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Math Help - BFGS algorithm

  1. #1
    Junior Member
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    BFGS algorithm

    Hello,

    Is this correct?

    Follow the BFGS algorithm and compute two iterates at (0,0)
    Where f({x_1},{x_2}) = {x^2} - {x_1}{x_2} + \frac{3}{2}x_2^2


    \nabla f({x_1},{x_2}) = (2{x_1} - {x_2},3{x_2} - {x_1})

    Where the BFGS formula is:



    {x^{(k + 1)}} = {x^{(k)}} - {t_k}D_k^{ - 1}\nabla f({x^{(k)}})

    and



    D = {x^{(k + 1)}} = {x^{(k)}} - {t_k}A\nabla f({x^{(k)}})



    \nabla f(0,0) = (0,0)

    Therefore (0,0) is a solution? I guess I was expected to do more? Is it correct that if \nabla f(0,0) = (0,0) I should stop?

    Thanks.
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  2. #2
    MHF Contributor
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    Re: BFGS algorithm

    Quote Originally Posted by mark090480 View Post
    Where f({x_1},{x_2}) = {x^2} - {x_1}{x_2} + \frac{3}{2}x_2^2
    Just for clarification, x^2 should be x_1^2 (that is what it should be to agree with your \nabla f).

    Quote Originally Posted by mark090480 View Post
    \nabla f(0,0) = (0,0)

    Therefore (0,0) is a solution? I guess I was expected to do more? Is it correct that if \nabla f(0,0) = (0,0) I should stop?

    Thanks.
    Since you are supposed to calculate two iterates, I suppose you should show that x^{(1)}=x^{(0)} = (0,0) and x^{(2)} = x^{(1)} = (0,0). This is, of course, trivial as you pointed out.
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  3. #3
    Junior Member
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    Re: BFGS algorithm

    OK, Good to know that is all that was expected. It seemed too easy and so I thought it couldn't be true
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