1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove that a function is the quadratic form associated to

  1. Feb 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Let G:R2[itex]\rightarrow[/itex]R be a C2 function such that G(tx,ty)=t2G(x, y). Show that:

    2G(x,y)=(x,y).HG(0,0).(x,y)t

    3. The attempt at a solution

    G is C2, so its Taylor expansion is:

    G(x,y) = G(0,0) + [itex]\nabla[/itex]G(0,0).(x,y) + [itex]\frac{1}{2}[/itex](x,y).HG(c).(x,y)t,

    where c lies on the line segment that goes from (0,0) to (x,y).

    Using that G(tx,ty)=t2G(x,y) I get that G(0,0) and the linear term equals 0.

    The problem is that I have HG(c) in the quadratic term, but I need HG(0,0).
     
  2. jcsd
  3. Feb 20, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If c is not (0,0) then what do you mean by "c" and "HG(c)"? You are taking the Taylor expansion at (0, 0), are you not?
     
  4. Feb 20, 2014 #3
    I can't increase the degree of the Taylor polynomial because G is C2, so the second degree term is the remainder written in matrix notation.

    HG(c) is the Hessian matrix of G evaluated at c, where c lies on the segment that goes from (0,0) to (x,y).
     
  5. Feb 20, 2014 #4
    I solved the problem deriving two times the function f(t) = G(tx, ty). Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove that a function is the quadratic form associated to
  1. Quadratic Form (Replies: 3)

  2. Quadratic Forms (Replies: 0)

  3. Quadratic forms (Replies: 3)

Loading...