# sol2 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 2...

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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....
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## This note was uploaded on 01/21/2012 for the course CAAM 330 taught by Professor Tompson during the Fall '09 term at UVA.

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sol2 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 2...

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