The fixed-point of a vector-valued function $f: \mathbb{R}^p \rightarrow \mathbb{R}^p$ is a value $\textbf{b}$ such that $f(\textbf{b}) = \textbf{b}$ (this value is not necessarily unique). Any ideas on how to compute the multivariate fixed point for the following function \begin{align*} f(\textbf{b}) = \Gamma^{-1}\textbf{z}\left(y - \frac{e^{\textbf{z}^\intercal \textbf{b}}}{c + e^{\textbf{z}^\intercal \textbf{b}}}\right) \end{align*} where $\textbf{z} \in \mathbb{R}^p$, $y \in \{0, 1\}$, $c \in \mathbb{R}$? You can see from one of my previous posts of a similar question, which is similar in nature and express something in terms of the Lambert $W$ function.
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1$\begingroup$ So, uh, what is "generalized" about your definition of a fixed point? And are you interested in a closed form expression for the fixed point? Because otherwise, Newton's method for $g(b) := f(b)-b$? $\endgroup$– HannesCommented Apr 5, 2018 at 18:01
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1$\begingroup$ Apologies, I would like a closed-form solution for the fixed point, if possible. Solving for this fixed-point is part of several layers of other optimizations going on, so I would like this to be as fast as possible. $\endgroup$– Tom ChenCommented Apr 5, 2018 at 18:26
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$\begingroup$ What does $\Gamma$ mean here? $\endgroup$– Pietro MajerCommented Apr 5, 2018 at 18:49
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$\begingroup$ $\Gamma \in \mathbb{R}^{p \times p}$ is positive-definite symmetric $\endgroup$– Tom ChenCommented Apr 5, 2018 at 19:02
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$\begingroup$ @Hannes Also, I corrected the generalized to just a fixed point. I realize what you mean, since a fixed point is just fixed point, but I meant "generalized" here in the context of multivariate. $\endgroup$– Tom ChenCommented Apr 5, 2018 at 19:06
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The right side is a scalar multiple of $\Gamma^{-1} {\bf z}$, and its dependence on $\bf b$ is only through ${\bf z} ^\intercal \bf b$, so this is really just a one-dimensional problem. Let ${\bf a} = \Gamma^{-1} {\bf z}$ and $k = {\bf z}^\intercal \bf a$. Then with ${\bf b} = x \bf a$, the fixed-point equation $f({\bf b}) = \bf b$ becomes $$x = y - \frac{e^{k x}}{c + e^{kx}}$$ AFAIK this does not have closed-form solutions (even using LambertW).
answered Apr 5, 2018 at 19:49