TITLE: Is there any mathematical way to describe what happens to a function when the graph is zoomed out infinitely far?
QUESTION [43 upvotes]: For example, if you zoom out very far on a graph of the function $y = x^3$, it appears like $x = 0$, or in general, if you zoom out on the graph $x^n$ for $n > 0$ it appears either like $x = 0$ with the restriction that for even bases $y \ge 0$.
However, if you look at the graph $\tan(x)$ it looks like an entire plane. Or, if you look at the graph $y = mx - \sin(x)$ it looks like $mx$. Or if you look at $ax\sin(x)$ it looks like two triangles where the larger $a$ is the larger the triangles are.
Is there some sort of general method for finding what a graph will look like given a function? Is there a collection of rules? Is there even any way of describing "two solid triangles" in the context of a single variable function?

REPLY [16 votes]: This can be codified in a mathematical way.
Let us define $m$ to be our magnification constant, in the sense that if we zoom out by a factor of $2$, we then see "two times more" of the graph. That is, if our "viewing window" is $[-a,a]$ on the $x$-axis (let us stay centered at the origin for simplicity), then we now see $[-2a,2a]$ or more generally $[-ma,ma]$. The same is true for the $y$ axis, again by this same factor $m$.
The graph will appear to be squished vertically by a factor of $1/m$, and in effect go $m$ times faster, compactifying the graph by a factor of $m$ in $x$.
Thus, say we have the graph of a function $f(x)$, and want to see what it would look like if we zoom out by factor $m$. Then we want to graph the new function
$$f_m(x) \stackrel{\text{def}}{=} \frac 1 m \cdot f \left(  mx\right)$$
You can play with a demo of this idea in Desmos here.

If you want to know what the graph will look like for large $m$, that's a matter of asymptotics. Generally, some idea can be garnered from the expression
$$f_\infty(x) \stackrel{\text{def}}{=} \lim_{m \to \infty} f_m(x) = \lim_{m \to \infty} \frac 1 m \cdot f(mx)$$
where $f_\infty$ is now a new function defined by this limiting process. Perhaps this function approaches another one pointwise and we can look at the graph of this $f_\infty$.
Take, for instance, the function
$$f(x) = 2x - \sin(x)$$
It easy to see that
$$f_m(x) = \frac 1 m \Big( 2mx - \sin(mx) \Big) = 2x - \frac 1 m \sin(mx)$$
and since $\sin(mx) \in [-1,1]$ for all $m,x \in \mathbb{R}$, that contribution gradually becomes "irrelevant": this is because when $m$ is large, we'll have a value in $[-1,1]$ divided by a large number, so it doesn't matter. Hence in the limit,
$$\lim_{m \to \infty} \left( 2x - \frac 1 m \sin(mx) \right) = 2x = f_\infty(x)$$
and of course, graphing these and trying large $m$ in the demo appears to confirm that: as $m$ increases, the blue graph approaches a line of slope $2$ which goes through the origin.


Another example you messed with observed behavior of basically becoming a straight, vertical line. This one is harder to codify, but it can be looked at in a slightly hand-wavier way.
Say a line pops up at $x=0$ for the zoomed-out function for a large magnification factor. (Not necessarily $f_\infty$, just imagine zooming out manually.) What this essentially can be seen as is, for inputs relatively near zero, the values of the function near that point increase far faster (when accounting for zoom) than the inputs might as they increase.
If we have a line elsewhere, at $x=x_0$, then in the zooming process the line no longer stays at $x_0$, but moves along: at a magnification of $m$, it will "appear" to be where $x_0/m$ was. Of course, for large $m$, it will appear to be at zero if it appears at all.
This suggests to me, somewhat crudely, that vertical line can appear at zero if the function grows faster than $\mathcal{O}(|x|)$ (since the derivative of $f_m$ is $f_m'(x) = f'(x)$ but we also want to account for growth in either direction, hence the modulus).
This at least seems to be true with the cases I tried (common roots and polynomials, exponentials), and the fact that $f(x) = x \ln(x)$ or even $f(x) = x \ln(\ln(x))$ seems to be exhibiting that behavior but taking extraordinarily large $m$ to do so seems to help corroborate the idea.
$f(x) = x \ln(\ln(x))$ is especially a powerful point because if you have absurd exponents in Desmos and make likewise absurd changes (e.g. $10^{200}$ to $10^{300}$ for $m$) you will see a nudge but it is so slow.


Finally, you also made a point about functions that "look like the entire plane," with $\tan(x)$ being the example.

The core criterion for this is periodicity, so you can smoothly cover the plane.

You also need to map to most of the reals. Not all of them need to be mapped to, but a good chunk do. I'm not entirely sure how to codify how much needs to be mapped to, however. I have this idea in my mind of a function that, as $m$ grows, it starts missing vertical stripes for whatever reason and doing so periodically. So there is some behavior, I believe, that needs to be nailed down further.


Some simple examples of this would be $\sec(x)$ and $\csc(x)$:

You can even go with "periodic extensions" if you don't want to do trigonometric functions, though they're a bit more annoying to implement in Desmos. Say for instance we define
$$f(x) = \begin{cases}
1/x & x \in (0,1] \\
1/(x-N) & x \in (N,N-1], \text{for } N \in \mathbb{Z} \setminus \{1\}
\end{cases}$$
(the idea being to take the segment of $1/x$ on $(0,1]$ and spread it everywhere). Then, playing with Desmos (e.g. limiting our $N$ to a range of $[-M,M] \cap \mathbb{Z}$ for simplicity and so it doesn't set my computer on fire), we can see something like this:

Of course, the fact that it doesn't hit the whole plane, and just the top half, is simply an artifact of the function being always positive. A similar idea comes with your other example of $f(x) = x \sin(x)$; the function is always stuck between $y = x$ and $y=-x$, and otherwise will generally hit every value eventually. Hence, zooming out looks like filling in the gaps between $y = x, y=-x$:

It all ultimately just depends on where your function's graph lives and what bounds it.

In summary:

To simulate the graph of a function $f$ zoomed out by factor $m$, graph $f_m(x) = f(mx)/m$.
To look at the "eventual shape" of the graph after large zooming, see if the limit $f_\infty(x) := \displaystyle \lim_{m \to \infty} f_m(x)$ exists pointwise, and see what it looks like.
Vertical lines seem to appear at $0$ for functions that grow "fast" enough, certainly faster than $\mathcal{O}(|x|)$ functions (which would just appear as lines), but anything even marginally faster seems to just eventually become a line.
A "space-" or "plane-filling" phenomenon occurs for functions which are periodic (and presumably nice-behaved in some way we probably need arcane examples to see violate). The shape that arises depends on the function and where it lives. Generally, unbounded functions can make some sort of shape though bounded functions obviously will just result in straight lines of little interest.