I ran across this comic, and it's gold. It is orginially published here The Comic

If I am correct, the first panel alone defines a self-referential loop if not a differential Equation:

$X$: Amount of Black to fill text
$Y$: Amount of ink to fill borderline
$Z$: Amount of black ink in graph

The amount of the image which would be black at first appears to be then $X+Y+Z$, but $Z$ also seems something as if bounded by $Z=X+Y+C+(\mathrm{something})Z$, where $A+B$ is that amount of black ink conveniently rendered outside of the graph, $C$ is an arbitrary amount of black ink added to the shaded part of the graph, and $Z$ is, of course, the amount of black required to draw this!

See, when you are shading the amount of black ink in the graph in the first panel: you can add as much as you want, but it must be related to how much is in the comic alltogether!

Though there are infinite amount of ways the graph could be shaded correctly, there is an infinite amount of ways the graph could be shaded incorrectly (here come the irrational numbers.) I'd like to try to prove this, but I have actual homework :P.

The only way to resolve this is to explain how Z varies with itself!

This creates a self-referential loop. It thus smells like a differential equation but I can't find it! How do we resolve this? (Is it the same thing as the population equation:$$ \frac {dP}{dt} = kP \rightarrow P(t)=Ce^{kt}$$ where P (the dependent variable) is the amount of ink in the panel and $t$, the dependent variable, is the amount of ink in the Graph? (Yes, I shamelessly ripped the population equation from my textbook and tried to applied it here. This is a question, not a statement.) Is it true, then, that amount of black ink in the graph and/or panel overall is bound by this equation: $A+B+C+P(t)$?

Also, if $X$,$Y$,and $C$ are allowed to vary does this not engender a 4-dimensional graph?

(I told you this comic is gold.)

That is just the first panel taken alone! Now Let's take the THIRD panel into consideration!

This panel looks a like Xeno's paradox in 2d, or a similar pattern to the horn of Gabriel!

As a Tangent, let us describe Xeno's paradox as such (in one dimension): an arrow is fired, once it reaches the halfway point has halfway to go. Once it reaches halfway of that halfway is has halfway to go again. Thus, there are an infinite number of half-ways! Just to be concise, let me express my opinion of something that at least kinda sorta reaches for its resolution: $$\int_1^\infty \frac{1}{x^2}\,dx = 1. $$ (Note this could also be discussed in terms of a similar series solution, such as $\sum_{i=1}^{n} \frac{1}{2^n} = \frac{\pi^2}{6}$, but actually trying to and resolving xeno's paradox is going way beyond the scope of this question.)

But, the point of this is that an infinite pattern can lead to a finite result (such as the area under curve $1\over{X^2}$). Now, look at the leftmost panel closely:

Third Panel

If the third panel plots the location of Black ink in the image, as it says, it must plot itself in miniature. This miniature, in turn, must include a plot of the comic; in this EXTRA-miniature plot, there must also be a plot of the comic, an infinite number of times.

In other words, the third panel must plot itself and infinite number of times.

Can it be shown that like the the amount of ink needed for these plots will so fast that a thus that a finite amount of ink is needed to draw an infinite number of these "reflected" third panels?

Let us try to compute the amount of ink recquired to draw this infinite number of self-mirrors.

Amount of Ink $\bf \propto$ Surface area of square

We can assume this because the area covered by ink will be a fixed proportion of the squares area if the thickness of the the ink varies with the size of the square, which it obviously must or the image cannot be drawn!

Let us say the sides of the small square are proportional (they must be if the plot is accurate) and that they are 1/5 the size of the first square. Therefore both the surface area and the amount of ink used will be $\frac{1}{(5^n)^2}$ including the first square if n=0. Now, let $A$ be the area of the first square. We already know that $$\sum_{i=1}^{n} \frac{1}{(5^n)^2} < (\sum_{i=1}^{n} \frac{1}{2^n} = \frac{\pi^2}{6})$$ therefore $A*(1 + \lim_{n \to +\infty}\sum_{i=1}^{n} \frac{1}{(5^n)^2})$ must be a finite number. Thus an infinite number of panels can mathematically be drawn with a finite amount of ink!

In other words, the added surface areas of a square, within a square, within a square, so on to infinity, is a finite combined area (remember the proof that the infinite series $\frac{1}{n^p}$ converges if $ p > 1$ - but I'll leave that to the reader.)

Does this first differential equation still apply? How would we unite these equations !?

I will not comment on the second panel, but I think it may be similar to the first.....

Thank you for reading all this!

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    $\begingroup$ Not sure what you're getting at with the differential equation, but you can pose this a fixed point problem: this comic is a fixed point of the function that takes any image/comic and produces a new comic with graphs describing it. Here's a neat reproduction of the comic in Mathematica: blog.wolfram.com/2010/09/07/self-description $\endgroup$ – Anthony Carapetis Sep 27 '13 at 5:16
  • $\begingroup$ I mean, I think the amount of ink that you shade as black on the graph is proportional to the amount of ink on the panel, but since the shaded portionis on the panel, it must be somewhat proportional to itself, and the only example that I know of of self-proportionality is the population model. I know when the only too you have is a hammer, every problem looks like a nail, but you I though this might still be it, for example, the population of ink molecules, or the population of an arbitary unit of surface area of ink. Thank you for the article! $\endgroup$ – user1833028 Sep 27 '13 at 5:19
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    $\begingroup$ But there's no "time" variable involved - it should just be a simple algebraic equation. If you consider the first panel alone, it'll look something like $\alpha x + c = x$, where $x$ is the amount of ink in the panel, $c$ is the amount of ink in the border/text and $\alpha$ is the ratio of the area of the graph disc to the area of the whole panel. This is just a linear equation, so you can easily find the unique solution. Things will get messier when you add in more panels (you'll get a system of equations) but it should be doable. $\endgroup$ – Anthony Carapetis Sep 27 '13 at 5:28
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    $\begingroup$ I looks like the Droste effect--- en.wikipedia.org/wiki/Droste_effect $\endgroup$ – Fred Kline Sep 27 '13 at 5:54
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    $\begingroup$ It would have been nice to credit the comic. It's xkcd.com/688 $\endgroup$ – Gerry Myerson Sep 27 '13 at 13:21

First off, when modeling a problem, it's wise to start from identifying your unknowns (step #0), rather than drawing parallels to other problems (that's maybe step #1, & it's contingent on step #0). In this case, the problem's self-referential aspect says 'implicit equation' & not 'differential equation' (which is a special class of implicit equations). Algebraic equations can also very well be implicit.

In this case, the unknown is clearly the amount of blank ink in the graph - let's call this $x$ & measure it in some units (e.g., as the ratio of the comic area colored black over the total area of the comic). Clearly, $x=x_1+x_2+x_3$ where $x_n$ denotes the amount of black ink in the $n-$th panel; there's no black outside these panels.

First, $x_1=p_1+\varepsilon_1 x$, where $p_1$ (offset) is the amount of black ink not contingent on $x$ - i.e., letters, border, & circle. Also, $\varepsilon_1$ is a scaling parameter equal to the amount of blank ink needed to color the entire disk black - clearly, $\varepsilon_1 < 1/3$ here.

Second, making the assumption that all three columns have the same base width, we have $x_2=p_2+\varepsilon_2 x$. Here, $\frac13\varepsilon_2$ is the amount of ink needed to color one full column; clearly, $\varepsilon_2<1/3$. This scaling parameter is the same for all three columns, because of the base width assumption. As an aside, note that varying that width from one column to the next would yield a sum $\varepsilon_{2,1} x_1+\varepsilon_{2,2} x_2+\varepsilon_{2,3} x_3$; then, we wouldn't be able to write a single eq. for $x$ but, rather, we should write three equations for $x_1,x_2,x_3$ - see also below. In other word, the eq. for $x$ wouldn't 'close.'

Third, the amount of ink in the third panel is $p_3+\varepsilon_3 x$, where $\varepsilon_3$ is the scaling factor corresponding to the the embedded comic (here also, $\varepsilon_3<1/3$).

All in all, the total amount of ink is $p+\varepsilon x$, where $p=p_1+p_2+p_3$ is the total offset and $\varepsilon=\varepsilon_1+\varepsilon_2+\varepsilon_3<1$ is the sum of scaling factors. Since this must equal $x$, we have $$ p+\varepsilon x = x , \quad\mbox{whence}\ x = p/(1-\varepsilon) . $$

One can play games with $p$ to arrive at that a solution cannot exist, but I'll stop here & remark, instead, that the linearity of the equation is inevitable on account of the comic being scale-invariant (scaling a valid/invalid comic up or down will yield a valid/invalid comic). Also, this offers a way of drawing the comic: determine $x$ and then $x_1,x_2,x_3$ by the corresponding equations; draw first panel using $x_1$; draw second panel using $x_1,x_2,x_3$; draw third panel without the embedded comic; scale down comic; embed in third panel; repeat ad infinitum.

P.S.: Thanks for suggesting this modeling problem, it'd make a great assignment! It's only the second elementary-yet-non-trivial modeling problem I can think of.


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