This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 2 Solutions Posted Wednesday 1 September 2010. (Corrected 6 Sept. 2010.) Due Wednesday 8 September 2010, 5pm. 1. [27 points: 4.5 points per problem] Consider the following sets of functions. Demonstrate whether or not each is a vector space (with addition and scalar multiplication defined in the obvious way). (a) { x R 2 : x 2 = x 3 1 } (b) { x R 3 : x 1 + 2 x 2 + 3 x 3 = 0 } (c) { f C [0 , 1] : f ( x ) 0 for all x [0 , 1] } (d) { f C [0 , 1] : max x [0 , 1] f ( x ) 1 } (e) { f C 1 [0 , 1] : f (0) = 0 } (f) { f C 2 [0 , 1] : f 00 ( x ) = 0 for all x [0 , 1] } Solution. (a) This set is not a vector space. The vector x = 1 1 is in the set, yet 2 x = 2 2 is not, since 2 6 = 2 3 = 8. (b) This set is a vector space. Suppose x and y are members of this set. Then x 1 + 2 x 2 + 3 x 3 = 0 and y 1 + 2 y 2 + 3 y 3 = 0. Adding these two equations gives ( x 1 + y 1 ) + 2( x 2 + y 2 ) + 3( x 3 + y 3 ) = 0 , and hence x + y is also in the set. Multiplying x 1 +2 x 2 +3 x 3 = 0 by an arbitrary constant R gives ( x 1 ) + 2( x 2 ) + 3( x 3 ) = 0 , and hence x is also in the set. (c) This set is not a vector space. The function f ( x ) = 1 for all x is in the set, but a scalar multiple, 1 f ( x ) = 1 for all x , takes negative values and thus violates the requirement for membership in the set. (d) This set is not a vector space. The function f ( x ) = 1 for all x is in the set, but a scalar multiple, 2 f ( x ) = 2 for all x , takes values greater than 1 and thus violates the requirement for membership in the set. (e) This set is vector space. If f and g are in the set, then f (0) = g (0) = 0, so d ( f + g ) dx (0) = f (0) + g (0) = 0 + 0 = 0 . Similarly, for any scalar , d ( f ) dx (0) = f (0) = 0 = 0 . (f) This set is a vector space. Suppose f and g are both members of this set. Then d 2 ( f + g ) dx 2 = d 2 f dx 2 + d 2 g dx 2 = 0 + 0 = 0 , and thus f + g is in the set. For any R note that d 2 ( f ) dx 2 = d 2 f dx 2 = 0 = 0 , and hence f is also in the set. Said another way: any member of the set must be a function of the form f ( x ) = + x , i.e., a line. The addition of two lines is still a line, and a scalar multiple of a line is also a line....
View
Full
Document
This note was uploaded on 01/21/2012 for the course CAAM 330 taught by Professor Tompson during the Fall '09 term at UVA.
 Fall '09
 Tompson

Click to edit the document details