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How it should be read? $\mathbf{x}:\mathbb{R}\to \mathbb{R}$

A value x mapped over the real field, should it be read like this?

FShrike
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Man
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    If the notation is being used in the standard manner, $x$ is a real valued function defined over the reals – FShrike May 29 '22 at 10:55
  • @FShrike $x$ was bold representing vector.. (Used x for position in physics – Man May 29 '22 at 10:57
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    The boldface font is \mathbf{} or \bf. If that is not what your edit intended, please correct me – FShrike May 29 '22 at 10:57
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    I would say that x is a function, not a "value", from R to R. That is, it takes a real number input and returns a real number. – George Ivey May 29 '22 at 10:58

2 Answers2

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Since (from your comment) $\mathbf{x}$ is the function that represents the position of a particle on a line, the mathematical formalism says simply that it's a real valued function of a real variable. The domain is time. If you wanted to refer to the independent variable by name you might write $\mathbf{x}(t)$.

Mathematicians would probably not call that function $\mathbf{x}$, nor call it a "value". I don't think a physicist would either.

Ethan Bolker
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It is nothing more than a function. In the context of you comment on your post, it could be interpreted as the position of a particle with respect to time, i.e. you could say that at time $t\in\mathbb{R}$, the particle is at position $\mathbf{x}(t)\in\mathbb{R}$. This is merely an interpretation, however, and it is important to realize that mathematically, there is nothing different between writing $\mathbf{x}:\mathbb{R}\to\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$, which you would undoubtedly recognize as a function. Indeed what you interpret the function as physically is up to you, as you use mathematics to model a physical system, but mathematically that does not matter.

Lorago
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