Is the function $f(x,y)=begin {cases} ( x^3(i+1)-y^3(1-i))/(x^2+y^2) & x^2+y^2>0\0 & x=y=0...












1












$begingroup$


Is the function $f(x,y)=begin {cases} frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} & x^2+y^2>0\ 0 & x=y=0 end{cases}$ continuous?



I would like to prove that thos function isn't continuous with help from two series and show that the limz doesn't equals the limf(z), z=(x,y)...sory for my English and my writting...I hope someone gets this (:






general-topology functions







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$endgroup$








  • 1




    $begingroup$
    What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
    $endgroup$
    – Devin Murray
    Oct 23 '13 at 12:10










  • $begingroup$
    Noramal topology, and yes the mepping is going so from R^2 to C.
    $endgroup$
    – banahova
    Oct 23 '13 at 14:29
















1












$begingroup$


Is the function $f(x,y)=begin {cases} frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} & x^2+y^2>0\ 0 & x=y=0 end{cases}$ continuous?



I would like to prove that thos function isn't continuous with help from two series and show that the limz doesn't equals the limf(z), z=(x,y)...sory for my English and my writting...I hope someone gets this (:






general-topology functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
    $endgroup$
    – Devin Murray
    Oct 23 '13 at 12:10










  • $begingroup$
    Noramal topology, and yes the mepping is going so from R^2 to C.
    $endgroup$
    – banahova
    Oct 23 '13 at 14:29














1












1








1


1



$begingroup$


Is the function $f(x,y)=begin {cases} frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} & x^2+y^2>0\ 0 & x=y=0 end{cases}$ continuous?



I would like to prove that thos function isn't continuous with help from two series and show that the limz doesn't equals the limf(z), z=(x,y)...sory for my English and my writting...I hope someone gets this (:






general-topology functions







share|cite|improve this question











$endgroup$




Is the function $f(x,y)=begin {cases} frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} & x^2+y^2>0\ 0 & x=y=0 end{cases}$ continuous?



I would like to prove that thos function isn't continuous with help from two series and show that the limz doesn't equals the limf(z), z=(x,y)...sory for my English and my writting...I hope someone gets this (:






general-topology functions




general-topology functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 23 '13 at 12:01









Ross Millikan

297k23198371




297k23198371










asked Oct 23 '13 at 11:55









banahovabanahova

135




135








  • 1




    $begingroup$
    What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
    $endgroup$
    – Devin Murray
    Oct 23 '13 at 12:10










  • $begingroup$
    Noramal topology, and yes the mepping is going so from R^2 to C.
    $endgroup$
    – banahova
    Oct 23 '13 at 14:29














  • 1




    $begingroup$
    What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
    $endgroup$
    – Devin Murray
    Oct 23 '13 at 12:10










  • $begingroup$
    Noramal topology, and yes the mepping is going so from R^2 to C.
    $endgroup$
    – banahova
    Oct 23 '13 at 14:29








1




1




$begingroup$
What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
$endgroup$
– Devin Murray
Oct 23 '13 at 12:10




$begingroup$
What topology do you have on your spaces? What are your spaces? Are you going from $mathbb{R}^2$ to $mathbb{C}$?
$endgroup$
– Devin Murray
Oct 23 '13 at 12:10












$begingroup$
Noramal topology, and yes the mepping is going so from R^2 to C.
$endgroup$
– banahova
Oct 23 '13 at 14:29




$begingroup$
Noramal topology, and yes the mepping is going so from R^2 to C.
$endgroup$
– banahova
Oct 23 '13 at 14:29










1 Answer
1






active

oldest

votes


















1












$begingroup$

I assume that the domain of $f$ is $Bbb{R}^2$ with normal topology and range of $f$ is $Bbb{C}$ with normal topology. Take $x=rcos t$, $y=rsin t$. Then if $(x,y)neq (0,0)$ then
$$frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} = frac{r^3((1+i)cos^3 t - (1-i)sin^3t)}{r^2}$$



Since $|(1+i)cos^3 t - (1-i)sin^3t|le 2sqrt{2}$, we get $|f(x,y)|le 2sqrt{2}r=2sqrt{2}sqrt{x^2+y^2}$ for $(x,y)neq (0,0)$. It forces $lim_{(x,y)to(0,0)} f(x,y)=0=f(0,0)$ so $f$ is continuous at the origin. It is easy to check that $f$ is continuous at another points on $Bbb{R}^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
    $endgroup$
    – banahova
    Oct 23 '13 at 14:27










  • $begingroup$
    @banahova Can I see the proof of discontinuity of $f$ at the origin?
    $endgroup$
    – Hanul Jeon
    Oct 23 '13 at 14:29










  • $begingroup$
    Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
    $endgroup$
    – banahova
    Oct 24 '13 at 9:46










  • $begingroup$
    could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
    $endgroup$
    – banahova
    Oct 25 '13 at 9:07










  • $begingroup$
    @banahova What is E? You want to say $varepsilon$?
    $endgroup$
    – Hanul Jeon
    Oct 25 '13 at 9:08











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

I assume that the domain of $f$ is $Bbb{R}^2$ with normal topology and range of $f$ is $Bbb{C}$ with normal topology. Take $x=rcos t$, $y=rsin t$. Then if $(x,y)neq (0,0)$ then
$$frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} = frac{r^3((1+i)cos^3 t - (1-i)sin^3t)}{r^2}$$



Since $|(1+i)cos^3 t - (1-i)sin^3t|le 2sqrt{2}$, we get $|f(x,y)|le 2sqrt{2}r=2sqrt{2}sqrt{x^2+y^2}$ for $(x,y)neq (0,0)$. It forces $lim_{(x,y)to(0,0)} f(x,y)=0=f(0,0)$ so $f$ is continuous at the origin. It is easy to check that $f$ is continuous at another points on $Bbb{R}^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
    $endgroup$
    – banahova
    Oct 23 '13 at 14:27










  • $begingroup$
    @banahova Can I see the proof of discontinuity of $f$ at the origin?
    $endgroup$
    – Hanul Jeon
    Oct 23 '13 at 14:29










  • $begingroup$
    Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
    $endgroup$
    – banahova
    Oct 24 '13 at 9:46










  • $begingroup$
    could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
    $endgroup$
    – banahova
    Oct 25 '13 at 9:07










  • $begingroup$
    @banahova What is E? You want to say $varepsilon$?
    $endgroup$
    – Hanul Jeon
    Oct 25 '13 at 9:08
















1












$begingroup$

I assume that the domain of $f$ is $Bbb{R}^2$ with normal topology and range of $f$ is $Bbb{C}$ with normal topology. Take $x=rcos t$, $y=rsin t$. Then if $(x,y)neq (0,0)$ then
$$frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} = frac{r^3((1+i)cos^3 t - (1-i)sin^3t)}{r^2}$$



Since $|(1+i)cos^3 t - (1-i)sin^3t|le 2sqrt{2}$, we get $|f(x,y)|le 2sqrt{2}r=2sqrt{2}sqrt{x^2+y^2}$ for $(x,y)neq (0,0)$. It forces $lim_{(x,y)to(0,0)} f(x,y)=0=f(0,0)$ so $f$ is continuous at the origin. It is easy to check that $f$ is continuous at another points on $Bbb{R}^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
    $endgroup$
    – banahova
    Oct 23 '13 at 14:27










  • $begingroup$
    @banahova Can I see the proof of discontinuity of $f$ at the origin?
    $endgroup$
    – Hanul Jeon
    Oct 23 '13 at 14:29










  • $begingroup$
    Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
    $endgroup$
    – banahova
    Oct 24 '13 at 9:46










  • $begingroup$
    could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
    $endgroup$
    – banahova
    Oct 25 '13 at 9:07










  • $begingroup$
    @banahova What is E? You want to say $varepsilon$?
    $endgroup$
    – Hanul Jeon
    Oct 25 '13 at 9:08














1












1








1





$begingroup$

I assume that the domain of $f$ is $Bbb{R}^2$ with normal topology and range of $f$ is $Bbb{C}$ with normal topology. Take $x=rcos t$, $y=rsin t$. Then if $(x,y)neq (0,0)$ then
$$frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} = frac{r^3((1+i)cos^3 t - (1-i)sin^3t)}{r^2}$$



Since $|(1+i)cos^3 t - (1-i)sin^3t|le 2sqrt{2}$, we get $|f(x,y)|le 2sqrt{2}r=2sqrt{2}sqrt{x^2+y^2}$ for $(x,y)neq (0,0)$. It forces $lim_{(x,y)to(0,0)} f(x,y)=0=f(0,0)$ so $f$ is continuous at the origin. It is easy to check that $f$ is continuous at another points on $Bbb{R}^2$.






share|cite|improve this answer









$endgroup$



I assume that the domain of $f$ is $Bbb{R}^2$ with normal topology and range of $f$ is $Bbb{C}$ with normal topology. Take $x=rcos t$, $y=rsin t$. Then if $(x,y)neq (0,0)$ then
$$frac{ x^3(i+1)-y^3(1-i)}{x^2+y^2} = frac{r^3((1+i)cos^3 t - (1-i)sin^3t)}{r^2}$$



Since $|(1+i)cos^3 t - (1-i)sin^3t|le 2sqrt{2}$, we get $|f(x,y)|le 2sqrt{2}r=2sqrt{2}sqrt{x^2+y^2}$ for $(x,y)neq (0,0)$. It forces $lim_{(x,y)to(0,0)} f(x,y)=0=f(0,0)$ so $f$ is continuous at the origin. It is easy to check that $f$ is continuous at another points on $Bbb{R}^2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 23 '13 at 13:15









Hanul JeonHanul Jeon

17.7k42781




17.7k42781












  • $begingroup$
    I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
    $endgroup$
    – banahova
    Oct 23 '13 at 14:27










  • $begingroup$
    @banahova Can I see the proof of discontinuity of $f$ at the origin?
    $endgroup$
    – Hanul Jeon
    Oct 23 '13 at 14:29










  • $begingroup$
    Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
    $endgroup$
    – banahova
    Oct 24 '13 at 9:46










  • $begingroup$
    could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
    $endgroup$
    – banahova
    Oct 25 '13 at 9:07










  • $begingroup$
    @banahova What is E? You want to say $varepsilon$?
    $endgroup$
    – Hanul Jeon
    Oct 25 '13 at 9:08


















  • $begingroup$
    I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
    $endgroup$
    – banahova
    Oct 23 '13 at 14:27










  • $begingroup$
    @banahova Can I see the proof of discontinuity of $f$ at the origin?
    $endgroup$
    – Hanul Jeon
    Oct 23 '13 at 14:29










  • $begingroup$
    Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
    $endgroup$
    – banahova
    Oct 24 '13 at 9:46










  • $begingroup$
    could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
    $endgroup$
    – banahova
    Oct 25 '13 at 9:07










  • $begingroup$
    @banahova What is E? You want to say $varepsilon$?
    $endgroup$
    – Hanul Jeon
    Oct 25 '13 at 9:08
















$begingroup$
I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
$endgroup$
– banahova
Oct 23 '13 at 14:27




$begingroup$
I think that's corect so, but some people have shown that this function isn't continuous oO..and yes, assumtion is about the topology and etc is corect...
$endgroup$
– banahova
Oct 23 '13 at 14:27












$begingroup$
@banahova Can I see the proof of discontinuity of $f$ at the origin?
$endgroup$
– Hanul Jeon
Oct 23 '13 at 14:29




$begingroup$
@banahova Can I see the proof of discontinuity of $f$ at the origin?
$endgroup$
– Hanul Jeon
Oct 23 '13 at 14:29












$begingroup$
Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
$endgroup$
– banahova
Oct 24 '13 at 9:46




$begingroup$
Sorry but I couldn't find it. This was my mini homework. I am not so good in english sorry I would better explain it in my own language.
$endgroup$
– banahova
Oct 24 '13 at 9:46












$begingroup$
could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
$endgroup$
– banahova
Oct 25 '13 at 9:07




$begingroup$
could I look at your proof throught the definition of continuity? Can I set up the E=2^(1/2)ω because of the metrik |(x,y)|=r<ω that implicates |f(x,y)|<E than there exist a ω=E/(2^(1/2)) ??? Have I understood this correctly?
$endgroup$
– banahova
Oct 25 '13 at 9:07












$begingroup$
@banahova What is E? You want to say $varepsilon$?
$endgroup$
– Hanul Jeon
Oct 25 '13 at 9:08




$begingroup$
@banahova What is E? You want to say $varepsilon$?
$endgroup$
– Hanul Jeon
Oct 25 '13 at 9:08


















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