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Converting function to Quadratic form

  1. Feb 13, 2010 #1
    1. Hi everyone. I'm looking for help with converting this function to quadratic form.

    2. The function is f(x1,x2)=(x2-x1)^4 + (12*x1*x2) - x1 + x2 - 3.

    The quadratic form I need to convert to is: f(x)=(1/2)x'Qx - x'b + h

    where x is a vector=[x1 x2]', '=transpose, Q and b are vectors, and h is the constant. Also, Q is symmetric and positive definite (Q=Q' > 0).

    3. The trouble I'm running into is that f(x1,x2) is 4th order, and the examples I have convert only 2nd order functions to the f(x) quadratic objective format above.

    I hit a wall when I expand the quadratic to:
    -b1 x1 + 0.5 a x1^2 - b2 x2 + 0.5 b x1 x2 + 0.5 c x1 x2 + 0.5 d x2^2
    where Q=[a, b; c, d] and b=[b1; b2] and try to figure out the variables.

    In case your interested, I'm looking for this info to solve a steepest descent problem, where the varying ak value is ak=g(k)'g((k))/( g(k)'Qg(k) ) where g(k)=Qx(k) - b.

    Thanks for your help!
    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 14, 2010 #2


    Staff: Mentor

    Here's something that might be helpful, although it has been many years since I did anything with quadratic forms.

    You have f(x1,x2)=(x2 - x1)4 + 12x1x2 - x1 + x2 - 3.

    Let u = (x2 - x1)2, and let v = x1x2.
    Then your function can be written as
    u2 + u + 12v - 3
    = u2 + u + 1/4 + 12 v - 13/4
    = (u + 1/2)2 + 12v - 13/4.
    This could be viewed as quadratic in u and v, but with a coefficient of 0 on the v2 term.

    Hope this helps.
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