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I'm trying to find the extreme values of the function $$ f(x) = \frac{a + \sum_{i=1}^n b_i x_i}{1 + \sum_{i=1}^n x_i^2} $$ where $x \in \mathbb{R}^n$, $b \in \mathbb{R}^n$, and $a \in \mathbb{R}$.

First, to find the critical points, I computed the $j$th partial derivative $$ \partial_{x_j} f(x) = \frac{b_j \left(1 + \sum_{i=1}^n x_i^2\right) - 2x_j \left(a + \sum_{i=1}^n b_i x_i\right)}{\left(1 + \sum_{i=1}^n x_i^2\right)^2} $$ (since all the partial derivatives exist and are continuous everywhere, the derivative of $f$ is equal to the Jacobian) and set it to zero to get $$ b_j \left(1 + \sum_{i=1}^n x_i^2\right) = 2x_j \left(a + \sum_{i=1}^n b_i x_i\right) $$ but now I'm stuck.

I tried solving for some special cases, in particular for the 1-D and 2-D cases.

In one dimension it's straightforward although tedious to find the two critical points. Classifying them, however, is brutal because computing the second derivative means using the quotient rule a second time.

In two dimensions, I can solve for $x_1$ in terms of $x_2$ and substitute, but the expressions become very large. Furthermore, when using the quadratic formula to solve for $x_1$, I get a constraint on $x_2$, $a$, $b_1$, and $b_2$. I was able to show that there exists a choice of $a$ and $b$ so that $x_1$ can in fact be solved for in terms of $x_2$.

I really don't think that this approach can be generalized though.

How else can I approach this problem?

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  • $\begingroup$ would you like to maximize your function with any means possible, or are you really looking for analytical solutions? $\endgroup$ – LinAlg Nov 20 '16 at 23:18
  • $\begingroup$ I'm looking for analytical solutions only. $\endgroup$ – Jacob Errington Nov 20 '16 at 23:28

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