Lagrange multipliers- no solution?

• Mar 6th 2013, 07:58 PM
Stereotomy
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?