0
$\begingroup$

An equation in my machine learning class says
$$ x= \begin{bmatrix} x_0 \\ x_1 \\ \vdots \\ x_n \\ \end{bmatrix} \in ℝ^{n+1} $$

I'm reading this as "x is the subset of numbers belonging to the set off all real numbers that are the square of the set total plus one" and I don't feel that's correct.

$\endgroup$
3
  • 1
    $\begingroup$ $ℝ^{2}$ means 2 dimensions, $ℝ^{3}$ means 3 dimensions, $ℝ^{4}$ means 4 dimensions, and so on. $\endgroup$
    – ndh
    Oct 22 '13 at 5:30
  • $\begingroup$ It's a set of all n-tuples of real numbers, i.e. that belong to $\mathbb R$. $\endgroup$
    – Kaster
    Oct 22 '13 at 5:30
  • 1
    $\begingroup$ x is an element in the set of all vectors of length $n+1$ where each "cell" in the vector has a real number $\endgroup$
    – DanielY
    Oct 22 '13 at 5:31
2
$\begingroup$

First of all $x$ isn't a subset of $\mathbb{R}^{n+1}$, it is an element of $\mathbb{R}^{n+1}$.

Let $X$ be a fixed set. Then $X^{n+1}$ is the set of ordered $(n+1)$-tuples of elements of $X$. That is, $$X^{n+1} = \{(x_0, x_1, \dots, x_n) \mid x_0, x_1, \dots, x_n \in X\}.$$ In the case where $X = \mathbb{R}$, these are just real $(n+1)$-dimensional vectors that we usually write as columns, as you have done.

$\endgroup$
2
$\begingroup$

For any set $X$, the notation $X^n$ denotes the set of all $n$-tuples of elements of $X$: the lists $(x_1,x_2,\ldots,x_n)$ where each $x_i$ is an (independently chosen) element of $X$. Therefore $\Bbb R^{n+1}$ denotes the set of all $n+1$-tuples of real numbers, which are here written vertically, and indexed starting from$~0$.

$\endgroup$
1
$\begingroup$

The equation simply states that $x$ is a $(n+1) \times 1$ vector. So $\mathbb{R}^{n+1}$ is the space where $(n+1) \times 1$ vectors lie.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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