Applying square root inside a function lowers its order when it seemingly should not












0












$begingroup$


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):



enter image description here



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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):



enter image description here



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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):



enter image description here



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







share|cite|improve this question











$endgroup$




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):



enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 19:27







daedsidog

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 21:26

























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


















  • $begingroup$
    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


















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