Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this question
I'm not sure I understand what you're trying to do. Are the other $x_j$ functions of $x_i$? – Jonathan Christensen Dec 19 '12 at 18:51
In statistics, we may use $\mathbf{x}_{(i)}$ to denote the vector with its $i$th entry removed. – Nicolás Kim Dec 19 '12 at 18:51
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 $j\not=i$, $j=1,\dots,n$." – JohnD Dec 19 '12 at 19:05
Yes. But now how do I say that the elements of $\mathbf{x}$ are functions of, or evaluated at, $x_i^*$? Maybe this: $$\mathbf{x}^*=\mathbf{x}(x_j(x_i^*))$$ – ben Dec 19 '12 at 19:11
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. – JohnD Dec 19 '12 at 19:17
up vote 1 down vote accepted

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 $\mathbf{x}^*=[x_j(x_{j^*})],\ j\not=j^*$.
share|cite|improve this answer

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 ${\bf x}\in{\mathbb R}^n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.