/sci/ - Science & Math

>>5688275
It might also be significant that they are hexadecimal digits. I'm not convinced myself though, because it is unlucky that coordinates, when written in hexa, would have only digits between A and F and no digit between 0 and 9, but for some reason no letter after F appears here.

It's most likely a coincidence, but who knows...

>>5688238
But 91 minutes makes no sense, does it? I think minutes and seconds stopped at 60 for navigation coordinates just as they do for time.

Another wild guess.

A famous instance of a triangle related to sailing is the Bermuda triangle or Devil's triangle. It happens to:
- Be on the east US coast,
- Touch Haiti, which is French-speaking, and be very close to Martinique and Guadeloupe, which are two French overseas departments.

Could be around that? It is kinda fitting that some kind of encrypted map would involve the Caribbean.

>>5687729
>are there infinitely many rational numbers between any two irrational numbers?
Yes. You can use roughly the same proof. If you give me two irrational numbers x and y with x<y:
- Call d the distance y-x between them,
- I choose n = ceil(1/d) (the smallest integer greater than or equal to 1/d),
- Now consider alpha = ceil(x*n) / n. We have that alpha is rational and that x*n <= ceil(x*n) < x*n+1. Therefore y <= alpha < (x*n+1)/n = x + 1/n <= x+1/d = x+(y-x) = y, so alpha is between x and y.

Now, consider a rational z between 0 and y-alpha. alpha+z is rational, and alpha+z is between x and y.

There are infinitely many rationals between 0 and y-alpha (every term of the 1/n sequence when n is greater than 1/(y-alpha), for instance), so there are infinitely many rationals between x and y.

>IF not, then does that mean that two irrational numbers can be "next" to each other, such that there are no other numbers between them?
>such that there are no other numbers between them
Consider two numbers x and y such that x<y. Then x < (x+y)/2 < y. There is always a real number between any two distinct real numbers.

>>5687451
Also, my reason to prefer the lines read from bottom to top is that starting from the East coast and moving West, it's land, so you won't see capes or anything. I prefer to imagine that we start from the North coast and follow it Westward, it makes more sense.

>>5687190
My decomposition of the message would be:

> The North coast,
> The third cape to the West,
> The East coast.

In three lines rather than four. Notice how the 2nd line "Le troisième cap" occupies the width of the sheet, so it may not be an intentional line-break.

Could these three lines refer to the rest of the "clues"?

The arrow could be a cardinal direction (either North because it's the most common reference, or West, referring to the "The third cape to the West" line).

I like the idea that the arrow means "West". Indeed, you start on the North coast. Imagine that the direction of the arrow is West ("West" is "to the right"). Then North is "downward". If you're on the North coast, "downward" is the sea. Now look at the dots around the arrow: they may indicate the sea, in which case the arrow is a cape, if you see what I mean.

>>5685591
> >>5685542
> proof by diagram bitch.

Works for me.

>>5685544
Nah, it doesn't have to be dense anywhere. It only has to have "barriers" closer and closer to "denseness" the further they are spaced, so that if you specify the a small width for your rectangle (e.g. 0.001), then borders with less than 0.001 between two stakes will be distant from each other by less than 1000, so the area of you rectangle will be 0.001 * [something less than 1000] < 1.

>>5685534
To make it work, I think I can just make it so that I don't just put points on the contours of my "squares" but also fill the regions between two squares using a grid with the same distance between its points as that in the exterior square. Not really worth plotting because it would be ugly and hard to read, but I think it works.

>>5685533
And I'm dumb... It obviously doesn't work ^^'

>>5685484
Here's mine (showing the (5,5)-(6,6) too).

>>5685471
> >>5685449
> was wrong
> >>5685461

Well erh, that rectangle definitely goes through many of my stakes.

>>5685463
> if ther are infinitely many irrational numbers between any two real numbers, which there are, then there's a non-zero radius between each coordinate that can be written in the form (rational number, rational number).

If you mean "There is a non-zero distance between any two rational points (x1,y1) and (x2,y2)", then yes, but it isn't relevant.

If you mean "There is a non-zero radius R(x,y) around any rational point (x,y) such that no rational point is within this distance R(x,y) of (x,y)", then it's wrong and it's really simple to see that it's wrong. Just tell me what R(0,0) should be. For any value r that you give me, I'll take the point <span class="math">\displaystyle\left( \frac{1}{\frac{1}{\lfloor r\rfloor}} , 0\right)[/spoiler], which is rational and is within distance r of (0,0).

>>5685444
It doesn't work. The radii may be as small as you want, but they cannot be zero. If I ask you to choose a given value of n, and you choose a finite one, then I can make a field. If I ask you to choose a value of n and you tell me "I'd rather take the limit when n is infinite", then there is no disk of non-zero radius around your stakes that does not contain any other stake.

See my post and that of >>5685441 and you'll notice a subtle difference: in your situation, the radius around any chosen point goes to 0 as you place more points. In our solution, the radius around any given point is fixed as we place more points (after the neighbors have been placed), and what goes to 0 is only the minimum of the radii, not any radius in particular.

>>5685421
It doesn't work. Rationals are dense in reals. It means that there is no open ball centered around any real (rational included) that doesn't include at least one rational number.

To be more concrete, tell me exactly what the radius is around the stake (0,0), within which no other stake is placed. You can't, but you should be able to do it.

Here's my attempt at being the king. Try and see if you can be the farmer by beating it.

I place stakes in the following manner:

For all <span class="math">n\in \mathbb{N}[/spoiler], I place the set of stakes <span class="math">S_n[/spoiler] on the square of side n centered at (0,0), where <span class="math">S_n[/spoiler] is obtained the following way, intuitively:
We regularly place stakes on <span class="math">S_n[/spoiler] which are close enough to each other so that a band that goes between two of these points will have a width at most <span class="math">\frac{1}{n+1}[/spoiler]. This way, any field for the farmer that goes between two successive points of <span class="math">S_n[/spoiler] will have to have length at least n+1 in order to have area 1. To have length n+1, it will need to go through the square centered in (0,0) and of side n+1. By induction, we then prove have that any valid field must go through the squares of any radius. The field therefore has infinite length, which is invalid. I win.

Let me know if it is unclear or if you think it doesn't work. I managed to convince myself. Also, I only use a countable amount of stakes, which I intuitively thought was not gonna be enough.

>>5675595
Damn, I'm pretty sure it was from this guy: http://www.umiacs.umd.edu/users/pvishalm/Research.html

But ITA doesn't require you to submit an article when you're invited, and the slides for his ITA talk aren't online, so I don't have anything on emotions.

There was a pretty awesome talk at ITA last February in which they projected image data in the "neutral emotion" face of people and the "emotion component". Let me see if I can find a link and maybe pictures.

>>5674619
Yeah, we only have a necessary condition. I gave it a thought earlier and I didn't manage to get to anything relevant.

The generic formal way to approach these questions is the following, OP:

1) Notice that y is obtained by iterating a function infinitely many times on some starting number. Here, we iterate <span class="math">f_x:a\mapsto x^a[/spoiler]. If we start with x=a, then we get the sequence x, x^x, x^(x^x), x^(x^(x^x)), and so on. Notice that f_x is continuous.

2) By definition of limits of sequences, this sequence only converges if you can make the difference between two successive terms arbitrarily small by considering further terms in the sequence. Because f is continuous, if there exists a sequence <span class="math">b_1,b_2,\dots[/spoiler] such that <span class="math">b_i[/spoiler] can be made arbitrarily close to <span class="math">f(b_i)[/spoiler], then the limit of that sequence is a fixpoint of f_x.

3) In the general case, we would try and find fixpoints of f_x for different values of x. Here, we know the fixpoint is 2, so we just have to find x so that the fixpoint is 2, by solving <span class="math">2=f_x(2)[/spoiler] (which is 2=x^2, as anon already mentioned).

4) There are two solutions to that equation (two x such that f_x has 2 as a fixpoint): sqrt2 and -sqrt2. Starting from here, you are forced to analyze both sequences (the one generated with x=sqrt2 and the one with x=-sqrt2), individually, to see whether they do converge to 2.

5) Proving that x=sqrt2 works is relatively easy. You have an increasing sequence defined by a recurrence relation, and a unique fixpoint, you can prove that it converges to 2 using several distinct theorems. Checking what happens for x=-sqrt2 is harder, I don't really know what the best answer would be to that (I think the series is divergent but I'm not sure how to prove it).

>>5674541
I gave a link. You didn't click it. Therefore you didn't know that it actually answered the question, and started flaming.

You were wrong. The link gave the answer to the question. The fact that the title of the link made you think it didn't is irrelevant.

You are an ass, you were wrong, and for some stupid e-pride reason you will refuse to admit you were wrong, which is fine because everybody knows you either know you're wrong, or are too stubbornly stupid to be worth talking to.

>>5674534
What the hell are you talking about?

> Let p > 2 be a prime number. Then
> (Supplement 1)
> x2 ≡ −1 (mod p) is solvable if and only if p ≡ 1 (mod 4).

This is exactly what>>5674497 says.

>>5674528
http://en.wikipedia.org/wiki/Law_of_quadratic_reciprocity

>>5674299
I'd give it a try every now and then if someone were to start posting them again.