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Unformatted text preview: True or False In order to be able to use the matrix M introduced in Exercise 69, as the hint suggests, we write this system in a somewhat unusual way, as [ w 1 w 2 ... w n ] f 1 ( a 1 ) f 2 ( a 1 ) f n ( a 1 ) f 1 ( a 2 ) f 2 ( a 2 ) f n ( a 2 ) f 1 ( a n ) f 2 ( a n ) f n ( a n ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright M = bracketleftbiggintegraldisplay 1 1 f 1 integraldisplay 1 1 f 2 integraldisplay 1 1 f n bracketrightbigg Since M is invertible (Exercise 69), we have the unique solution [ w 1 w 2 ... w n ] = bracketleftbiggintegraldisplay 1 1 f 1 integraldisplay 1 1 f 2 integraldisplay 1 1 f n bracketrightbigg M 1 . Now if f is any polynomial in P n 1 and f = n summationdisplay j =1 c j f j , then integraldisplay 1 1 f = n summationdisplay j =1 c j integraldisplay 1 1 f j = n summationdisplay j =1 c j n summationdisplay i =1 w i f j ( a i ) = n summationdisplay i =1 w i n summationdisplay j =1 c j f j ( a i ) = n summationdisplay i =1 w i f ( a i ), as claimed. 4.3. 72 If we work with the basis f 1 ( t ) = 1 , f 2 ( t ) = t , and f 3 ( t ) = t 2 of P 2 , then we have to solve the system w 1 + w 2 + w 3 = 2 w 1 + w 2 = 0 w 1 + w 2 = 2 3 (see Exercise 71). The solution is w 1 = w 3 = 1 3 and w 2 = 4 3 , so that integraldisplay 1 1 f = 1 3 f ( 1) + 4 3 f (0) + 1 3 f (1) for f in P 2 . This is what you get when you apply Simpsons Rule (with two subintervals) to f ; note that Simpsons Rule gives the exact value of the integral in this case. True or False Ch 4.TF. 1 F; A basis of R 2 3 is bracketleftbigg 1 bracketrightbigg , bracketleftbigg 1 bracketrightbigg , bracketleftbigg 1 bracketrightbigg , bracketleftbigg 1 bracketrightbigg , bracketleftbigg 1 bracketrightbigg , bracketleftbigg 1 bracketrightbigg , so it has a dimension of 6. Ch 4.TF. 2 T; check with Definition 4.1.3c. Ch 4.TF. 3 T; The linear transformation T ( ax + b ) = a + ib is an isomorphism from P 1 to C , with the inverse T 1 ( a + ib ) = ax + b . Ch 4.TF. 4 T; by Theorem 4.2.4c. Ch 4.TF. 5 T; This fits all properties of Definition 4 . 1 . 2. 211 Chapter 4 Ch 4.TF. 6 F; The transformation T could be: T ( f ) = bracketleftbigg bracketrightbigg , in which case the kernel would be all of P 6 and the dimension of the kernel would be 7. Ch 4.TF. 7 T; We are looking at P 6 , with a basis 1 ,t,t 2 ,t 3 ,t 4 ,t 5 ,t 6 , which has seven elements. Ch 4.TF. 8 True, we can check both requirements of Definition 4.2.1. Ch 4.TF. 9 T; check the three properties listed in Definition 4.1.2. Ch 4.TF. 10 T; by Definition 4.2.1....
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This note was uploaded on 04/19/2011 for the course MTH 203 taught by Professor Staff during the Spring '08 term at Grand Valley State University.
 Spring '08
 Staff

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