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WHOAguitarninja
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Aye...title should say in R^2, sorry about that.
I'm hitting somewhat of a wall in my understanding of a theorem (or rather a special case of a theorem). The theorem as stated in the book is as follows.
The Inverse FUnction Theorem in the Plane
Let O be an open subset of the plane R^2 and suppose that the mapping F: O\rightarrowR^{2} is continuously differentiable. Let (x, y) be a point in O at which the derivative matrix DF(x,y) is invertible.
Then there is a neighborhood U of the point (x,y) and a neighborhood V of its image F(x,y) such that F: U\rightarrowV is one to one and onto.
The theorem goes on to talk about the inverse functions, but that's not where I'm getting stuck. My problem is this. Consider the function F(r,\theta) = (r*cos\theta, r*sin\theta).
The determinant of the derivative matrix of this function is just r, so the theorem seems that it should only break down at r=0. However consider the point (r, 2*pi)It does not seem to me that it's 1-1 in a neighborhood around this point, which seems to contradict the theorem.
Where am I misunderstanding things. Is the mapping actually 1-1 here? It seems not to me as it seems to me if you just let theta go to infinity it circles the same ring in the image. I understand that if you restrict it to a ring then it invalidates the theorem as the neighborhood then isn't open due to the ring being thin. But I don't see any way to get around the looping problem.
I'm hitting somewhat of a wall in my understanding of a theorem (or rather a special case of a theorem). The theorem as stated in the book is as follows.
The Inverse FUnction Theorem in the Plane
Let O be an open subset of the plane R^2 and suppose that the mapping F: O\rightarrowR^{2} is continuously differentiable. Let (x, y) be a point in O at which the derivative matrix DF(x,y) is invertible.
Then there is a neighborhood U of the point (x,y) and a neighborhood V of its image F(x,y) such that F: U\rightarrowV is one to one and onto.
The theorem goes on to talk about the inverse functions, but that's not where I'm getting stuck. My problem is this. Consider the function F(r,\theta) = (r*cos\theta, r*sin\theta).
The determinant of the derivative matrix of this function is just r, so the theorem seems that it should only break down at r=0. However consider the point (r, 2*pi)It does not seem to me that it's 1-1 in a neighborhood around this point, which seems to contradict the theorem.
Where am I misunderstanding things. Is the mapping actually 1-1 here? It seems not to me as it seems to me if you just let theta go to infinity it circles the same ring in the image. I understand that if you restrict it to a ring then it invalidates the theorem as the neighborhood then isn't open due to the ring being thin. But I don't see any way to get around the looping problem.