{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Differential Equations Solutions 21

Differential Equations Solutions 21 - 31 CHALLENGE 5.20(a...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
31 CHALLENGE 5.20. (a) Since U Σ V x = b , we have x = V Σ 1 U b . If we let c = U b , then α 1 = c 1 1 , and α 2 = c 2 2 . (b) Here is one way to look at it. This system is very ill-conditioned. The condition number is the ratio of the largest singular value to the smallest, so this must be large. In other words, σ 2 is quite small compared to σ 1 . For the perturbed problems, Ax ( i ) = b E ( i ) x ( i ) , so it is as if we solve the linear system with a slightly perturbed right-hand side. So, letting f ( i ) = U E ( i ) x ( i ) , the computed solution is x ( i ) = α ( i ) 1 v 1 + α ( i ) 2 v 2 , with α ( i ) 1 = ( c 1 + f ( i ) 1 ) 1 , α ( i ) 2 = ( c 2 + f ( i ) 2 ) 2 . From the figure, we know that f ( i ) must be small, so α ( i ) 1 α 1 . But because σ 2 is close to zero, α ( i ) 2 can be quite different from α 2 , so the solutions lie almost on a straight line in the direction v 2 . CHALLENGE 5.21. Some possibilities: Any right eigenvector u of A corresponding to a zero eigenvalue satisfies
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.
  • Fall '11
  • Dr.Robin
  • Singular value decomposition, Orthogonal matrix, right singular vector, singular vector, largest singular value, eigenvalue satisfies Au

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern