How can I prove non-geometrically that there is a bijective correspondence between polar and cartesian...












1












$begingroup$


We have a function $f: mathbb{R}^2 rightarrow mathbb{R}^2$ as $f(x,y) = (sqrt{x^2 + y^2}$, $tan^{-1}left(frac{y}{x}right))$ which takes a Cartesian pair $(x,y)$ to its polar form, and a function $g: mathbb{R}^2 rightarrow mathbb{R}^2$ as $g(r,theta) = (rcostheta, rsintheta)$, which takes a polar pair $(r, theta)$ to its Cartesian representation.



I'm having trouble proving that there's a bijective correspondence between the two representations. It is easy to show, for instance, that $f circ g = text{id}_{mathbb{R}^2}$, but not quite as easy to show the other direction.



Any hints or help would be appreciated.



--



Edit: I have also tried messing around with injectivity and surjectivity between the two functions, so that the one-way composition-identity equality suffices to prove that the two functions are inverses. But I run into the same problems, which the below answer resolves by using trig identities. Is there a way to circumvent this?






coordinate-systems polar-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you restricted $r$ and $theta$ at all?
    $endgroup$
    – pjs36
    Feb 21 '15 at 20:42










  • $begingroup$
    Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:44






  • 1




    $begingroup$
    There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
    $endgroup$
    – Matt Samuel
    Feb 21 '15 at 20:45










  • $begingroup$
    The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:45










  • $begingroup$
    Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:53
















1












$begingroup$


We have a function $f: mathbb{R}^2 rightarrow mathbb{R}^2$ as $f(x,y) = (sqrt{x^2 + y^2}$, $tan^{-1}left(frac{y}{x}right))$ which takes a Cartesian pair $(x,y)$ to its polar form, and a function $g: mathbb{R}^2 rightarrow mathbb{R}^2$ as $g(r,theta) = (rcostheta, rsintheta)$, which takes a polar pair $(r, theta)$ to its Cartesian representation.



I'm having trouble proving that there's a bijective correspondence between the two representations. It is easy to show, for instance, that $f circ g = text{id}_{mathbb{R}^2}$, but not quite as easy to show the other direction.



Any hints or help would be appreciated.



--



Edit: I have also tried messing around with injectivity and surjectivity between the two functions, so that the one-way composition-identity equality suffices to prove that the two functions are inverses. But I run into the same problems, which the below answer resolves by using trig identities. Is there a way to circumvent this?






coordinate-systems polar-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you restricted $r$ and $theta$ at all?
    $endgroup$
    – pjs36
    Feb 21 '15 at 20:42










  • $begingroup$
    Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:44






  • 1




    $begingroup$
    There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
    $endgroup$
    – Matt Samuel
    Feb 21 '15 at 20:45










  • $begingroup$
    The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:45










  • $begingroup$
    Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:53














1












1








1





$begingroup$


We have a function $f: mathbb{R}^2 rightarrow mathbb{R}^2$ as $f(x,y) = (sqrt{x^2 + y^2}$, $tan^{-1}left(frac{y}{x}right))$ which takes a Cartesian pair $(x,y)$ to its polar form, and a function $g: mathbb{R}^2 rightarrow mathbb{R}^2$ as $g(r,theta) = (rcostheta, rsintheta)$, which takes a polar pair $(r, theta)$ to its Cartesian representation.



I'm having trouble proving that there's a bijective correspondence between the two representations. It is easy to show, for instance, that $f circ g = text{id}_{mathbb{R}^2}$, but not quite as easy to show the other direction.



Any hints or help would be appreciated.



--



Edit: I have also tried messing around with injectivity and surjectivity between the two functions, so that the one-way composition-identity equality suffices to prove that the two functions are inverses. But I run into the same problems, which the below answer resolves by using trig identities. Is there a way to circumvent this?






coordinate-systems polar-coordinates







share|cite|improve this question











$endgroup$




We have a function $f: mathbb{R}^2 rightarrow mathbb{R}^2$ as $f(x,y) = (sqrt{x^2 + y^2}$, $tan^{-1}left(frac{y}{x}right))$ which takes a Cartesian pair $(x,y)$ to its polar form, and a function $g: mathbb{R}^2 rightarrow mathbb{R}^2$ as $g(r,theta) = (rcostheta, rsintheta)$, which takes a polar pair $(r, theta)$ to its Cartesian representation.



I'm having trouble proving that there's a bijective correspondence between the two representations. It is easy to show, for instance, that $f circ g = text{id}_{mathbb{R}^2}$, but not quite as easy to show the other direction.



Any hints or help would be appreciated.



--



Edit: I have also tried messing around with injectivity and surjectivity between the two functions, so that the one-way composition-identity equality suffices to prove that the two functions are inverses. But I run into the same problems, which the below answer resolves by using trig identities. Is there a way to circumvent this?






coordinate-systems polar-coordinates




coordinate-systems polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 21 '15 at 21:04







jwalk

















asked Feb 21 '15 at 20:41









jwalkjwalk

1266




1266












  • $begingroup$
    Have you restricted $r$ and $theta$ at all?
    $endgroup$
    – pjs36
    Feb 21 '15 at 20:42










  • $begingroup$
    Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:44






  • 1




    $begingroup$
    There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
    $endgroup$
    – Matt Samuel
    Feb 21 '15 at 20:45










  • $begingroup$
    The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:45










  • $begingroup$
    Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:53


















  • $begingroup$
    Have you restricted $r$ and $theta$ at all?
    $endgroup$
    – pjs36
    Feb 21 '15 at 20:42










  • $begingroup$
    Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:44






  • 1




    $begingroup$
    There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
    $endgroup$
    – Matt Samuel
    Feb 21 '15 at 20:45










  • $begingroup$
    The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:45










  • $begingroup$
    Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
    $endgroup$
    – copper.hat
    Feb 21 '15 at 20:53
















$begingroup$
Have you restricted $r$ and $theta$ at all?
$endgroup$
– pjs36
Feb 21 '15 at 20:42




$begingroup$
Have you restricted $r$ and $theta$ at all?
$endgroup$
– pjs36
Feb 21 '15 at 20:42












$begingroup$
Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
$endgroup$
– jwalk
Feb 21 '15 at 20:44




$begingroup$
Yes, I also have assumed that $rge 0$ and $-pi < theta < pi$.
$endgroup$
– jwalk
Feb 21 '15 at 20:44




1




1




$begingroup$
There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
$endgroup$
– Matt Samuel
Feb 21 '15 at 20:45




$begingroup$
There is no bijective correspondence if you allow $r=0$. If $r=0$ then $theta$ can be anything without changing the point.
$endgroup$
– Matt Samuel
Feb 21 '15 at 20:45












$begingroup$
The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
$endgroup$
– copper.hat
Feb 21 '15 at 20:45




$begingroup$
The function is not defined for $x=0$. You need to use something like $operatorname{arg}$ (or $operatorname{atan2}$) instead, and you need to remove $(0,0)$ from the domain.
$endgroup$
– copper.hat
Feb 21 '15 at 20:45












$begingroup$
Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
$endgroup$
– copper.hat
Feb 21 '15 at 20:53




$begingroup$
Note that the way $f_2$ is defined does not distinguish between $(x,y)$ and $-(x,y)$.
$endgroup$
– copper.hat
Feb 21 '15 at 20:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Substitute $r= sqrt{x^2+y^2}$ and $theta = arctan(frac{y}{x})$ into $g$ and use the Relations for inverse trigonometric functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:51











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Substitute $r= sqrt{x^2+y^2}$ and $theta = arctan(frac{y}{x})$ into $g$ and use the Relations for inverse trigonometric functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:51
















2












$begingroup$

Substitute $r= sqrt{x^2+y^2}$ and $theta = arctan(frac{y}{x})$ into $g$ and use the Relations for inverse trigonometric functions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:51














2












2








2





$begingroup$

Substitute $r= sqrt{x^2+y^2}$ and $theta = arctan(frac{y}{x})$ into $g$ and use the Relations for inverse trigonometric functions.






share|cite|improve this answer









$endgroup$



Substitute $r= sqrt{x^2+y^2}$ and $theta = arctan(frac{y}{x})$ into $g$ and use the Relations for inverse trigonometric functions.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 21 '15 at 20:46









kryomaximkryomaxim

2,3621825




2,3621825












  • $begingroup$
    Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:51


















  • $begingroup$
    Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
    $endgroup$
    – jwalk
    Feb 21 '15 at 20:51
















$begingroup$
Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
$endgroup$
– jwalk
Feb 21 '15 at 20:51




$begingroup$
Amazing! Though, that sure was not worth spending hours trying to figure out. Thanks!
$endgroup$
– jwalk
Feb 21 '15 at 20:51


















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