# Smooth Mapping of Unit Circle

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.

## Answers and Replies

HallsofIvy
Homework Helper
Both $x^2+ y^2= 1$ and $x^4+ y^2= 1$ 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).

mathwonk
Homework Helper
2020 Award
is there a smoothness problem at x=0? (in answer #1)

mathwonk
Homework Helper
2020 Award
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.

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.

mathwonk