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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?

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  • $\begingroup$ Seems like (and I could be mistaken) when you zoom out the graph, the shape reflects $~\displaystyle \frac{y}{x} = \frac{f(x)}{x}.~$ When (for example) $~\displaystyle f(x) = x^n,~$ you have that $~\displaystyle \frac{f(x)}{x} = x^{n-1}.$ $\endgroup$ Commented Feb 16, 2022 at 5:33
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    $\begingroup$ The question seems imprecise to me, but at least for the case that the graph is that of $F(x, y) = 0$ for some polynomial $F$ in $x, y$ you might be interested in reading about homogenization of a polynomial as a subset of $2$-dimensional projective space and inspecting the part of that subset "at infinity". For example, for $y = x^3$, we can set $F(x, y) = x^3 - y$, homogenize to produce the polynomial $\hat{F}(x, y, z) = x^3 - y z^2$, and identify the (affine) $xy$-plane with $z = 1$. The part of the solution set of $\hat F$ "at infinity" (i.e., where $z = 0$) is just $\{x = 0, z = 0\}$. $\endgroup$ Commented Feb 16, 2022 at 5:38
  • $\begingroup$ @TravisWillse apologies if I don't fully understand your response, some of the notation/shorthand you are using is beyond me, I in fact only really know notation introduced in highschool math as I am currently in calculus 12. I will however take you up on your recommendation and read about homogenization of a polynomial $\endgroup$
    – MilesZew
    Commented Feb 16, 2022 at 5:56
  • $\begingroup$ @MilesZew The topic is "classical", and in particular it doesn't use a lot of modern machinery that takes a lot of time to learn. The other keyword you'll want to look up is "projective variety". $\endgroup$ Commented Feb 16, 2022 at 6:00
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    $\begingroup$ The apparent solid triangles in your $a \,x\sin(x)$ example is not actually solid or space-filling. When you zoom out, you seem to be seeing the width of the curve being meaningful rather than negligible $\endgroup$
    – Henry
    Commented Feb 17, 2022 at 12:35

2 Answers 2

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Zooming out a graph $f(x,y) = 0$ amounts to scaling the coordinates equally; introduce a parameter $t > 0$ and plot $f(tx, ty) = 0$. As $t\to 0$, the picture resembles zooming into the graph, while $t \to \infty$ correponds to zooming out.

In the case of $y = x^3$ or $y - x^3 = 0$, one has $(ty) - (tx)^3 = 0$ or $y = t^2x^3$. One also sees that in the case of a linear function $y = ax$, the graph looks the same at all zoom scales since the $t$ parameter cancels out.

In the example you mentioned, $y = x - \sin x$ corresponds to $ty = tx - \sin tx$ or $y = x - \frac 1t \sin tx$. Then as $t \to \infty$, the sin term drops off and the graph approaches $y = x$.


What the graph "becomes" or ultimately looks like as $t \to \infty$ can be easily described for some simple cases, like the examples above. For a more complicated example, like $y = x \sin x$, it is trickier to describe what is happening to the graph as $t \to \infty$.

For $y = x \sin x$ specifically, as you observed it resembles two large triangles, which together can be described as the set of points $T = \{(x,y) : -|x| \leq y \leq |x|\} \subset \mathbb R^2$. To argue that the graph $ty = tx \sin tx$ somehow "approaches" or "converges to" $T$ as $t \to \infty$ requires some notion of what it means for a family of subsets of $\mathbb R^2$ to "approach" or "converge to" another subset. There are many ways to interpret this idea, and some will be more useful than others in certain contexts.

My first guess to formalize this idea might go something like this: let $G_t = \{(x,y) : f(tx,ty) = 0\}$ be the graph of the desired function at the scale $t$. We will say the family of subsets $G_t$ "approaches" or "converges to" the subset $S \subset \mathbb R^2$ if it holds that $(x,y) \in S$ if and only if there is a sequence $(t_n)_n$ such that $t_n \to \infty$ as $n \to \infty$, as well as a sequence of points $(x_n,y_n) \in G_{t_n}$ such that $(x_n,y_n) \to (x,y)$ as $n\to \infty$.

With this definition, we can prove that the graph of $y = x \sin x$ approaches the two-triangles set $T$. Here is the argument:

Let $(x,y) \in T$ be arbitrary, ie $-|x| \leq y \leq |x|$ or equivalently $-1 \leq y/x \leq 1$. There is therefore a value $\theta > 0$ such that $y/x = \sin(\theta + 2\pi n)$ for all $n$, or equivalently $y = x \sin (\theta + 2\pi n)$.

So let $(t_n)_n$ be the sequence such that $t_n = \frac1x(\theta + 2\pi n)$ for each $n$ (in the case that $x \geq 0$; in the case that $x < 0$, replace $t_n$ by $-t_n$), and let $(x_n,y_n)_n$ be the constant sequence $(x,y)$. Then clearly $(x_n,y_n)_n \to (x,y)$ as $n\to \infty$, and moreover $y_n = x_n \sin (t_n x_n)$ for each $n$, ie $(x_n,y_n) \in G_{t_n}$ for each $n$.

In the converse case, suppose $(x,y) \not\in T$, ie $y/x < -1$ or $y/x > 1$. Then we want to show there is no sequence of times $t_n$ and points $(x_n,y_n) \in G_{t_n}$ such that $t_n \to \infty$ and $(x_n,y_n) \to (x,y)$ as $n \to \infty$. We can use a little topology to argue this.

Note that $G_t \subset T$ for every $t > 0$, and $T$ is a closed subset of $\mathbb R^2$. This means there can be no sequence of points from $T$ converging to $(x,y)$, if $(x,y) \not\in T$.

This completes the proof, and so we can say that the graph of $y = x\sin x$ does indeed converge to the subset $T$ of two triangles as you zoom out infinitely far (relative to the definition above).


There may be a way to interpret what I have outlined above using the idea of a limit set from the field of dynamical systems.


Coming up with some theorem about what will happen to a general graph $f(x,y) = 0$ seems like a tall order, but there are a few things you can quickly say. Any asymptotes anywhere in the domain will be pushed to the line $x = 0$. Aside from that, the picture $\lim_t G_t$ depends on the limiting behavior of $y = f(x)$ as $x \to \infty$; whether it oscillates or converges to some value or if it grows without bound. It also matters the growth rate, specifically whether it is faster or slower than a linear function.

The linear functions are (more or less) the only functions which are preserved by zooming in and out, so if $y = f(x)$ grows any faster than a linear function (all linear functions) as $x \to \infty$, this will also cause the graph to smoosh against the line $x = 0$ as $t \to \infty$. Conversely, anything which grows slower than a linear function (eg, $y = \sqrt x$ will be smooshed against the line $y = 0$ in the limit.


There is another, more compact definition for the set $S = \lim_t G_t$ if we write it in terms of the distance from a point to a set. Given a closed subset $A \subset \mathbb R^2$ and a point $(x,y) \not\in A$, we can define the distance

$$d\big((x,y), A\big) = \inf \big\{d\big((x,y), (a_1,a_2)\big) : (a_1, a_2) \in A\big\}$$

where the distance between the points $(x,y)$ and $(a_1, a_2)$ is the usual Euclidean distance formula. This says that the distance from the point $(x,y)$ to the set $A$ is whatever the smallest/minimal distance is between $(x,y)$ and any point of $A$. Using this notion, we can restate the definition of what we mean when we say the family of subsets $G_t$ converges to another subset. Namely,

$$(x,y) \in \lim_{t\to\infty} G_t \iff \lim_{t\to\infty} d\big((x,y), G_t\big) = 0$$

Using this definition, here is a proof that the graph of $y = x^2$ converges to the half-line $\{(0,y) : y \geq 0\}$ as one zooms out infinitely far.

In this case, $G_t = \{(x,y) : ty = (tx)^2\}$ or equivalently $y = tx^2$. So pick an arbitrary $y \geq 0$, in which case for each $t > 0$ one can let $x = \sqrt{y/t}$. For this $x$ and $y$ it holds that $(x,y) \in G_t$, and therefore

$$d\big((0,y), G_t\big) \leq x = \sqrt{y/t}.$$

Using the squeeze theorem, we then see that $d\big((0,y), G_t\big) \to 0$ as $t \to \infty$. This demonstrates that $\{(0,y) : y \geq 0\} \subset \lim_t G_t$. For the reverse inclusion, we would need to argue that any point $(x,y)$ with $x \neq 0$ or $y < 0$ must stay "bounded away" from the set $G_t$ for all sufficiently large $t$.

We can make use of a simple lemma: if $A_t$ is a descending family of closed subsets such that $G_t \subset A_t$ for each $t$, then $\lim_t G_t \subset \bigcap_t A_t$. This is true because, for any point $(x,y) \not\in \bigcap_t A_t$, it holds that $(x,y) \not\in A_t$ for all sufficiently large $t$ - and, because each $A_t$ is closed, any point outside of $A_t$ is a positive distance away from $A_t$.

Say, for instance, $(x,y) \not\in A_{t_0}$ and observe $G_t \subset A_t \subset A_{t_0}$ for all $t \geq t_0$. We also have, because $A_{t_0}$ is closed, that $(x,y) \not\in A_{t_0}$ implies $d\big((x,y),A_{t_0}\big) > 0$. Hence

$$d\big((x,y), G_t\big) \geq d\big((x,y), A_{t_0}\big) > 0$$

for all $t \geq t_0$, and therefore $d\big((x,y),G_t\big) \not\to 0$ as $t \to \infty$.

We can apply this lemma to our problem here in the following way. For each $t > 0$, let $A_t = \{(x,y) : y \geq 2\sqrt{t}|x| - 1 \text{ and } y \geq 0\}$. Observe that $(A_t)_t$ is a descending family of subsets and $G_t \subset A_t$ for each $t$. By the lemma above, we then have that

$$\lim_{t\to\infty} G_t \subset \bigcap_{t>0} A_t = \{(0,y) : y \geq 0\}$$

This completes the proof that the graph of $y = x^2$ converges to the half-line $x = 0$, $y \geq 0$ as one zooms out infinitely far. See below for an illustration of $G_t$ and $A_t$ for a particular $t$ value. The idea is that we overapproximate the graph $G_t$ by a sort of triangular cone shape (also implicitly restricting to the upper half-plane).

The reason why $\bigcap_t A_t = \{(0,y) : y \geq 0\}$ is clear from the picture; as $t \to \infty$, the slope of the boundary goes to infinity as well, and so it will "sweep past" any point $(x,y)$ with $x \neq 0$, in which case $(x,y)$ has no hope of belonging to the intersection.

enter image description here


There is a subtle difference between the two definitions of the set $\lim_t G_t$ that I presented above. In the first definition, $(x,y) \in \lim_t G_t$ when the set $G_t$ is close to $(x,y)$ at infinitely many times $t$, but the distance between $G_t$ and $(x,y)$ may also be large for infinitely many times. In the second definition, $(x,y) \in \lim_t G_t$ when $G_t$ is close to $(x,y)$ for all sufficiently large $t$, which is a stronger condition. This doesn't change the validity of either of the proofs I presented above, as showing convergence in the second sense will imply convergence in the first sense (but not the other way around).

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  • $\begingroup$ While it make take me time to fully understand some of the terms and notations here, this has been very useful in understanding how to express the idea of infinite zoom. When initially formulating this question I had some sense that I would have to do with limits at infinity and therefore growth rate. If I'm understanding correctly, a key point here seems to be that any equation of the form ty = (...) has a y term with an exponent of one and therefore will grow faster than with anything with a degree less than one and slower than anything with a degree more than one. (cont. in next comment) $\endgroup$
    – MilesZew
    Commented Feb 16, 2022 at 7:42
  • $\begingroup$ Therefore if you have a right hand side that is growing to infinity and a left hand side that is growing at an irrelevant rate you get y = $\infty$ (y = x as x-> $\infty$) which means y immediately shoots off to infinity, which approaches x = 0 $\endgroup$
    – MilesZew
    Commented Feb 16, 2022 at 7:46
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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.

enter image description here


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.

enter image description here


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 $\csc(x)$:

enter image description here

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:

enter image description here

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 $y = x, y=-x$:

enter image description here

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.
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    $\begingroup$ Very interesting. I like how you mentioned how periodic functions seem to fill space. Trying to think about this, you might say that periodic have some gap between each period that approaches 0 as m approaches infinity. Perhaps this could be formalized more if you were to construct an equation for the size gaps between periods depending on the zoom level. $\endgroup$
    – MilesZew
    Commented Feb 16, 2022 at 7:59
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    $\begingroup$ That's simple enough. In a function of period $p$, we know a value will repeat at most every $p$. Hence, $$f(x) = f(x+p)$$ Passing that to $f_m$ yields $$f_m(x) = \frac 1 m f (mx) = \frac 1 m f \left( m \left( x + \frac p m \right) \right)$$ $\endgroup$ Commented Feb 16, 2022 at 8:36
  • $\begingroup$ alright, so in that case, could you use this equation to show that, for the $y = xsinx$ function, since every point has an infinite number of points to the right of it (or left for negative values), and the distance between these points gets arbitrarily small as m approaches infinity, it forms a straight line at the points y value starting from the points x value? and then to show it "fills" the space between $y = x$ and $y = -x$ you show that as the zoom level approaches infinity, the peaks and troughs of the function approach $y = x$ and $y = -x$ $\endgroup$
    – MilesZew
    Commented Feb 17, 2022 at 1:39
  • $\begingroup$ Actually, I think an easier example would be $tanx$. in a somewhat similar way to $xsinx$ every point has an infinite number of points to the right and left of it, and so if the size between those points gets arbitrarily small in the limit it would form a line to the right and the left. Then, knowing that a branch (is it called branch?) of the function $tanx$ has points at every y value, we can see that it "fills" every point on the plane. Is this intuition formally true? $\endgroup$
    – MilesZew
    Commented Feb 17, 2022 at 1:48
  • $\begingroup$ Rereading your answer this seems to be exactly what you're saying, though less verbosely, is what I'm saying basically what you're suggesting? $\endgroup$
    – MilesZew
    Commented Feb 17, 2022 at 1:51

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