Smooth Mapping of Unit Circle

In summary, the conversation discusses the smooth mapping between the unit circle and the curve, and possible solutions for this problem. Riemann's mapping theorem is mentioned as a potential approach, and HallsofIvy provides a bijection that addresses the smoothness issue at x=0.
  • #1
2
0
Hi, I have been told that in R^2 the unit circle {(x,y) | x^2 + y^2 = 1} is smoothly mappable to the curve {(x,y) | x^4 + y^2 = 1}.

Can someone please tell me what this smooth map is between them? I can only think of using the map (x,y) --> (sqrt(x), y) if x is non-negative and (sqrt(-x), y) if x is negative. Thanks for any help.
 
Physics news on Phys.org
  • #2
Both [itex]x^2+ y^2= 1[/itex] and [itex]x^4+ y^2= 1[/itex] loop around the origin. Draw the line from the origin through a point on the circle. Where that ray crosses the second graph is s(x,y).
 
  • #3
is there a smoothness problem at x=0? (in answer #1)
 
  • #4
The advantage of Halls' answer seems to be that he is projecting along a direction that never becomes tangent to the circle. I.e. #1 projects horizontally, and #2 projects radially. Still it is not so trivial for me to prove #2 is actually smooth, as the equation I am getting for r is still undefined at x=0, although it seems to extend.

An abstract approach is Riemann's mapping theorem, with extension to the boundary, that apparently gives an analytic map.
 
  • #5
HallsofIvy: Thanks, that's a nice bijection. I clearly need to think more geometrically for this type of problem.

mathwonk: I don't think there is a problem at x=0 (for Halls' map) because you can just define r to be 1 for x=0 and then it is smooth on S1.
 
  • #6
well you have prove it is smooth.
 

Suggested for: Smooth Mapping of Unit Circle

Replies
10
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
15
Views
1K
Replies
5
Views
1K
Back
Top