20e-lecture13

20e-lecture13 - Lecture 13 Wednesday April 29th...

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Lecture 13 - Wednesday April 29th [email protected] Key words: Systems of equations, Implicit differentiation Know how to do implicit differentiation, how to use implicit and inverse function theorems 13.1 Examples Recall the two main theorems of the last section: Implicit Function Theorem Let f ( x,y ) : R n + m R m where x R n and y R m and let U be an open ball containing a point a R n . Suppose that det( f ) 6 = 0 on U when f is treated only as a function of y . Then there is a differentiable function g : R n R m such that f ( x,g ( x )) = 0 in some open ball containing a . Inverse Function Theorem Let f : R n R n be defined on an open ball around a point a R n and suppose f has continuous partial derivatives on this ball. If f is non- singular on this ball, then there exists a unique function g : R n R n such that f ( g ( x )) = x in some open ball containing a . We now show how to use these theorems. In the first example, we can explicitly find the inverse of the function, without using the implicit function theorem. Example 1. The function f ( x,y ) = ( x + y,x - y ) has an inverse on U = R : we have to solve u + v = x and u - v = y . Clearly this means u = x + y 2 and v = x - y 2 . In other words, the function g ( x,y ) = (( x + y ) / 2 , ( x - y ) / 2) is an inverse of f ( x,y ). What this means is that the equations u + v = x and u - v = y are solvable for x and y in terms of u and v . However for most functions f ( x,y ) there is no hope of computing an inverse. The inverse function theorem nevertheless tells us when an inverse exists, even though it may be impossible to find explicitly. Let’s use the inverse function theorem to show that the function f in the last example has an inverse. 1
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Example 2. In the last example the Jacobian matrix is f ( x,y ) = ± 1 1 1 - 1 . The determinant is - 2, so this function f is invertible (i.e. the equations u + v = x and u - v = y are solvable for x and y in terms of u and v . Next we consider f ( x,y ) = ( x 2 + y 2 ,xy ). Then f ( x,y ) = ± 2 x 2 y y x . This is non-singular everywhere except at (0
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20e-lecture13 - Lecture 13 Wednesday April 29th...

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