Math 250a Problem Set 5 Solutions

# Math 250a Problem Set 5 Solutions - Mathematics 250 Fall...

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Unformatted text preview: Mathematics 250 Fall 2006 Selected solutions, Assignments #4 and #5 HH 2.7.9: Find a number &gt; 0 such that the set of equations x + y 2 = a y + z 2 = b z + x 2 = c has a unique solution near 0 when | a | , | b | , | c | &lt; . Response: We think of the left hand side of these equations as defining a function from R 3 to itself: f ( x y z ) = x + y 2 y + z 2 z + x 2 . Then we are talking abaut solving the equations f ( x y z ) = a b c . We can apply Kantorovichx Theorem to this question. For example, we can try to solve this equation by using the Newton-Raphson Method, starting at p o = . For this, we need to know f (0) and DF . This allows us to compute r o = r (0) = || DF- 1 ||| F (0) | . Then we need to know the Lipschitz constant for DF on the ball B r o (0), and to check whether the estimate required by Kantorovichs Theorem is valid. We compute DF = 1 2 y 1 2 z 2 x 0 1 . In particular, DF is the identity, so its inverse is also the identity, and this has norm equal to 1. Next we look at DF p- DF q . if p = p 1 p 2 p 3 and q = q 1 q 2 q 3 , then from the above formula for Df , we see that Df p- Df q = 0 2( p 2- q 2 ) 0 0 2( p 3- q 3 ) 2( p 1- q 1 ) 0 0 = 2 d 1 d 2 d 3 0 1 0 0 1 1 0 0 , (1) where d j = p j- q j . Although it is typically difficult to compute exactly the norm of an operator, In this case we can do it exactly, using the factorization of equation (1). We take as known that the norm of a diagonal matrix is the maximum of the absolute values of the diagonal entries. Also, the matrix R = 0 1 0 0 1 1 0 0 simply permutes the entries of a vector ( p 1 p 2 p 3 p 2 p 3 p 1 ). Therefore, R preserves the length of all vectors; we call it an isometry. Let A be any matrix. and let R be an isometry. Then we have | ( AR )( v ) | = | A ( R ( v )) | || A || | R ( v ) | = || A ||| v | . Hence, from the definition of norm, we see that || AR || || A || . Similarly, writing A = ( AR ) R- 1 , we can show that also || A || || AR || . Hence || AR || = || A || . Combining this result with the factorization in equation (1) and our formula for the operator norm of diagonal matrices, we conclude that || Df p- Df q || = 2 max j {| p j- q j |} 2 | p- q | . 1 It follows that the Lipschitz constant for Df on any ball B r (0) is 2. We now use this in the estimates for the Kantorovich Theorem. Since the Lipschitz constant is inde- pendent of domain, we only need the second estimate, which is that M || Df- 1 x o || 2 | f ( x o )- y | 1 2 . Since the Lipschitz constant M is equal to 2, and || Df- 1 x o || = 1, this estimate comes down to...
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## Math 250a Problem Set 5 Solutions - Mathematics 250 Fall...

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