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ma01c09rec5-14 - MA1C ANALYTIC RECITATION 1 Simply...

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MA1C ANALYTIC RECITATION 5/14/09 1. Simply Connected Your book talks about simply connected regions and defines them to be regions “with no holes.” This is that is necessary for the theorems and stuff in there, but people spend a lot of time in real life thinking about things related to being simply connected. Given two continuous paths f, g : [0 , 1] R n (the R n can be replaced with any topological space, as I defined awhile ago) such that f (0) = g (0) and f (1) = g (1), they are said to be path homotopic if there exists a continuous map H : [0 , 1] × [0 , 1] R n such that: H ( t, 0) = f ( t ) H ( t, 1) = g ( t ) H (0 , s ) = f (0) = g (0) H (1 , s ) = f (1) = g (1) You can think of H as interpolating between f and g smoothly. We denote “ f is path homotopic to g ” by f g . Let’s denote the constant map at a point x by e x , i.e. e x ( t ) = x . We say that f is null homotopic if f e x for some x . Now we can define simply connected: a set S R n is simply connected if every path f : [0 , 1] S is null homotopic. You may recall that I defined the fundamental group of a space as the space of loops “up to deformation.” That deformation is homotopy, so the correct definition of the fundamental group is the set of loops based at a point, where two loops are considered the same if they are homotopic. The operation on loops is composition. Try to think about why the fundamental group of the torus is Z × Z , where composition of loops corresponds to addition of vectors in Z 2 . 2. Green’s Theorem Suppose you have some region R R 2 with a single (piecewise smooth) boundary curve C parameterized counterclockwise by c . Let F ( x, y ) = ( P ( x, y ) , Q ( x, y )) be a vector field where P and Q are continuously differentiable scalar fields defined in an open set containing R . Then ZZ R ∂Q ∂x - ∂P ∂y dxdy = Z C Pdx + Qdy That’s green’s theorem. It can be helpful both ways, but it often turns out that the double integral is easier. 2.1. Example. Find the line integral of R C F · dc , where F ( x, y ) = ( x 2 + y, 3 y 2 - x ), and C is the unit square R , traversed once counterclockwise. To do this, we compute ∂Q ∂x = - 1 and ∂P ∂y = 1. By Green’s theorem, then Z C F · dc = ZZ R - 1 - 1 dxdy = - 2 We could even do a more complicated region–the beauty of Green’s theorem is that you can wind up with a constant integrand. Say R
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