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Unformatted text preview: Foundations of Computational Math I Exam 1 Takehome Exam Open Notes, Textbook, Homework Solutions Only Calculators Allowed Tuesday 19 October, 2010 Question Points Points Possible Awarded 1. Basics 15 2. Bases and Orthogonality 20 3. Factorization 30 Complexity 4. Backward Stability 30 5. Conditioning and 25 Backward Error Total 120 Points Name: Alias: to be used when posting anonymous grade list. 1 Problem 1 (15 points) Each question below has a brief answer and justification. 1.a . (5 points) Explain the idea of a “hidden bit” in a floating point system with base β = 2 and the benefit achieved by using it. 1.b . (5 points) Can the idea of a “hidden bit” be usefully generalized to a floating point system with base β 6 = 2? 1.c . (5 points) Suppose you have a problem whose condition number is κ ≈ 10 5 . Given that you want at least 2 digits of accuracy in the solution how many decimal digits would you recommend be used in the floating point system used to solve the problem? Solution: The hidden bit is taken to be a 1 for normalized floating point numbers in an FP system with β = 2 and 0 for subnormal floating point numbers. The benefit is that it allows t bits of precision while actually storing only t 1. It does not work with β 6 = 2 since in that case there is more than one value for the first digit in a normalized FP number, i.e., 0 , 1 , . . . , β 1. The relative forward error is related to the uncertainty in the problem data by the condition number, i.e.,  f ( d + e ) f ( d )   f ( d )  ≤ κ  e   d  For the forward error to be smaller than 10 2 with κ = 10 5 we need the uncertainty in the data to be 10 7 . So we need to compute with at least 7 digits. 2 Problem 2 (20 points) Consider the vector space R 3 and the subspace S of dimension 1 given by S = span [ v 1 ] , v 1 = 1 1 1 2.a . Determine a basis { v 2 , v 3 } of the subspace S ⊥ where the vectors v 2 and v 3 are not orthogonal. 2.b . Derive from the basis { v 2 , v 3 } a second basis { q 2 , q 3 } of the subspace S ⊥ where the vectors q 2 and q 3 are orthonormal vectors. Solution: By definition S ⊥ is a subspace with dimension 2 comprising all vectors w ∈ R 3 such that w T v 1 = 0. Given the structure of v 1 any vector that has a single 1 element and a single 1 element is in S ⊥ . So we can take, for example, v 2 = 1 1 and v 3 = 1 1 Clearly, v 2 and v 3 are linearly independent due to nonzero structure so they are a basis for S ⊥ . Also note that v T 2 v 3 = 1 so they are not orthogonal as desired. To derive { q 2 , q 3 } from { v 2 , v 3 } where q T 2 q 3 = 0, q T 2 q 2 = q T 3 q 3 = 1 we must use the results of the homework problem that considered the least squares approximation of one vector, x , by another, y , and produced an orthogonal decomposition of x , i.e., min α k x αy k 2 x = z + α min y, α min = y T x/y T y Applying this result yields z = v 3 v T 2 v 3 v T 2 v 2 v 2 =...
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This note was uploaded on 07/25/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '11 term at University of Florida.
 Spring '11
 Gallivan

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