Results 1 to 4 of 4

Thread: Numerical Analysis Question?

  1. #1
    Join Date
    Aug 2007
    Posts
    2

    Default Numerical Analysis Question?

    Find a linear tranformation of x=x(u) that will take the interval 0<=u<=1 onto the interval -1<=u<=1.

    Show how this is done.

    If this helps: Original equation is e^x using the Maclaurin expansion:

    1+x+(x^2/2)+(x^3/6)+(x^4/24)+(x^5/120)+(x^6/720)
    And how do you find the inverse tranformation u=u(x) that takes the interval -1<=x<=1 onto the interval 0<=u<=1
    and the original should interval should read
    0<=u<=1 onto the interval -1<=x<=1.

  2. #2
    Join Date
    Aug 2007
    Posts
    1

    Default

    The way I do it:

    Use a sketch. Locate the original interval on x and the desired interval on y using the endpoints. Now you have 2 points that describe a line. Just find the equation of the line.

    Or, if you're good,

    p1: (0,-1)
    p2: (1,1)

    slope is 2/1

    at x-0, y=-1, thus b=-1

    equation is y=2x -1

    See how that works?

  3. #3
    Join Date
    Aug 2007
    Posts
    1

    Default

    Are you sure such a transformation exists?

    A linear transformation is a function L such that L(ax + by) = aL(x) + bL(y) for all x, y in your set of "vectors" and all a, b in your set of "scalars."

    If your field of scalars includes the reals (which is typically does), then if we had a linear transformation L, we would have L(k) = kL(1) for all 0 <= k <= 1. In particular, L is always either nonpositive or nonnegative for all 0 <= k <= 1, so it is certainly not onto -1 <= u <= 1.

    Am I missing something here? Do you mean a Moebius transformation or a conformal mapping or some other kind of function?

  4. #4
    Join Date
    Jun 2007
    Posts
    7

    Default

    How 'bout U = 2u - 1 ? ☺


    Doug

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •