Recent content by ironman2

Got it, thanks!

So, the function is locally 1:1 if the Jacobian is zero on the given interval? Can you give me some more hints on how to proceed to prove that it is globally 1:1?

Sequence like Taylor-series? I don't quite understand... Nvm the 0,0 logic, I was assuming it's a disk centered on the origin, which obviously it's not.

Since f is continuous, would lim r->0 f(xr,yr) = (a,b)? I'm thinking a,b = 0,0 since its the center...

Inside the disk? I thought of using a, b as x,y since a,b were inside the disk and somehow relating a, b to r... but can't seem to do it.

Ok, so if we take f1 = sin u/cosv and f2 = sin v/cos u, the determinant of the Jacobian becomes 1 - tan2vsin2u... which is a mess. I don't get the bigger picture from here. What do I from the Jacobian?

Oh my bad, ofcourse this isn't a linear map, so the determinants methods is ruled out. No, I'm not familiar with Inverse Function Theorem... can you tell me what it is and how can I use it here?

Homework Statement If Dr is a closed disk of radius r centered at (a,b) find lim r->0 (1/pir2) \int\intfdA over Dr. The Attempt at a Solution From mean value equality, \int\int fdA = f(x,y)A(D) where A(D) is the area of the region which here is pir2. So the lhs becomes lim r->0 f(x,y)...

Ok, so I've figured out that I can prove that T is one-to-one if the determination of the transformation matrix A is not zero. How do I find the transformation matrix given that sin and cos are involved? Also, I'm guessing that if I can build an inverse function that expresses (u,v) as some...

Homework Statement D* = {(u,v) | u>0, v>0, u + v < pi/2} and D = {(x,y) | 0<x<1, 0<y<1} The map T : R2 -> R2 is given by (x,y) = T(u,v) = (sin u/cos v, sin v/cos u). To prove that T is a bijectionHomework Equations and The Attempt at a Solution I'm kinda stuck finding the limits for u and v...