sol4 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 4...

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

View Full Document Right Arrow Icon
CAAM 336 · DIFFERENTIAL EQUATIONS Problem Set 4 · Solutions Posted Wednesday 15 September 2010. Due Wednesday 22 September 2010, 5pm. 1. [20 points] The equation x 1 + x 2 + x 3 = 0 defines a plane in R 3 that passes through the origin. (a) Find two linearly independent vectors in R 3 whose span is this plane. (b) Find the point in this plane closest (in the standard Euclidean norm, k z k = z T z ) to the vector v = 1 0 1 by formulating this as a best approximation problem. (You may use MATLAB to invert a matrix.) Solution. (a) Since two linearly independent vectors determine a plane, we simply need to find two linearly independent vectors that satisfy x 1 + x 2 + x 3 = 0. One can do this by inspection, for example, and find 1 - 1 0 , 0 1 - 1 . However, it would be nice to have an orthogonal basis for this space. To do that, pick one vector, say the first vector given above; set the second vector to be ( α,β,γ ) T . We would like the this vector to be in the plane: α + β + γ = 0 and to be orthogonal to the first vector: 1 - 1 0 T α β γ = α - β + 0 = 0 . This gives two equations in three unknowns, which will be satisfied if β = α and γ = - 2 α for any α , i.e., we have the vector α α - 2 α . With α = 1, we have two orthogonal vectors whose span is the desired plane: x = 1 - 1 0 , y = 1 1 - 2 . (b) The closest point in the plane to the vector v is found solving the usual best-approximation problem matrix equation: ± x T x x T y y T x y T y ²± c 1 c 2 ² = ± x T v y T v ² ,
Background image of page 1

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

View Full DocumentRight Arrow Icon
that is, ± 2 0 0 6 ²± c 1 c 2 ² = ± 1 - 1 ² . The orthogonality of the vectors x and y make this an easy problem to solve: c 1 = 1 / 2 , c 2 = - 1 / 6 . Thus, the best approximation to v = (1 , 0 , 1) T is the vector b v = c 1 x + c 2 y = 1 / 2 - 1 / 6 - 1 / 2 - 1 / 6 1 / 3 = 1 / 3 - 2 / 3 1 / 3 . We can verify this answer by checking (1) that b v is in the desired plane: 1 / 3 - 2 / 3+1 / 3 = 0, and (2) verifying that the error v - b v = 2 / 3 - 2 / 3 2 / 3 is orthogonal to the two basis vectors x and y for the plane, ( v - b v ) T x = ( v - b v ) T y = 0. 2. [25 points] Recall that a linear operator P is a projection from the vector space V to the vector space V provided P 2 = P , that is, P ( Pf ) = Pf for all f V . Consider V = C [ - 1 , 1] with the usual inner product ( u,v ) = Z 1 - 1 u ( x ) v ( x ) dx, and the two linear operators P e and P o the project a function onto their even and odd parts. That is, ( P e f )( x ) = f ( x ) + f ( - x ) 2 , ( P o f )( x ) = f ( x ) - f ( - x ) 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the 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.

Page1 / 7

sol4 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 4...

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