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Suppose I have a vectorvalued function of four variables, $\mathbf{F}:\mathbb{R}^4\rightarrow \mathbb{R}^3$, such that $\mathbf{F}(x,y,z,t)$.

Now suppose $x,y,z$ are function of $t$, such that $\mathbf{F}(x(t),y(t),z(t),t)$. Is $\mathbf{F}$ still a function of four variables, $\mathbb{R}^4\rightarrow \mathbb{R}^3$?

A similar question:

Suppose I instead have a vectorvalued function of three variables, $\mathbf{F}:\mathbb{R}^3\rightarrow \mathbb{R}^3$, such that $\mathbf{F}(x,y,z)$.

Now if $\mathbf{F}(x(t),y(t),z(t))$, is $\mathbf{F}$ still a function of three variables, $\mathbb{R}^3\rightarrow \mathbb{R}^3$?

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  • $\begingroup$ $\mathbf{F}$ is defined as a function $\mathbb{R}^4\to\mathbb{R}^3$. The definition doesn't change. But you may think of $t\mapsto F(x(t),y(t),z(t),t)$ which is a function $\mathbb{R}\to\mathbb{R}^3$ and a function in one variable. $\endgroup$ – Mundron Schmidt Jun 30 '17 at 10:58
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    $\begingroup$ Related: Can we add variables and create new functions? <> There's a serious implicit issue of mathematical grammar here: the phrases "such that $\mathbf{F}(x(t), y(t), z(t))$" and "if $\mathbf{F}(x(t), y(t), z(t))$" contain no verb. That is, "$\mathbf{F}(x(t), y(t), z(t))$" is neither a condition (as required by "such that...") nor a statement (as required by "if..."). Based on your earlier question, I think this is not merely an incidental aspect of the question. $\endgroup$ – Andrew D. Hwang Jun 30 '17 at 11:08
  • $\begingroup$ @AndrewD.Hwang What is the correct mathematical grammar in this question? English is not my native language. $\endgroup$ – JDoeDoe Jun 30 '17 at 11:53
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    $\begingroup$ I'd be inclined to write, "Now suppose $x$, $y$, $z$ are functions of $t$, and define $g(t) = \mathbf{F}(x(t),y(t),z(t),t)$." The equality sign acts as a verb; "$g(t) = \mathbf{F}(x(t),y(t),z(t),t)$" is a logical condition. (By contrast, "such that $\mathbf{F}(x(t),y(t),z(t),t)$" is analogous to writing "such that shoes" or "such that green".) Doing so should help clarify that $\mathbf{F}$ is a function of four variables, but $g$ is a function of just one variable (as John Hughes explains). $\endgroup$ – Andrew D. Hwang Jun 30 '17 at 12:10
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    $\begingroup$ Similarly, writing "if $g(t) = \mathbf{F}(x(t),y(t),z(t),t)$, ..." clearly separates the two functions $g$ (a function of one variable) and $\mathbf{F}$ (a function of three variables) whose identification seems to be the source of confusion. :) $\endgroup$ – Andrew D. Hwang Jun 30 '17 at 12:11
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$F$ is still a function of four variables. When you write the thing you wrote, you're really defining something new implicitly (in a way that physicists often do, and without renaming, which tends to confuse lots of other folks).

For instance, I'd write things this way:

Given \begin{align} F& :\Bbb R^4 \to \Bbb R^3 \\ x& :\Bbb R \to \Bbb R \\ y& :\Bbb R \to \Bbb R \\ z& :\Bbb R \to \Bbb R \\ \end{align} let \begin{align} G& :\Bbb R \to \Bbb R^3: t \mapsto F(x(t), y(t), z(t), t). \end{align} Now it's evident that $G$ is a function of one variable, arrived at by composing several functions.

As I said, a physicist will sometimes write $F(t) = F(x(t), y(t), z(t), t)$ (or something like that), using the name "F" for both the 4-argument function $F$ and the one-argument function $G$. A typical example of this is "We have a time-varying electric field $E$ on all of 3-space, and a particle whose trajectory is given by functions $x,y,z$ of time. Then the electric field at time $t$ [implicit meaning: the electric field, at the point where the particle happens to be at time $t$] is $E(t) = E(x(t), y(t), z(t), t)$. And the derivative of this is just $E'(t) = \ldots$." I always find things like this baffling

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  • $\begingroup$ Thanks! First you write $F :\Bbb R^4 \to \Bbb R$, shouldn't it be $F :\Bbb R^4 \to \Bbb R^3$? And also, is it neccesary to separately specify the domain and codomain of $x,y,z$? I mean, aren't they "included" in $\Bbb R^4$? $\endgroup$ – JDoeDoe Jun 30 '17 at 15:59
  • $\begingroup$ Yes -- I failed to see "vectorvalued" in your original note, and though "real valued" (even though my example used the electric FIELD). I'll edit. $\endgroup$ – John Hughes Jun 30 '17 at 17:05
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I'm going to focus on the 4-variable case. The 3-variable case is similar.

In general, a function of $n$ variables is a function that takes $n$ arguments. In particular, a function with domain $\Bbb R^n$ is a function of $n$ variables.

On its own, $F$ is a function of 4 variables because it is a map from $\Bbb R^4$ to $\Bbb R^3$.

What you are looking at is what happens when $F$ is composed with another function. Let $\varphi: \Bbb R \to \Bbb R^4$ such that $t \to (x(t), y(t), z(t), t)$. Then $\varphi$ is a function of one variable (in this case $t$) which outputs vectors in $\Bbb R^4$. Then, the function $F(x(t), y(t), z(t), t)$ is just the function composition $(F \circ \varphi)(t)$. This function is a map from $\Bbb R$ to $\Bbb R^3$:

$$ \Bbb R \underbrace{\mapsto}_{\varphi} \Bbb R^4 \underbrace{\mapsto}_{F} \Bbb R^3 $$

So $F$ (which corresponds to $F(x,y,z,t)$) is a function of $4$ variables, but $F \circ \varphi$ (which corresponds to $F(x(t),y(t),z(t),t)$ ) is a function of one variable.

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The number of variables is the number of elements you need to change to change the value of your function. In your case, knowing that $x,y,z$ are functions of $t$, you are implicitly saying that $\bf F$ can be viewed as a function of $t$ only too.

In symbols: you have a function of four variables $$\mathbf{F}:\bf{R}^4\longrightarrow \mathbf{R}^4$$ when the variables are $x,y,z,t$. But you also have the functions $x,y,z:\mathbf{R}\longrightarrow \mathbf{R}$ defined by $t\mapsto x(t),y(t),z(t)$ respectively, so you have the function $$\mathbf{R}\longrightarrow \mathbf{R}^4$$ defined by $t\mapsto (x(t),y(t),z(t),t)$. The composite $$\mathbf{R}\longrightarrow \mathbf{R}^4\overset{\mathbf{F}}{\longrightarrow} \mathbf {R}^4$$ is defined by $$t\mapsto (x(t),y(t),z(t),t)\mapsto \mathbf{F}(x(t),y(t),z(t),t)$$ and realises $\mathbf{F}$ as a function of one single variable.

Edit: about derivatives

With the above notations and assuming that $\mathbf{F}$ (that I will now call $F$) has the minimum required regularity (for instance, it is differentiable partially with respect to all the variables) you can take the partial derivatives: $$\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{ \partial F}{\partial z}, \frac{\partial F}{\partial t}$$ Moreover, if even your functions $t\mapsto x(t),y(t),z(t)$ are differentiable with respect to $t$, you can define the derivatives $ \frac{dx}{d t}, \frac{d y}{d t}, \frac{dz}{dt}$; formally speaking, you are doing the same also for the last variable $t$, considering the identity function $t\mapsto t$ whose derivative is $1$. When you want to consider the total derivative of $F$ with respect to $t$, you are just differentiating the composite function $t\mapsto (x(t),y(t),z(t),t)\mapsto F(x(t),y(t),z(t),t)$ and this, recalling the derivative rule for composite functions, amounts to consider: $$ \frac{d F}{d t} = \frac{\partial F}{\partial x} \frac{dF}{dt}+ \frac{\partial F}{\partial y} \frac{d y}{dt}+ \frac{\partial F}{\partial z} \frac{d z}{d t} + \frac{\partial F}{\partial t}\cdot 1$$

In particular, note that $\partial F/\partial t\neq dF/dt$.

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  • $\begingroup$ Thanks! I guess the partial derivative of $\mathbf{F}(x(t),y(t),z(t),t)$ is nonsense? So I can only take the ordinary derivative wrt $t$, $\frac{d}{dt}\mathbf{F}(x(t),y(t),z(t),t)$? What is the proper prime notation for this, $\mathbf{F}(x'(t),y'(t),z'(t),t')$, $\mathbf{F}'(x(t),y(t),z(t),t)$ or something else? $\endgroup$ – JDoeDoe Jun 30 '17 at 16:10
  • $\begingroup$ Dear @JDoeDoe, partial derivatives are perfectly defined if there is enough regularity for $F$; please see my edit $\endgroup$ – Caligula Jul 3 '17 at 9:37

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