I am working through a proof and have got stuck at a certain point. Where I have reached is the following: $$ q(\mathbf{x}) = \frac{|\mathbf{\Sigma}|^2}{(\mathbf{x}-\mathbf{x}_0)'\mathbf{\Sigma}'\mathbf{\Sigma}(\mathbf{x}-\mathbf{x}_0)+|\mathbf{\Sigma}|^2} $$ in which $\mathbf{x}$, $\mathbf{x}_0$ are $m \times 1$ vectors, $\mathbf{\Sigma}$ is $m \times m$ matrix and $|\mathbf{\Sigma}|$ is the determinant of $\mathbf{\Sigma}$, which is in fact the Jacobian matrix of a system of $m$ functions $\mathbf{f}(\mathbf{x}) = [f_i(\mathbf{x}),...,f_m(\mathbf{x})]'$, $f_i(\mathbf{x}):\mathbb{R}^m \rightarrow \mathbb{R}$.
Related work strongly suggests that the above reduces to $$ q(\mathbf{x}) = \frac{1}{(\mathbf{x}-\mathbf{x}_0)'(\mathbf{x}-\mathbf{x}_0) + 1}. $$ Can anyone provide the remaining step(s)? Or alternatively advise if this is not possible.
EDIT: On reflection, it would be adequate to demonstrate (if it is possible to do so) that $$ ||q(\mathbf{x})||_2 = \frac{1}{(\mathbf{x}-\mathbf{x}_0)'(\mathbf{x}-\mathbf{x}_0) + 1}. $$.
The best simplification i can see is rewriting the equation as
$$ q(x) = \frac{1}{1+\big\|\frac{Σ}{|Σ|}(x-x_0)\big\|_2^2} $$
This is generally equal to $\frac{1}{1+\|x-x_0\|_2^2}$ if and only if $Σ$ is orthogonal. Apart from that you can write it as a geometric series or derivative of arctan, if $|\det(Σ)|>ρ(Σ)$.
$$ q(x) = \frac{1}{1+\big\|\frac{Σ}{|Σ|}(x-x_0)\big\|_2^2} = \sum_{k=0}^{\infty} (-1)^k \Big\|\frac{Σ}{|Σ|}(x-x_0)\Big\|_2^{2k} = \frac{\arctan(\big\|\frac{Σ}{|Σ|}(x-x_0)\big\|)}{\big\|\frac{Σ}{|Σ|}(x-x_0)\big\|} $$