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Math Help - Lagrange multipliers- no solution?

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    Lagrange multipliers- no solution?

    I'm investigating optimal solutions to equations of the form:

    f(\{x_i\}) = \sum\limits_k^M (A_k + \sum\limits_i^N B_{ik}x_i)^{\frac{1}{2}}

    Subject to the constraint:

    g(\{x_i\}) = \sum\limits_i^N x_i^2 = 1

    A_k and B_{ik} are also constrained independently of \{x_i\} to keep the expression real, but I don't think this is important.
    This gives the Lagrangian:

    \Lambda(\{x_i\}, \lambda) = \sum\limits_k^M (A_k + \sum\limits_i^N B_{ik}x_i)^{\frac{1}{2}} - \lambda(\sum\limits_i^N x_i^2 - 1)

    So taking the derivative with respect to x_a gives:

    \frac{d\Lambda(\{x_i\}, \lambda)}{dx_a} = \frac{1}{2}\sum\limits_k^M (A_k + \sum\limits_i^N B_{ik}x_i)^{-\frac{1}{2}}B_{ak} - 2 \lambda x_a = 0

    Is this correct so far? Because it doesn't seem like this series of equations has a solution in general.
    Oddly, it also seems like making the change of variables  x_i -> y_i^2 does give a solution. Following the above method, the equations are:

    \frac{d\Lambda(\{y_i\}, \lambda)}{dy_a} = \sum\limits_k^M (A_k + \sum\limits_i^N B_{ik}y_i^2)^{-\frac{1}{2}}B_{ak}y_a - 4 \lambda y_a^3 = 0

    Which can be solved by setting all but one y_i to 0, and the remaining one to 1.

    Am I making some mistake here?
    Last edited by Stereotomy; March 6th 2013 at 09:00 PM. Reason: Fixing formatting
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