Confusion about function composition

Question

I've gone through many math courses thinking that a function $f(x)=x^2$ could mean $f(anything) = anything^2$, however I realized that that isn't always true.

For example, suppose we have a function $f(x) = kx$ where $f$ represents the opposing force. We can't just substitute $t$ for time into that equation, as that isn't physically true. It could be the case that $x=2t$. In that case, $f(t) = 2kt$.

But suppose that $t$ is actually a function of another variable $p$. Let $t(p)= p^2$. If I were to naively try to do function composition to get $f(t(p))$, I would get $f(t(p)) = kp^2$, ignoring the relationship between $x$ and $t$. The correct answer would be $f(t(p)) = 2kp^2$.

What is really going on when I do the above function composition? Is there something more complex going on?

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Update:

Even after reading @peek-a-boo's post, I am still confused about a couple of points. I guess my original question might not have been as clear as I wanted it to be.

Say we have a spring that follows Hooke's law $F_s(x) = -kx$, where F is the spring's force in newtons, and x is spring's compression in meters. It seems to me (I might be wrong) that it is incorrect to say that $F_s(t) = -kt$, where $t$ is time. My reasoning for that is that it isn't necessarily the case that the force at a certain compression is equal to the force at the same time.

Question 1: Is my reasoning correct? Is it wrong to "plug in" $t$ into $F_s(x) = -kx$?

My understanding that you could "plug in" $t$ in the sense that it is a dummy variable, but you can't in the sense of time. You would have to account for the fact that $x$ is a function of $t$.`

Question 2: If my reasoning is correct (you can ignore this question if it is not), why is it not correct to plug in $t$ here?

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Now let us define a new function F:R→R as F(x)=kx (I’m explicitly using a different letter than f because the f above is different from this F). If you now fix a number t and decide to consider the number x=2t, then F(x)=F(2t)=k(2t)=2kt.

As for this, I see that "plugging in" $2t$ into $F(x)$ leads to the physically correct functions: $F(x)=F(2t)=k(2t)=2kt$, but what if I wanted $F(t)$ to be a function where I could get the force from just inputting a time $t$? What would $F(t)$ be?

Answer 1

however I realized that that isn't always true.

Wrong! Whoever/whatever made you change your mind about your previous sentence, ignore them/that (for now).

If $f:\Bbb{R}\to\Bbb{R}$ is the function defined for each $x\in\Bbb{R}$ as $f(x)=x^2$, then $f(@)=@^2$, and $f(t)=t^2$ and $f(\ddot{\smile})=\ddot{\smile}^2$. Anyone telling you otherwise is just trying to skip a whole bunch of steps because they’re impatient/lazy. If $t$ is a real number and you consider the real number $x=2t$, then $f(x)=f(2t)=(2t)^2=4t^2$. I have said this before (e.g here and here) and I’ll say it again: math does not care what your favourite letter is.

Now let us define a new function $F:\Bbb{R}\to\Bbb{R}$ as $F(x)=kx$ (I’m explicitly using a different letter than $f$ because the $f$ above is different from this $F$). If you now fix a number $t$ and decide to consider the number $x=2t$, then $F(x)=F(2t)=k(2t)=2kt$.

Now, let us introduce three new functions

• $X:\Bbb{R}\to\Bbb{R}$, defined as $X(t)=2t$.
• $T:\Bbb{R}\to\Bbb{R}$ defined as $T(p):= p^2$.

Then, you can consider all sorts of compositions: $F$ alone, or $F\circ X$ or $F\circ T$ or $F\circ X\circ T$. These will all give us different functions (if $k\neq 0$): for each $x,t,p\in\Bbb{R}$, we have \begin{align} F(x)&=kx,\quad\text{and}\quad (F\circ X)(t)=2kt,\quad\text{and}\quad (F\circ X\circ T)(p)=2kp^2. \end{align} And just to really drive home my point, we also have for each $x,t,p\in\Bbb{R}$ that \begin{align} F(t)&=kt,\quad\text{and}\quad (F\circ X)(p)=2kp,\quad\text{and}\quad (F\circ X\circ T)(x)=2kx^2, \end{align} and we also have that for each $\zeta\in\Bbb{R}$, \begin{align} F(\zeta)&=k\zeta,\quad\text{and}\quad (F\circ X)(\zeta)=2k\zeta,\quad\text{and}\quad (F\circ X\circ T)(\zeta)=2k\zeta^2. \end{align}

I strongly suggest you appreciate and keep this distinction between the functions (conceptually, and also notationally… atleast until you know for sure how to translate back and forth between a precise and imprecise statement). In Physics books they will often blur the distinction in the notation of the functions, and require the reader to do more mental gymnastics*, where they place emphasis on the ‘variables’ and the choice of letters, and instead will speak of

• force as a function of position
• force as a function of time
• force as a function of momentum,

or whatever else the context might necessitate, and write these respectively as $F(x)$ or $F(t)$ or $F(p)$ even though they’re not talking about the function $F$ alone! They’re talking about $F$ followed by various compositions.

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_{*: Of course I understand why one does use such notation, because after a while it does indeed get a little tiring to keep having to introduce new notation for functions related by ‘obvious’ compositions. But note that as a student you always have to be strict for yourself, and not rely on the ‘sloppiness’ of the teachers. It’s kind of like how in school we learn grammar, and in essays we’re teachers will penalize improper grammar, but in everyday communication, we don’t adhere to these rules strictly.}