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Thread: 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 $\displaystyle (0,0)$
    Where $\displaystyle f({x_1},{x_2}) = {x^2} - {x_1}{x_2} + \frac{3}{2}x_2^2$


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

    Where the BFGS formula is:



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

    and



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



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

    Therefore (0,0) is a solution? I guess I was expected to do more? Is it correct that if $\displaystyle \nabla f(0,0) = (0,0)$ 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 $\displaystyle f({x_1},{x_2}) = {x^2} - {x_1}{x_2} + \frac{3}{2}x_2^2$
    Just for clarification, $\displaystyle x^2$ should be $\displaystyle x_1^2$ (that is what it should be to agree with your $\displaystyle \nabla f$).

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

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

    Thanks.
    Since you are supposed to calculate two iterates, I suppose you should show that $\displaystyle x^{(1)}=x^{(0)} = (0,0)$ and $\displaystyle x^{(2)} = x^{(1)} = (0,0)$. 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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