I have a sphere $S^2$, how can I define it?
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I'm not sure how to say, using proper mathematical notation, a sphere $S^2$ is equal to the basic unit sphere $1=x^2 + y^2 + z^2$. Just looking for a quick answer, thank you.
notation
asked Aug 9 at 19:05
John Miller
926
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up vote
1
down vote
favorite
I'm not sure how to say, using proper mathematical notation, a sphere $S^2$ is equal to the basic unit sphere $1=x^2 + y^2 + z^2$. Just looking for a quick answer, thank you.
notation
asked Aug 9 at 19:05
John Miller
926
How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm not sure how to say, using proper mathematical notation, a sphere $S^2$ is equal to the basic unit sphere $1=x^2 + y^2 + z^2$. Just looking for a quick answer, thank you.
notation
asked Aug 9 at 19:05
John Miller
926
I'm not sure how to say, using proper mathematical notation, a sphere $S^2$ is equal to the basic unit sphere $1=x^2 + y^2 + z^2$. Just looking for a quick answer, thank you.
notation
asked Aug 9 at 19:05
John Miller
926
asked Aug 9 at 19:05
John Miller
926
asked Aug 9 at 19:05
John Miller
926
asked Aug 9 at 19:05
John Miller
926
926
How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10
add a comment |Â
How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10
How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10
How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Using set builder notation:
$$S=(x,y,z)in Bbb R^3mid x^2+y^2+z^2=1$$
If it is clear that the ambient space is $Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply
$$S=x^2+y^2+z^2=1$$
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
 |Â
show 5 more comments
up vote
0
down vote
Yes for $(x,y,z) in mathbbR^3$ the cartesian equation for a sphere centered at the origin and with radius $R$ is
$$x^2+y^2+z^2=R^2$$
and more in general with center in $C=(a,b,c)$
$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
Note that the equation is derived from Pytagorean Theorem, that is
$$R= sqrt(x-a)^2+(y-b)^2+(z-c)^2$$
answered Aug 9 at 19:07
gimusi
65.9k73684
add a comment |Â
up vote
0
down vote
There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).
See introduction here.
answered Aug 9 at 19:35
mvw
30.6k22251
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Using set builder notation:
$$S=(x,y,z)in Bbb R^3mid x^2+y^2+z^2=1$$
If it is clear that the ambient space is $Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply
$$S=x^2+y^2+z^2=1$$
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
 |Â
show 5 more comments
up vote
3
down vote
Using set builder notation:
$$S=(x,y,z)in Bbb R^3mid x^2+y^2+z^2=1$$
If it is clear that the ambient space is $Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply
$$S=x^2+y^2+z^2=1$$
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
 |Â
show 5 more comments
up vote
3
down vote
up vote
3
down vote
Using set builder notation:
$$S=(x,y,z)in Bbb R^3mid x^2+y^2+z^2=1$$
If it is clear that the ambient space is $Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply
$$S=x^2+y^2+z^2=1$$
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
Using set builder notation:
$$S=(x,y,z)in Bbb R^3mid x^2+y^2+z^2=1$$
If it is clear that the ambient space is $Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply
$$S=x^2+y^2+z^2=1$$
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
answered Aug 9 at 19:09
Arnaud Mortier
19.2k22159
19.2k22159
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
 |Â
show 5 more comments
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
Would it not be $x,y,z in Bbb R$?
– Tyler6
Aug 9 at 19:10
2
2
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $Bbb R^3$.
– Arnaud Mortier
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
@Tyler6 there is no difference between saying $(x,y,z)inBbb R^3$ and saying $xin Bbb R$ and $yinBbb R$ and $zin Bbb R$. Saying $x,y,zinBbb R$ is just shorthand for the second. $~x,y,zinBbb R,~x^2+y^2+z^2=1$ is the same set as written in the post above.
– JMoravitz
Aug 9 at 19:11
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector
– Tyler6
Aug 9 at 19:12
1
1
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
@Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$.
– Arnaud Mortier
Aug 9 at 19:13
 |Â
show 5 more comments
up vote
0
down vote
Yes for $(x,y,z) in mathbbR^3$ the cartesian equation for a sphere centered at the origin and with radius $R$ is
$$x^2+y^2+z^2=R^2$$
and more in general with center in $C=(a,b,c)$
$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
Note that the equation is derived from Pytagorean Theorem, that is
$$R= sqrt(x-a)^2+(y-b)^2+(z-c)^2$$
answered Aug 9 at 19:07
gimusi
65.9k73684
add a comment |Â
up vote
0
down vote
Yes for $(x,y,z) in mathbbR^3$ the cartesian equation for a sphere centered at the origin and with radius $R$ is
$$x^2+y^2+z^2=R^2$$
and more in general with center in $C=(a,b,c)$
$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
Note that the equation is derived from Pytagorean Theorem, that is
$$R= sqrt(x-a)^2+(y-b)^2+(z-c)^2$$
answered Aug 9 at 19:07
gimusi
65.9k73684
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes for $(x,y,z) in mathbbR^3$ the cartesian equation for a sphere centered at the origin and with radius $R$ is
$$x^2+y^2+z^2=R^2$$
and more in general with center in $C=(a,b,c)$
$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
Note that the equation is derived from Pytagorean Theorem, that is
$$R= sqrt(x-a)^2+(y-b)^2+(z-c)^2$$
answered Aug 9 at 19:07
gimusi
65.9k73684
Yes for $(x,y,z) in mathbbR^3$ the cartesian equation for a sphere centered at the origin and with radius $R$ is
$$x^2+y^2+z^2=R^2$$
and more in general with center in $C=(a,b,c)$
$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$
Note that the equation is derived from Pytagorean Theorem, that is
$$R= sqrt(x-a)^2+(y-b)^2+(z-c)^2$$
answered Aug 9 at 19:07
gimusi
65.9k73684
answered Aug 9 at 19:07
gimusi
65.9k73684
answered Aug 9 at 19:07
gimusi
65.9k73684
answered Aug 9 at 19:07
gimusi
65.9k73684
65.9k73684
add a comment |Â
add a comment |Â
up vote
0
down vote
There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).
See introduction here.
answered Aug 9 at 19:35
mvw
30.6k22251
add a comment |Â
up vote
0
down vote
There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).
See introduction here.
answered Aug 9 at 19:35
mvw
30.6k22251
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).
See introduction here.
answered Aug 9 at 19:35
mvw
30.6k22251
There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).
See introduction here.
answered Aug 9 at 19:35
mvw
30.6k22251
answered Aug 9 at 19:35
mvw
30.6k22251
answered Aug 9 at 19:35
mvw
30.6k22251
answered Aug 9 at 19:35
mvw
30.6k22251
30.6k22251
add a comment |Â
add a comment |Â
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How can you define it? You just have defined it!
– Dietrich Burde
Aug 9 at 19:10