3e-spring2011-chapter_4_tf

3e-spring2011-chapter_4_tf - True or False In order to be...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 6

3e-spring2011-chapter_4_tf - True or False In order to be...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online