them. Here, my simplification should not alter the order of the function, or so I thought...
Why is this happening?
real-analysis
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
I have these two functions:
$f(x) = xsqrt{8-x^2}$
$g(x) = sqrt{8x^2 - x^4}$
If I factor out $x^2$ out of the root in $g(x)$, I will get:
$g(x) = sqrt{x^2(8-x^2)}$
If I attempt to apply the square root on $x^2$, I will be left with $g(x) = pm xsqrt{8-x^2}$, which is either identical to $f(x)$, or its opposite. Both of these results contradict the original function graph ($f(x)$ is red while $g(x)$ is blue. I offset them by a small value to make them distinct, as they were previously superimposed):
I know that if you "externally" simplify equations, you lower their order, and therefore alter them. Here, my simplification should not alter the order of the function, or so I thought...
Why is this happening?
real-analysis
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
$endgroup$
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
6
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28
add a comment |
$begingroup$
I have these two functions:
$f(x) = xsqrt{8-x^2}$
$g(x) = sqrt{8x^2 - x^4}$
If I factor out $x^2$ out of the root in $g(x)$, I will get:
$g(x) = sqrt{x^2(8-x^2)}$
If I attempt to apply the square root on $x^2$, I will be left with $g(x) = pm xsqrt{8-x^2}$, which is either identical to $f(x)$, or its opposite. Both of these results contradict the original function graph ($f(x)$ is red while $g(x)$ is blue. I offset them by a small value to make them distinct, as they were previously superimposed):
I know that if you "externally" simplify equations, you lower their order, and therefore alter them. Here, my simplification should not alter the order of the function, or so I thought...
Why is this happening?
real-analysis
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
$endgroup$
I have these two functions:
$f(x) = xsqrt{8-x^2}$
$g(x) = sqrt{8x^2 - x^4}$
If I factor out $x^2$ out of the root in $g(x)$, I will get:
$g(x) = sqrt{x^2(8-x^2)}$
If I attempt to apply the square root on $x^2$, I will be left with $g(x) = pm xsqrt{8-x^2}$, which is either identical to $f(x)$, or its opposite. Both of these results contradict the original function graph ($f(x)$ is red while $g(x)$ is blue. I offset them by a small value to make them distinct, as they were previously superimposed):
I know that if you "externally" simplify equations, you lower their order, and therefore alter them. Here, my simplification should not alter the order of the function, or so I thought...
Why is this happening?
real-analysis
real-analysis
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
edited Nov 29 '18 at 19:27
daedsidog
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
asked Nov 29 '18 at 19:11
daedsidogdaedsidog
29017
29017
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
6
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28
add a comment |
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
6
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
6
6
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that :
$$g(x) = sqrt{x^2(8-x^2)} = |x|sqrt{8-x^2} = begin{cases} xsqrt{8-x^2} ; ; ;, ; xgeq 0 \ -xsqrt{8-x^2}, ; x<0 end{cases}$$
Whereas $f(x) = xsqrt{8-x^2}$ for all $x in mathbb R$.
Thus, they are not defined in the same way and this way $g(x)$ will always be positive (which is easy to see anyway, since a rooted value is $geq 0$).
Credit to Peter Szilas as well for mentioning the same thing, I hadn't seen the comments earlier when answering.
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
$endgroup$
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
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%2f3019056%2fapplying-square-root-inside-a-function-lowers-its-order-when-it-seemingly-should%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that :
$$g(x) = sqrt{x^2(8-x^2)} = |x|sqrt{8-x^2} = begin{cases} xsqrt{8-x^2} ; ; ;, ; xgeq 0 \ -xsqrt{8-x^2}, ; x<0 end{cases}$$
Whereas $f(x) = xsqrt{8-x^2}$ for all $x in mathbb R$.
Thus, they are not defined in the same way and this way $g(x)$ will always be positive (which is easy to see anyway, since a rooted value is $geq 0$).
Credit to Peter Szilas as well for mentioning the same thing, I hadn't seen the comments earlier when answering.
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
$endgroup$
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
add a comment |
$begingroup$
Note that :
$$g(x) = sqrt{x^2(8-x^2)} = |x|sqrt{8-x^2} = begin{cases} xsqrt{8-x^2} ; ; ;, ; xgeq 0 \ -xsqrt{8-x^2}, ; x<0 end{cases}$$
Whereas $f(x) = xsqrt{8-x^2}$ for all $x in mathbb R$.
Thus, they are not defined in the same way and this way $g(x)$ will always be positive (which is easy to see anyway, since a rooted value is $geq 0$).
Credit to Peter Szilas as well for mentioning the same thing, I hadn't seen the comments earlier when answering.
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
$endgroup$
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
add a comment |
$begingroup$
Note that :
$$g(x) = sqrt{x^2(8-x^2)} = |x|sqrt{8-x^2} = begin{cases} xsqrt{8-x^2} ; ; ;, ; xgeq 0 \ -xsqrt{8-x^2}, ; x<0 end{cases}$$
Whereas $f(x) = xsqrt{8-x^2}$ for all $x in mathbb R$.
Thus, they are not defined in the same way and this way $g(x)$ will always be positive (which is easy to see anyway, since a rooted value is $geq 0$).
Credit to Peter Szilas as well for mentioning the same thing, I hadn't seen the comments earlier when answering.
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
$endgroup$
Note that :
$$g(x) = sqrt{x^2(8-x^2)} = |x|sqrt{8-x^2} = begin{cases} xsqrt{8-x^2} ; ; ;, ; xgeq 0 \ -xsqrt{8-x^2}, ; x<0 end{cases}$$
Whereas $f(x) = xsqrt{8-x^2}$ for all $x in mathbb R$.
Thus, they are not defined in the same way and this way $g(x)$ will always be positive (which is easy to see anyway, since a rooted value is $geq 0$).
Credit to Peter Szilas as well for mentioning the same thing, I hadn't seen the comments earlier when answering.
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
edited Nov 29 '18 at 21:26
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
answered Nov 29 '18 at 19:21
RebellosRebellos
14.6k31247
14.6k31247
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
add a comment |
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
Rebellos.Thanks, very small matter, enjoy reading your posts:)Greetings.
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:25
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
$begingroup$
@PeterSzilas Thanks mate ! What did you mean by "very small matter" ? Greetings from Greece !
$endgroup$
– Rebellos
Nov 29 '18 at 21:26
1
1
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
$begingroup$
Rebellos.$|x|=sqrt{x^2}$,is not such a big contribution:)Greetings, from Larnaca(Cyprus)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:32
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%2f3019056%2fapplying-square-root-inside-a-function-lowers-its-order-when-it-seemingly-should%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
$begingroup$
It is evident that $f(0)=0$, contradicting what the graph is showing. Thus I would conclude that the graph is incorrect. It seems that the graph of $f$ has been lifted a bit, but the reason why this has been done is a mistery.
$endgroup$
– Crostul
Nov 29 '18 at 19:15
6
$begingroup$
$sqrt{x^2}=|x|$. $f(x) not =g(x).$
$endgroup$
– Peter Szilas
Nov 29 '18 at 19:16
$begingroup$
The graph is incorrect because I offset it by a slight margin. The functions were superimposed.
$endgroup$
– daedsidog
Nov 29 '18 at 19:16
$begingroup$
@PeterSzilas That makes it all clear, but shouldn't you post that as an answer?
$endgroup$
– daedsidog
Nov 29 '18 at 19:21
$begingroup$
deadsidog.Small matter:)
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:28