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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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    I'm not sure I understand what you're trying to do. Are the other $x_j$ functions of $x_i$?2012-12-19
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    In statistics, we may use $\mathbf{x}_{(i)}$ to denote the vector with its $i$th entry removed.2012-12-19
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    Do you mean there is some (fixed) index $i$ that you want to avoid, but want to range over all the others from say $1$ to $n$? If so, then say something like, "Let $i$ be fixed. Then $\mathbf{x}=\mathbf{x}(x_j)$, for all $j\not=i$, $j=1,\dots,n$."2012-12-19
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    Yes. But now how do I say that the elements of $\mathbf{x}$ are functions of, or evaluated at, $x_i^*$? Maybe this: $$\mathbf{x}^*=\mathbf{x}(x_j(x_i^*))$$2012-12-19
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    Does $\mathbf{x}(x_j)$ mean that the $j$th component of $\mathbf{x}$ is $x_j$? Because this is easily confused with a function evaluation.2012-12-19
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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 ${\bf x}\in{\mathbb R}^n$.