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In the function below the boldfaced $\mathbf{x}$ is the vector of "all the other $x$'s besides $x_i$, evaluated at $x_i^*$." Does the following notation convey that? If not, or if there is a better way to notate this, I appreciate the advice.

$f^* =f(x_i^* , \mathbf{x}) \:\:\:\:; \:\:\:\:\mathbf{x}=\mathbf{x}(x_i^*)$

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    @ben: Was the answer helpful?2012-12-23

2 Answers 2

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Maybe one of these is what you want:

  1. Let $\mathbf{x}=[x_j]$, $j=1,\dots n$, and $j^*\in\{1,\dots,n\}$ be fixed. Set $\mathbf{x}^*=[x_j],\ j\not=j^*$.
  2. Let $\mathbf{x}=[x_j]$, $j=1,\dots n$, and $j^*\in\{1,\dots,n\}$ be fixed. Set $\mathbf{x}^*=[x_j(x_{j^*})],\ j\not=j^*$.
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It seems that for some reason you want to expose the $i$th coordinate variable. One way to do this is defining ${\bf x}_{(i)}':=(x_1,\ldots,x_{i-1},x_{i+1},x_n)$ and introducing the special notation $(x_i,{\bf x}_{(i)}')$ for arbitrary vectors ${\bf x}\in{\mathbb R}^n$.