What causes the heuristic 'if you're not with us, then you're against us' so persistent? Rumination on some discourse leaden reasons that lead to polarization of political views.

One of the reasons that (general) political view declarations are important, and persist, is because they play a key role in aiding the interpretation of what is being said. Of course what is said, often strongly implies such a declaration, but not always; that's why it becomes useful to declare "what side one is on", even when one wishes to remain neutral (until one's understanding of the issue at hand has matured), unless one is prepared to have their utterances treated with suspicion. This is a more general phenomenon, not only restricted to political views, but I believe that politics makes the phenomenon more pronounced. Much more could be said about this, but here I'll just give an example, which illustrates the idea. I will use an abstract example (to avoid the risk of creating a distraction).

Let the context be one where there is a stark divide in views regarding some issue in it. Now, say there's a proposition 'A' that expresses an attitude regarding that issue in the mentioned context. That is, some would think, and feel very strongly that A, whereas others would think, and feel very strongly that not A.

EXAMPLE 1
Person 1: "A"
Undeclared: "Of course"

EXAMPLE 2
Person 1: "Not A"
Undeclared: "Of course"

In both cases, it's very likely that Person 1 (or person 2) will inquire "what do you mean?" And why are they doing this? Was the Undeclared's statement not clear enough? It was, but coming from an Undeclared it may be taken as meaning the opposite of what it says, e.g. being a sarcastic remark; the existence of such remarks, being common where political views clash, adds to the ambiguity of the actual intended content of undeclared's statements. The degree of this phenomenon, I suspect, would be a function of the context's scope, i.e. given some context, what is the extent of the Undeclared's lack of declaration.

Needless to say, this leaves those who wish to maintain neutrality (which is more often than not the wise attitude, I believe) in an uncomfortable position of their opinions being either notoriously misunderstood or treated with suspicion, so there exists a pressure to declare oneself politically, even if one is neutral. In a "war of words", which often is the form of political discourse outside of echo chambers, being declared facilitates unpacking the non-explicit, intended content of the actual statements that are made. So enhancing one's clarity is a tempting reward for the mere cost of an instrumental declaration (even when it is inconsistent with one's actual view). Consequently, this mechanism inadvertently facilitates the proliferation and sustainment of extreme views, at the cost of the more balanced ones.

Bracia

Bajeczka kórą wymyśliłem lata temu (w 2003!)---wersja tekstowa poniżej nie byłaby możliwa bez udziału i zdolności literackich przyjaciela, Jerzego Rybińskiego, alias 'Stryjaga'.

BRACIA

Pewnego razu, w pewnym miejscu świata,
Brat Duży przybywa do Wielkiego Brata
I prośbę przedstawia chyląc się mu do nóg
By ten w kwestii wzrostu Dużemu dopomógł
- zechciał bratu nadać kilkanaście cali…
by się podzielił…by ciut użyczył
Duży by tego bardzo sobie życzył.

Wielkiemu mimo zdziwienia koncept nagły się jawi,
Tedy tając obawę czy Duży to strawi rzecze:
- „Nie tylko wzrost mój jest wielki przecie…
Z miłości do was, co chcecie zrobię…
Zatem postawcie sobie na głowie
Ten dzban wysmukły…o tam stojący
Lecz uważajcie, aby niechcący nie zbić!”

„Dzięki niemu bowiem…od Dużych większy o łokieć,
z Wielkimi się zrównacie.
- Wniosek z tego drogi bracie,
bez ochyby się znajdziecie
w gronie Wielkich – w Wielkim Świecie.”

Duży Wielkiego dobrocią wzruszony,
z wdzięcznością pyta:
- Co ze swej strony mógłby dla brata czynić w odpłacie?
Wielki odpowie:
„Mój drogi bracie, ja bym zaledwie o drobiazg Cię prosił,
abyś codziennie mi wodę przynosił.”

A valuable lesson.

Mother to child---'If you learn how to tell the time from an analogue clock by the end of the week, you shall be rewarded'. Within a week the child returns, demonstrating mastery in the ability of telling the time from a clock, and reminds the mother about the reward that was promised. 'You have already received your reward'---answeres the mother.

There are uncountably many binary sequences whose limiting relative frequency is x.

This is a proof of an elementary result in probability theory that I gave, prompted by a colleague's inquiry, and motivated by the fact that the question had initially stumped a University of Queensland professor of the relevant field of mathematics.

There are uncountably many binary sequences whose limiting relative frequency is x. by Mariusz Popieluch on Scribd

Link to PDF of proof.

There's no "hard way" of solving a problem --- an anecdote about John von Neumann as told by Eugine Wigner.

The following problem can be solved either the easy way or the hard way.

Two cyclists 40 miles apart are riding toward each other on a straight track; each one is going at a speed of 20 miles per hour. A swallow starting above one of one of them flies back and forth between them at a rate of 50 miles per hour. It does this until the cyclists meet. What is the total distance the swallow has flown?

The swallow actually flies back and forth an infinite number of times before the cyclists meet, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the cyclists are 40 miles apart and each cyclist is going 20 miles an hour, it takes one hour for the cyclists to meet. Therefore the swallow was flying for one hour. Flying at a rate of 50 miles per hour, it must have flown 50 miles. That's all there is to it.

When this problem was posed to John von Neumann by Max Born, Neumann immediately replied, "50 miles!"
"It is very strange," said Born, "but nearly everyone tries to sum the infinite series."
"What do you mean, strange?" asked Von Neumann. "That's how I did it!"

Source: John von Neumann Documentary starting at approximately 18 minutes into the film.
Based on the version of the anecdote from "Math Jokes".

Best of imperfect worlds.

Lev somewhat dissatisfied with his previous cosmic project, decided to embark on a more careful enterprise of world creation, and turn down the perfection parameter from maximum. This time he decided to play with the parameters of fundamental values, and observe how they'd influence the hedonistic dynamics. The idea was to run a simulation where hugs are set as having principal value, and consequently become the sought after currency -- in other words, hugs in that world were to be the sole wealth determinant. But in what sense 'hugs'? -- one may ask. Receiving them of course! And naturally they'd have to be of genuine sincerity; neither bought nor forced in any way. That is, hugs have value only if they're sincere and welcomed. But then how does one accumulate such wealth, given that hugs are such ephemeral phenomena? Surely, one can't be in possession of a great number of hugs. The only way one can accumulate wealth of this kind, that is, become a prosperous hugee, is to guarantee and maintain the existence of those willing to give those hugs, i.e. huggers (aka ready-to-hug beings). In other words, one can attain a high flux of hugs, and wealth would be interpreted as maintaining a high hug-flux. This can be done in many ways of course, and Lev calibrated the simulation with no limit on the degrees of freedom concerning valid hug-acquisition. Naturally those who attain the ability to reach and maintain a high hug-flux steady state, aka hugagogues  are sought after as raw models for guidance, whose wisdom would guarantee the attainment of wealth in that world. Lev has hypothesized that such a world would be among those that are the closest to being perfect without giving rise to any absurd consequences, which inevitably accompany perfection.

Explicit solution (formula) to the "truth table" recurrence relation.

If anyone has done any introductory logic, or has been introduced to representing sets in terms of their characteristic function, then the following binary matrix will look familiar. Each row is a distinct combination of the elements of some set. This is the matrix for a three element set. Call it matrix A.

1 1 1 ...
0 1 1
1 0 1
0 0 1
1 1 0
0 1 0
1 0 0
0 0 0
:
:

Some of the most common applications/interpretations of this matrix are:

i) For any natural number N (of columns), the rows of the matrix (of which there is 2^N) represent the set of all functions N→{0,1}., were N is a finite subset of the Natural numbers.

ii) The rows of a table thus generated, exhaust all truth value assignments to a propositional variable appearing in a formula of Propositional Calculus.

iii) Equivalently each row can be interpreted as the image of the characteristic function of a power set of some set S (we don't need to assume the axiom of countable choice since here we're dealing with finite sets). In other words each row of the matrix corresponds to a distinct combination of the elements of S. In fact the map f : ℘(S)→ ROWS (of the matrix) is a bijection: f(X)=r iff r(i)=1 iff i∈X, where we identify the matrix column indecies with the wellordering of the elements of S.

So clearly this matrix is a big deal, and subsequently the result here is an important one, since it basically compresses the entire matrix to a simple function of its rows and columns.

The way to generate the matrix in a way to ensure that all combinations are exhausted, is to follow the obvious pattern for column generation --- single iteration, double iteration, quadruple iteration, etc. In general, each column i has a 2^i-1  iteration of 1's and 0's (starting with 1's in each column).

Below is the proof for the explicit formula of such matrices recurrence relations, i.e. as a function of the matrix'  row and column number. That is, given only the row and column numbers, the formula gives the value that appears in the matrix on those 'coordinates'. That is, the formula takes the following functional form:

                                                                       θ : 2ⁿ×n → {0,1}

EXAMPLE: θ(k,i), where the row number is k = 7 and column number is i = 2, i.e. θ(7,2) = 0.

Now for the proof, which I have only given a sketch of in a previous post. Subsequently I lost the proof, and only had the formula, which bothered me, so I re-proved it last week. So here we go. Most of the pattern recognition which underlies the solution is obviously to be found in the matrix itself, and the diagram below intends to capture and make salient those patterns which may not be so obvious at first glance. The integer numbered columns, i.e. columns 1, 2, 3, 4,..., are the actual columns of the matrix A. 

To the right of each actual matrix column i, I list k(mod 2ⁱ), which is the first pattern that ought to be observed. Can you see how a single cycle of k(mod 2ⁱ) matches the length of the pair of iterated 1's and 0's, in column i? The next key observation, and perhaps the crucial one, is that the pattern of out actual matrix columns is basically a function of k(mod 2ⁱ). More precisely, note how each i'th actual column has the same pattern as k(mod 2ⁱ)−k(mod 2^i-1), with the exception of being slightly 'misaligned' and having some product of 2 in the place there 1's ought to be. I have indicated the k(mod 2ⁱ)−k(mod 2^i-1) rows with a yellow heading. Have a look for a while, and the pattern* should 'jump out'. Once you're convinced that it is so, all we need to do is (i) to rectify the 'misalignment', and (ii) the multiples of 2 integers instead of 1's. By noting that the 'misalignment' is also a function of column number, we see that adding 2^i-1−1 to k, does the trick. Finally divide all values in each i'th column by 2^i-1 and we're basically done. That is, our formula can be expressed as the function:

This formula is good enough and does the trick, but using the identity below, relating the mod and floor functions, it can make it more concise.

Giving the formula its final form. I guess it could be rendered even more concise and elegant with further algebraic fiddling, but I shall leave it there.

*Admittedly, a more rigorous proof is required than merely conjecturing that this pattern ought to hold in general. I may get around to it soon. But it does seem obvious that the proof is correct, since no surprises will arise in the relationship of the values of k(mod 2ⁱ) and k(mod 2^i-1) as both k and i increase.

If we switch the order of 1's and 0's in the table, i.e. if we let 0's precede 1's, like so (call this matrix B):

000...
100
010
110
001
101
011
111
:
:
Then the same reasoning yields a more elegant formula: