Can you solve Ted's given problem sets?

don't put words in my mouth. I haven't even read it yet.

let's focus on one problem at a time. Curious if we can get solutions to problems in OP

Meanwhile, here's a problem for you. Let the unit interval be divided into [math] n [/math] equal subintervals [math] I_{1}, I_{2}, ... , I_{n} [/math]. Let [math] P_{k} [/math] be any parallelogram in the complex plane such that one side of [math] P_{k} [/math] is [math] \{x+0i : x \in I_{k}\} [/math] and another side of [math] P_{k} [/math] is [math] \{x+1i : x \in J_{k}\} [/math], where [math] J_{k} [/math] is any subinterval of [math] [0,1] [/math] of length [math] \frac{1}{n} [/math]. Let [math] R_{k} [/math] be the region enclosed by [math] P_{k} [/math] and let

[math] A_{n} ( P_{1}, P_{2}, ..., P_{n} ) = [/math] measure of the set [math] \bigcup\limits_{k=1}^{n} R_{k} [/math].

In other words, [math] A_{n} ( P_{1}, ..., P_{n} ) [/math] is the total area covered by all of the parallelograms [math] P_{1}, ..., P_{n} [/math]. Let [math] \alpha (n) [/math] be the minimum value of [math] A_{n} ( P_{1}, ..., P_{n} ) [/math] as [math] P_{1}, ..., P_{n} [/math] range over all possible choices of the parallelograms.

1. Prove that [math] \alpha (n) \rightarrow 0 [/math] as [math] n \rightarrow \infty [/math]

2. Do there exist constants [math] c_{-}, c_{+} > 0 [/math](?)* such that [math] \frac{c_{-}}{log \; n} \leq \alpha (n) \leq \frac{c_{+}}{log \; n} [/math] for all [math] n [/math]?

3. Does the sequence [math] \alpha (n) [/math] decrease monotonically?

*At a glance, it is not perfectly clear whether the "greater than zero" comment is meant with respect to both c's, or just c_+. Ed.

Since we can approximate triangles to an arbitrary degree using the polygons, we can simply work in triangles instead. This easily gets [math] \alpha (n) [/math] below 1/2, but I'm working on how to get it to 0.

I meant parallelograms, not polygons

Just to make sure I understand the posing of this question, it works for a 2x2 grid right? Namely you put matches on all interior gridlines

Great thanks

how does one get in contact with ted kaczynski?

Both must be postive to make an interesting question

If C_+ is negative the inequality is trivially false, and if C_- is negative it's trivially true

You can send him mail. He's in prison.

bop.gov/inmateloc/

Go to INMATE, FIND AN INMATE, then put THEODORE KACZYNSKI in. You can call him, e-mail him, or send him mail.

Forewarning, he doesn't correspond with stupid.

Exactly so, yes. One thing that is slightly confusing in the original problem's wording is the language in a), which suggests something like a 1-1 relationship between match-side-slots and squares to be "covered". But all that a) really means is: "put the matches in the legal spots according to the language of the problem." Obviously any one match placed counts toward exactly two different squares, since "edge" placements are not allowed. Hint-hint.

Moreover, in the situation where n=2, covering all four of the "interior gridlines" precisely covers all of the "legal space" (four slots) available in this case. It is easy to see that any solution for any n must have all four of its "corner squares" covered in similar wise as a prerequisite of solution.

So let's consider n=3. In the crude skecth-related, a solution must immediately have its corner-squares covered, which (always) forces us to set red matches at the eight appropriate slots. In this case, this has the happy effect of immediately satisfying all the side squares. And yet, in order to cover the center square, any choice.of placement would immediately give three matches to a side square, forcing a non-solution. Thus by inspection we can immediately say that the problem has a solution when n = 2, and does not have a solution when n = 3.

Hint: the odd-even thing is important.

Yes, this is basic crypology right?

People send him math homework. Like a /SQT/ but to Ted.

Yes. But if you ask him stupid questions (if he even responds) he'll tell you he doesn't have time to waste on you and to not write him any further

...

Of course he does.

>guy's main claim to fame is sending mail-pieces to people with bombs in them, killing a few

>becomes infamous, people become interested in him

>several people write him and request that he write back, that is, send them mail-pieces

No, it doesn't matter that he's in prison and that the stuff is obviously screened back-and-forth. The irony, the humor, the stupidity of it all is just too delicious not to comment on. In fact it's double-stupid because either you're on the enhanced watch-list or you get a Happy Birthday surprise.

I was thinking of this. I think that they probably thoroughly check his mail though. I'm also not sure that he has interest in targeting anymore, or if he had interest in targeting intellectuals in the first place.

People have been mailing him for a long time and I don't think he's bombed anyone while in prison.

some journalist that wrote him
youtube.com/watch?v=BurUfcV0GR4

Yes, basically in the limit you are sending each point on y=0 to a single point on y=1. If you choose 2 points on y=1, you still get a limiting area of 1/2 though.

I don't see it.

Which I why I pre-empted this pedantry. The irony alone suffices.

>Awards

I see how to prove the first step

Detail your proof.

I'm working on it. I assume this problem has been solved and sent back to him though right? OP isn't the person mailing him is he?

I'll show you what he has to say about each of the problems. To my knowledge he wrote to this author but don't know if there were any further correspondence. I doubt the person he gave the problems to even followed up

...

...

...

Unibomber

I'm kind of rusty because I dropped out, but here's my best. It's messy I know. Tell me if there's something wrong.

To prove [math] \alpha(n) \rightarrow \infty [/math] as [math] n \rightarrow \infty [/math]

Take all [math] P_k [/math] that satisfy [math] a < I_k < b [/math] where [math] 0\leq a,b \leq 1 [/math] and make them all share the same [math] J_k [/math]. By restricting this [math] J_k [/math] to a decreasing interval between [math] 0 [/math] and [math] 1 [/math] we can make our collection of [math] P_k [/math] approach a triangle. Without loss of generality we may now begin working with triangles in order to find the lower limit. Label these as [math] T_k [/math] having point [math] p_k [/math] on [math] y=1 [/math].

Now group all triangles [math] T_k [/math] and [math] T_{k+1} \ \ [/math] such that [math] p_k < p_{k+1} \ [/math] and [math] p_{k+1} \ - p_k = \frac{1}{n} [/math]. Relabel these pairs as [math] P_k [/math] and rescale and [math] I_k [/math] and [math] n [/math] appropriately. We now have parallelograms [math] P_k [/math] missing half their area. We may use these rectangles without loss of generality.

Repeat our process of forming triangles out of these new rectangles and turning them back into rectangles, each time reducing the area of the rectangles by 1/2 indefinitely. This means there's no lower limit on [math] \alpha(n) [/math]

I am unsure myself. Hopefully someone here can verify your proof.

There's some steps and details that should be clarified, but I think it's the right idea. For instance, if we simply say "without loss of generality" and just start using our new shape, we can't know if it decreases monotonically. Only what the end goal will be.

Also I'm a bit fuzzy on exactly how much area the rectangles lose at each step of the process because I'm not sure how to account for intersecting yet.

This is not a proof.

This area of math is not familiar to me. Hopefully someone here can verify. Ted says he can prove the first stated problem.

Yea, I know it's messy, but the thinking isn't fallacious.

Maybe it's easier if I just explain my thinking.
It's possible to group the parallelograms such that they approach a triangle as n goes to infinity with arbitrary accuracy. Without loss of generality we can simply start using triangles instead of parallelograms. Now take two adjacent traingles and make their tip the same distance apart as their base. We now have a "parallelogram" missing half it's area. Take this to be our new parallelogram. Now do the same process: using this new parallelogram and let n go to infinity to have it approach a triangle. Our new triangle will have half the area of the first. Now use this new triangle as described before to turn it back into a rectangle, this time only having 1/4th the area we originally started out with. Repeat this process indefinitely to get rectangles of arbitrarily small area.

It's very awkward to try and phrase mathematically. Ideally you'd just want to say what goes where for each n, but it's not clear to me an easy way of describing that.

Bump

Am I missing something?

Suppose you can do this for [math] n \geq N [/math] where [math] N \in \mathbb{n} [/math]. Then you can do it for [math] n +2 [/math].

Proof: lay matches appropriately for the four corners. Now lay matches in between each square that are on the sides.

What we'll have is all outer squares are touching exactly 2 matches. Now we have a [math] n \cross n [/math] square in the middle, which by assumption, we can fill without putting any matches on the outside of the box.

This shows if we can solve it for one odd/even number, we can solve it for all odd/even numbers greater than that one. The question is how to prove that there are NO solutions at all for odd/even no matter how large [math] n [/math] is.

First line of the proof is messed up. Just suppose you can do it for some [math] n \in \mathbb{N} [/math]

By rearanging the triangles into rectangles, you are changing the problem.

I say this problem is bullshit, since one parallelogram has area 1/n, and the others have at least 1/(2n)

[eqn]
\frac{\tan(t)}{\tan(t)+\cot(t)}=
\frac{\tan^2(t)}{\tan^2(t)+1}=
\frac{\tan^2(t)}{\sec^2(t)}=
\sin^2(t)
[/eqn]

/SQT/ but ask Ted Style

>where Jk is any subinterval of [0,1] of length 1n1n

Im kinda a noob at this hobby math, can some one explain to me what this means?

nice tex

>give your teacher an F if she doesn't solve a problem you give her

Because you got trips I'll explain it in layman terms.

The scenario that is described is as follows:

in the xy plane from x=0 to 1 with y=0, cut the distance up into n equal pieces. These will be the bases of n parallelograms. The top of the parallelograms will be on the line y=1 between x=0 and x=1. So the bottoms don't move, but the tops are allowed to move around.

Now there's many different ways you can choose the parallelograms and depending on how you choose them, they may cover an area of 1 or they may cover an area of less than 1 due to overlapping.

The first proposal 1) states that as n goes to infinity, it's possible to make the parallelograms cover an arbitrarily small amount of area by overlapping them in some clever way.

Threadly reminder to all: Kaczynski's doctoral dissertation, /Boundary Functions/ (NOT TO BE CONFUSED WITH OTHER WRITINGS HAVING THAT SAME PHRASE IN THEIR TITLE), is (apparently) only availble at three university libraries which are concentrated around the DC area. Michigan apparently has lost its print copy and may be stingy/strict with microfilm.

If you live near one of the libraries in the link, please consider making a field trip and having the dissertation, such as you find it, copied off and posted to Veeky Forums. So far as I am aware, it is not available elsewhere on the internet, and you would be doing a service to the history of math by securing a curiosity for future consumption.

worldcat.org/title/boundary-functions/oclc/34661830

Pic related is Kaczynski's mathematical bibliography as far as I've managed to push it. the dissertation link is a FRAGMENT, not the whole thing:

search.proquest.com/docview/288225414

Have you tried contacting the libraries that have paper copies and asking if they could scan it for you?

It's admittedly unlikely but given that it's a rare document and you aren't violating any private rights/paywalls it's worth asking.

What's the proof for this one? I've been searching everywhere but I can't find it.

what the fuck is that U notation?

he's asking you to prove the fundamental theorem of algebra.

i did this today in math c

a reunion ? wtf is up with this board ?

I did this a month or two ago... it's mostly simple algebra. As I recall, the solution turns out to have a unique answer only because there is an implied constraint of not having a fraction of a penny to start with.

In my mind, I keep flipping back and forth between "this is bullshit" and "this is brilliant"

union.

>It's possible to group the parallelograms such that they approach a triangle as n goes to infinity
Yes, but all of these triangles will have their base on the bottom (the real line).

I think you are wanting there to be triangles whose base is on the top (the line x+i where x is real). I don't see how you then make a parallellogram using two triangles unless one has its base on the bottom and the other has its base on the top.

No, both are at the bottom. Example, take two adjacent triangles with base length 1/n. Make the one on the left have a top located at 1/4. Take the one on the right have a top located at 1/4 + 1/n. Then they will overlap perfectly with a paralellogram except that they will be missing half the area in a v shape.

I used my argument because it's the only way I could describe it. However ideally you'd want to describe a process or a specific shape you want the parallelograms to approach as n goes to infinity. What I hoped to show with my argument is that such a shape does exist which they can approach, and have the area go to zero.

No, I meant what I said. By triangles making a parallelogram I meant a shape that can overlap a parallelogram, but is missing half its area in a v shape.

I now see a flaw which convinces me this argument needs refinement as can't be used as stands. Specifically imagine n rectangles forming a triangle. Now if we remove half of the area of each rectagle we cannot know how much area is lost in the entire shape. This is because for areas that overlap. The same amount is not removed per rectangle in areas that overlap as it is for areas that don't overlap. Hence 1/2 of the area is not lost each time the rectangles are turned into triangles and back into parallelograms.

The reason I found this counterexample or suspected it existed was because I realized that a decrease in area can only be accompanied with an increase in overlap, which my argument seems to neglect entirely.

However on the other hand I see a way to show that the way I use limits of shapes is in fact sound reasoning. This can be done by simply saying "we can approach this shape and make the area 1/2. Or we could approach this shape and make the area of the parallelograms 1/4, or we could approach this shape and make the area 1/8, etc..."

una-

What's Uncle Ted's postal address?

I'll send him a postcard from Korea.

Suppose we divide the top edge of the square into n equal parts {J_k} just like we did on the bottom. Now take a random permutation p of 1...n and make a parallelogram from I(k) and J(p(k)). The total area of all these is 1, but you'd need to subtract areas that are covered exactly twice and after that subtract 2 times the areas that are covered exactly 3 times, then 3 times the areas that are covered exactly 4 times , etc. maybe if the permutation scrambles enough, these subtraction get arbitrarily close to 1 as n goes to infinity

If n equals 2^m, maybe pair up I_k with J_j, where j is the integer you get by reversing the order of the bits in the m-bit representation of k?

I'm pretty convinced I have an argument which shows that 1) is not in fact true. Maybe Ted made a mistake? But what are the chances of that.

Odds are low.

Yes, I keep thinking it's impossible. But then I start to have doubts. Back and forth. Bump for interest. Anybody heard of this problem?

I'd try to hack some code to do this, see if you can get an example below 1/2, but it's a daunting task with all those intersecting parallelograms.

I see my mistake now. I was trying to use arguments along the lines of "in order for the area of the square to be 0, it must be 0 in a rectangle of width 1 and height [math] \delta [/math]. Scaling [math] \delta [/math] to 1, this means that if the area in the whole square can be made 0, we're allowed to restrict the angles the rectangles can go up at to an arbitrary degree and still make it be 0.

The mistake is that restricting them to an arbitrary degree doesn't matter as n will get so large it won't change anything.

Make two triangles and intersect them. The area is less than 1/2 as both triangles have area 1/2, but are also intersecting which takes some more area away.

how can I contact him? can someone give me his address for physical or electronic mail?

nice idea user, bump for support

if you know any London libraries that have it I can maybe try to do something

Ted is a genius.

The area would be 1/2+1/2-intersection>=1/2

With two triangles each of base 1/2 the area would be 1/4 + 1/4 - intersection.

bump for actual math discussions

>base 1/2
oh yes I see what you mean now thanks