Theory of drawing (multivariable) functions (I am new)












1












$begingroup$


My question is a bit different. When is it possible to draw a function? E.g a 4-dimensional vector space is no more imaginable. If we a function from $mathbb R^2$ to $mathbb R$, we can draw it 3- dimensional. If we use polar coordinates we could at least draw the codomain in $mathbb R^2$. So whats the theory behind it ? Until which point can i really visualize it? and when is it no more possible, would appreciate a long answer or at least link to an explanation






calculus analysis education







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$endgroup$












  • $begingroup$
    Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:45










  • $begingroup$
    i am talking about parametric equation. the codamain can be drawn on a plane.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 19:51






  • 1




    $begingroup$
    You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:53










  • $begingroup$
    polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:07










  • $begingroup$
    But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 20:21
















1












$begingroup$


My question is a bit different. When is it possible to draw a function? E.g a 4-dimensional vector space is no more imaginable. If we a function from $mathbb R^2$ to $mathbb R$, we can draw it 3- dimensional. If we use polar coordinates we could at least draw the codomain in $mathbb R^2$. So whats the theory behind it ? Until which point can i really visualize it? and when is it no more possible, would appreciate a long answer or at least link to an explanation






calculus analysis education







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:45










  • $begingroup$
    i am talking about parametric equation. the codamain can be drawn on a plane.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 19:51






  • 1




    $begingroup$
    You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:53










  • $begingroup$
    polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:07










  • $begingroup$
    But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 20:21














1












1








1





$begingroup$


My question is a bit different. When is it possible to draw a function? E.g a 4-dimensional vector space is no more imaginable. If we a function from $mathbb R^2$ to $mathbb R$, we can draw it 3- dimensional. If we use polar coordinates we could at least draw the codomain in $mathbb R^2$. So whats the theory behind it ? Until which point can i really visualize it? and when is it no more possible, would appreciate a long answer or at least link to an explanation






calculus analysis education







share|cite|improve this question











$endgroup$




My question is a bit different. When is it possible to draw a function? E.g a 4-dimensional vector space is no more imaginable. If we a function from $mathbb R^2$ to $mathbb R$, we can draw it 3- dimensional. If we use polar coordinates we could at least draw the codomain in $mathbb R^2$. So whats the theory behind it ? Until which point can i really visualize it? and when is it no more possible, would appreciate a long answer or at least link to an explanation






calculus analysis education




calculus analysis education






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 19:59









Andrei

11.6k21026




11.6k21026










asked Dec 16 '18 at 19:34









Ömer F. YiÖmer F. Yi

114




114












  • $begingroup$
    Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:45










  • $begingroup$
    i am talking about parametric equation. the codamain can be drawn on a plane.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 19:51






  • 1




    $begingroup$
    You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:53










  • $begingroup$
    polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:07










  • $begingroup$
    But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 20:21


















  • $begingroup$
    Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:45










  • $begingroup$
    i am talking about parametric equation. the codamain can be drawn on a plane.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 19:51






  • 1




    $begingroup$
    You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 19:53










  • $begingroup$
    polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:07










  • $begingroup$
    But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 20:21
















$begingroup$
Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
$endgroup$
– Noah Schweber
Dec 16 '18 at 19:45




$begingroup$
Can you clarify what you mean by "If we use polarcoordinates we could at least draw the Codomain in R^2."?
$endgroup$
– Noah Schweber
Dec 16 '18 at 19:45












$begingroup$
i am talking about parametric equation. the codamain can be drawn on a plane.
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 19:51




$begingroup$
i am talking about parametric equation. the codamain can be drawn on a plane.
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 19:51




1




1




$begingroup$
You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
$endgroup$
– Noah Schweber
Dec 16 '18 at 19:53




$begingroup$
You can do that with rectangular coordinates too. Can you give me an example of a function you need polar coordinates to draw?
$endgroup$
– Noah Schweber
Dec 16 '18 at 19:53












$begingroup$
polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 20:07




$begingroup$
polar coordinates are not what im asking for. this was only an example. to answer your question e^ix.
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 20:07












$begingroup$
But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
$endgroup$
– Noah Schweber
Dec 16 '18 at 20:21




$begingroup$
But that's no easier to draw in polar than rectangular (assuming $x$ is allowed to be an arbitrary complex number).
$endgroup$
– Noah Schweber
Dec 16 '18 at 20:21










1 Answer
1






active

oldest

votes


















0












$begingroup$

A function $f$ from $X$ to $Y$ should for these purposes be thought of as a subset of the Cartesian product of $X$ and $Y$ - that is, if we can first picture $Xtimes Y$ we can picture $f$ by thinking of it as a "shape" living in $Xtimes Y$.



The crucial obstacle now is the dimension. If $X=mathbb{R}^m$ and $Y=mathbb{R}^n$, then $Xtimes Y$ is basically the same as $mathbb{R}^{m+n}$ - this is why trying to graph a function from $mathbb{R}^2$ to $mathbb{R}^2$ seems to force one into four-dimensional space.



Changing coordinate systems doesn't really help, unless you adopt a truly evil coordinate system (in which case your picture is ruined anyways): for example, even the shift from rectangular to polar coordinates (contrary to what you seem to claim) keeps the dimension of the plane the same (we still use two numbers to describe a given point). Once $X$ and $Y$ have some reasonable notion of geometry, we've really determined what $Xtimes Y$ is going to look like, and in particular how complicated it will be. Switching from one "reasonable description" to another won't result in a drastic change, although it may improve some aspects of the picture (e.g. make it easier to figure out the bounds in some integral).



This is all a first step into topology, and more specifically the study of manifolds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:10










  • $begingroup$
    can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
    $endgroup$
    – Ömer F. Yi
    Dec 17 '18 at 1:17











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

A function $f$ from $X$ to $Y$ should for these purposes be thought of as a subset of the Cartesian product of $X$ and $Y$ - that is, if we can first picture $Xtimes Y$ we can picture $f$ by thinking of it as a "shape" living in $Xtimes Y$.



The crucial obstacle now is the dimension. If $X=mathbb{R}^m$ and $Y=mathbb{R}^n$, then $Xtimes Y$ is basically the same as $mathbb{R}^{m+n}$ - this is why trying to graph a function from $mathbb{R}^2$ to $mathbb{R}^2$ seems to force one into four-dimensional space.



Changing coordinate systems doesn't really help, unless you adopt a truly evil coordinate system (in which case your picture is ruined anyways): for example, even the shift from rectangular to polar coordinates (contrary to what you seem to claim) keeps the dimension of the plane the same (we still use two numbers to describe a given point). Once $X$ and $Y$ have some reasonable notion of geometry, we've really determined what $Xtimes Y$ is going to look like, and in particular how complicated it will be. Switching from one "reasonable description" to another won't result in a drastic change, although it may improve some aspects of the picture (e.g. make it easier to figure out the bounds in some integral).



This is all a first step into topology, and more specifically the study of manifolds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:10










  • $begingroup$
    can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
    $endgroup$
    – Ömer F. Yi
    Dec 17 '18 at 1:17
















0












$begingroup$

A function $f$ from $X$ to $Y$ should for these purposes be thought of as a subset of the Cartesian product of $X$ and $Y$ - that is, if we can first picture $Xtimes Y$ we can picture $f$ by thinking of it as a "shape" living in $Xtimes Y$.



The crucial obstacle now is the dimension. If $X=mathbb{R}^m$ and $Y=mathbb{R}^n$, then $Xtimes Y$ is basically the same as $mathbb{R}^{m+n}$ - this is why trying to graph a function from $mathbb{R}^2$ to $mathbb{R}^2$ seems to force one into four-dimensional space.



Changing coordinate systems doesn't really help, unless you adopt a truly evil coordinate system (in which case your picture is ruined anyways): for example, even the shift from rectangular to polar coordinates (contrary to what you seem to claim) keeps the dimension of the plane the same (we still use two numbers to describe a given point). Once $X$ and $Y$ have some reasonable notion of geometry, we've really determined what $Xtimes Y$ is going to look like, and in particular how complicated it will be. Switching from one "reasonable description" to another won't result in a drastic change, although it may improve some aspects of the picture (e.g. make it easier to figure out the bounds in some integral).



This is all a first step into topology, and more specifically the study of manifolds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:10










  • $begingroup$
    can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
    $endgroup$
    – Ömer F. Yi
    Dec 17 '18 at 1:17














0












0








0





$begingroup$

A function $f$ from $X$ to $Y$ should for these purposes be thought of as a subset of the Cartesian product of $X$ and $Y$ - that is, if we can first picture $Xtimes Y$ we can picture $f$ by thinking of it as a "shape" living in $Xtimes Y$.



The crucial obstacle now is the dimension. If $X=mathbb{R}^m$ and $Y=mathbb{R}^n$, then $Xtimes Y$ is basically the same as $mathbb{R}^{m+n}$ - this is why trying to graph a function from $mathbb{R}^2$ to $mathbb{R}^2$ seems to force one into four-dimensional space.



Changing coordinate systems doesn't really help, unless you adopt a truly evil coordinate system (in which case your picture is ruined anyways): for example, even the shift from rectangular to polar coordinates (contrary to what you seem to claim) keeps the dimension of the plane the same (we still use two numbers to describe a given point). Once $X$ and $Y$ have some reasonable notion of geometry, we've really determined what $Xtimes Y$ is going to look like, and in particular how complicated it will be. Switching from one "reasonable description" to another won't result in a drastic change, although it may improve some aspects of the picture (e.g. make it easier to figure out the bounds in some integral).



This is all a first step into topology, and more specifically the study of manifolds.






share|cite|improve this answer









$endgroup$



A function $f$ from $X$ to $Y$ should for these purposes be thought of as a subset of the Cartesian product of $X$ and $Y$ - that is, if we can first picture $Xtimes Y$ we can picture $f$ by thinking of it as a "shape" living in $Xtimes Y$.



The crucial obstacle now is the dimension. If $X=mathbb{R}^m$ and $Y=mathbb{R}^n$, then $Xtimes Y$ is basically the same as $mathbb{R}^{m+n}$ - this is why trying to graph a function from $mathbb{R}^2$ to $mathbb{R}^2$ seems to force one into four-dimensional space.



Changing coordinate systems doesn't really help, unless you adopt a truly evil coordinate system (in which case your picture is ruined anyways): for example, even the shift from rectangular to polar coordinates (contrary to what you seem to claim) keeps the dimension of the plane the same (we still use two numbers to describe a given point). Once $X$ and $Y$ have some reasonable notion of geometry, we've really determined what $Xtimes Y$ is going to look like, and in particular how complicated it will be. Switching from one "reasonable description" to another won't result in a drastic change, although it may improve some aspects of the picture (e.g. make it easier to figure out the bounds in some integral).



This is all a first step into topology, and more specifically the study of manifolds.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 19:45









Noah SchweberNoah Schweber

122k10149284




122k10149284












  • $begingroup$
    a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:10










  • $begingroup$
    can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
    $endgroup$
    – Ömer F. Yi
    Dec 17 '18 at 1:17


















  • $begingroup$
    a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
    $endgroup$
    – Ömer F. Yi
    Dec 16 '18 at 20:10










  • $begingroup$
    can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
    $endgroup$
    – Ömer F. Yi
    Dec 17 '18 at 1:17
















$begingroup$
a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 20:10




$begingroup$
a brilliant answer. so for example R^2 to R^2 is no more possible to draw for us as it is 4 dimensional. So all the things are a bit complicated if you abstract the knowledge and the whole aspect of drawing function becomes very difficult when talking about manifolds ?
$endgroup$
– Ömer F. Yi
Dec 16 '18 at 20:10












$begingroup$
can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
$endgroup$
– Ömer F. Yi
Dec 17 '18 at 1:17




$begingroup$
can i say that DOUBLE integral of f(x,y) dx dy is same as integral f(x)dx * integral f(y) dy genereally ? And another question is i have seen in many literature if you should integrate over R (real numbers) some choose to integrate from 0 to infinity doesn't it have to be from -infinity to +infinity ? could you also help me up with that ? thank you
$endgroup$
– Ömer F. Yi
Dec 17 '18 at 1:17


















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