/sci/ - Science & Math

why is it that calculus always culminates in helping a red neck nigger farmer maximizing his fence? I mean every concept in calculus can be used to help nigger redneck farmers build their fences. calculus is a useless idea there i said it.... in the real world if a farmer wants to maximize the area of his fence he'll just look at it and make it a good decision, he won't use integrals and derivatives to optimize his fence area

>>5685311

Why do you hate farmers?

>>5685318
i don't hate farmers at all. i run half of my yard as a garden and grow all kinds of shit. its just that the height of problem solving seems to be making fences for mules and malicious kings

>>5685324

Government bureaucracy is always a problem. In real life and in math books.

The King can just make the circles infinitely small (but not zero) and cover everywhere on the plane.

>>5685336
this. At the moment the question is trivial.

>>5685343
Infinitesimals are not real numbers. The x-y plane implies R^2.

>>5685260
i don't think the king can. Pretty sure the farmer wins this one. Working on a proof or something.

>>5685347
Just take the limit then.

>>5685350
just kidding I think it's the king now.

>>5685357
I would guess the King, otherwise this question will be a bitch to prove.

Gimme 5 mins.

>>5685347
If he works a chessboard then it becomes Cantor's diagonal argument and the king wins

why doesn't the king put a steak at every coordinate that is (rational number, rational number)

Then the farmer would have make his field in a rectangle with infinitessimal length, and so the area of his field would be poorly defined.

>>5685368
Stakes infinitely close together.

Conceptually the King wins easily, I don't understand why this is even a problem. Just choose an arbitrarily small radius for each of the stakes and the area left is arbitrarily small. Am I reading the question wrong?

first of all it is hard to quess who will win.
If the farmer must take a fixed point(say 0,0) as corner the king will win. However is the king takes the elements of Q, with denominator at most n, the farmer will win.

If the king will win, it must be possible to construct a set of points, such that each rectangle contains one of these point.
If the farmer wins, we must prove, that the kings points become infinitely close together, (so we need to try and find some limit point)

>>5685372
but between every rational number there must be infinitely many irrational numbers.
So the radius between two stakes must be non-zero, but the length of any rectangle that can squeeze through the stakes must be infinitely small but non-zero.

hmm what's the area of a rectangle with sides (irrational infinity)^-1 * (irrational infinity)

I can't think of a scenario where the farmer doesn't win

The answer is that the farmer can win. For instance, if the King puts all of his stakes on the x axis, the farmer can just fence off a unit square somewhere in the top half of the plane.

>>5685382
You're doing it wrong. The radius between any two given stakes is non-zero, yes, but that doesn't mean that you can describe a finite circle around a given stake with no other stakes in it.

The question states that the King <preemptively> places stakes at <infinitely> many points in the <plane>.

How would the farmer compete with that?

>>5685392
by making a very thin rectangle.

>>5685390
yes it does. just because I can't say what value that radius has doesn't mean that radius isn't non-zero.

all the question says is
each point where he places a stake must have a circle of non-zero radius around it that contains no other stakes.
> a circle of non-zero radius

>>5685398
Give an example of a radius and a point in Q^2 where there's no other points inside the circle.

>>5685407
>Constructivist pls go

>>5685407
I did: >>5685368

>>5685414
troll confirmed, kthx

>>5685260
I think it goes something like this (I am going to leave out a lot of edge cases):

Assume rectangle has area exactly 1.

Let rectangle have size (x,1/x)

We want to "block" all such rectangles, and we construct a different grid for each of the (infinite) cases 2^a < x < 2^(a+1) for an integer a.

To block such a rectangle we construct a grid of the form (m*2^(a-1), n*2^(-a-1)) for integers m and n. For example when a=0 this will be a grid of the form (m/2, n/2).

Now we have infinitely many grids, indexed by a, that we need to fit together while satsifying the property given in the question. I don't have time to do that but it seems like it is possible. We "fit" the grids together by translating each grid by some constant, so the actual set of points is
((m*2^(a-1)+x_a, n*2^(-a-1)+y_a)
as a,m,n range over the integers, for some constants ...,x_{-1},x_0,x_1,... and ...,y_{-1},y_0,y_1,....

I might have found it:
Look at the square 2^n by 2^n, divide in squares of 2^-n by 2^-n. Gives a grid/checkerboard. King must have at least 1 stake at each row or column.
By just looking at the rows, optimal way would be starting at (1,1) and following the partern move k to the right and 1 up. After 2^n/k steps he is (1,2^n/k+1). By minizing this for the distance between all the points, k should be around \sqrt{2^n}, which means that the distance between the points becomes \sqrt{2^n}\times 2^-n, which is 2^{-n/2}. So at this point, ALL the circles cant be larger then 2^{-n/2}

Because the farmer can pick arbitrary small rectangles. we can take the limit n->0, which shows that the farmer wins, because the king the radii around the stakes goes to 0

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.

>>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.

The farmer can pick the rectangle with coordinates (5,5),(5,6) (6,6)(6,5), without passing through any of your poinst

>>5685458
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).

Looks like someone needs to re-take first year analysis.

>>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.

>>5685449
was wrong
>>5685461

However, the distance between the squares of "length" n, is 1/2. So between square 10 and square 11, we have room for a rectangle of width 1/4 and height 4, without passing through any of the squares centered around 0. (see >>5685444 for my idea of a win for the farmer)

>>5685467

so the other subtle difference was i'm trying to prove that the farmer wins. (because the kings set must be dense to block all the rectangles even the ones not containing 0)

>>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).

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

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

The complete solution, based on my reasoning here >>5685441 is to place stakes at all points of the form (2^m,2^n) where m and n are integers such that m+n>=-1. Pic related.

>>5685467
I agree

>>5685484
Correction, that should be m+n >= -2.

>>5685463
This is obviously bullshit. ANY 2 points will have a non-zero radius between them you div, otherwise they're not different points. ANY set of points will satisfy your "property".

>>5685480
i messed up the references
my own post
>>5685461 was wrong, but for example the rectangle [-2,2]x[10.1;10.6] has area one and lies in the space between square 10 and square 11.

>>5685449
>Also, I only use a countable amount of stakes, which I intuitively thought was not gonna be enough.

Your intuition seems pretty bum, bro. Any solution will use a countable number of points.

I think
<div class="math">
\left.\left\{\left(m, \frac{n}{|2m+3|}\right)
\right|m,n \in \mathbb{Z}\right\}
\cup \left.\left\{\left(\frac{n}{|2m+3|}, m\right)
\right|m,n \in \mathbb{Z}\right\}
</div>
works for the king.

>>5685488
So all the points (a 2^n, b 2^m) with n+m=-2. and a,b integers? Around zero this set is dense, which means that there is no nonzero radius s.t. there are no other stackes contained.
If we pick a radius around 0, s.t. only 1 stake is included, this leaves enough space to make a rectangle of area 1.

>>5685484
A small typo as well. Here is the final answer plus proof:

Place a stake at all points of the form (m*2^a, n*2^(-a-1)) where m,n,a are integers. I won't bother proving this set of points has the required properties. The grid is essentially the same as the one I posted.

Now to prove the farmer cannot win against this grid:

Farmer chooses rectangle at (p,q) with radius (x,y).

Let a be such that 2^a <= x <= 2^(a+1).

Therefore x*y <= y*2^(a+1)
Therefore 1 <= y*2^(a+1)
Therefore y >= 2^(-a-1)

Since x >= 2^a there must be an integer of the form m*2^a between p and p+x. Call this w.

Since y >= 2^(-a-1) there must be an integer of the form n*2^(-a-1) between q and q+y. Call this z.

The point (w*2^a, z*2^(-a-1)) lies in the farmers rectangle.

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

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

>>5685521
Yeah I fucked that up, here too >>5685527 it actually needs some gray-coding style thing so that the "lines" where the points are infinitely dense end up at x=infinity and y=infinity instead of x=0 and y=0

>>5685527
>won't bother proving this set of points has the required properties

Your set contains 0. Each circle of positive radius around zero contains another stake. (n=0,m=1, a is small enough) so it does not satisfy the property

>>5685527
>I won't bother proving this set of points has the required properties.
> /sci/

>>5685537
you are right, I had a working solution, then I thought of this "simpler" version but I forgot about 0. Will post the actual solution in a minute.

>>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.

I still think the farmer can win. If im right
>>5685444
shows the set of points the king picks must be dense somewhere if he wishes to block all the squares.

>>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.

>>5685541
that'll work, I also realize where i went wrong in my first "farmer will win" solution.

>>5685518
In fact |2m+2| would work as well.

Without loss of generality, assume the farmer's plot is longer than it is tall. Then the rectangle crosses at least one x=m line. Each x=m line contains stakes at intervals of 1/|2m + 2|. The height must therefore be less than 1/|2M + 2| where M identifies the x=m line crossed with the largest |m|. That means the width of the rectangle must be larger than 2|M| + 2, and the number of lines crossed must be at least 2|M| + 2. But the number of y=m lines with |m| <= |M| is only 2|M| + 1, and we have a contradiction.

i am confused. which one is the final solution that y'll came up with?

>>5685537
>>5685521
Final correction to >>5685527
Use points of the form ((m + 1/2)*2^a, (n+1/2)*2^(-a-1)).

Proof is as before, including not proving the property.

>>5685542
proof by diagram bitch.

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

Works for me.

>>5685582
And for the circles property, for those points where only one coordinate is an integer, a circle of radius 1/|2m + 3| will work. If both coordinates are integers, choose the smaller of 1/|2x + 3| and 1/|2y + 3|.

>>5685591
Are there infinite points in that diagram (in theory, not plotted)?

>>5685591
Proof of property:

Fix a. Note that of all the points of the form ((m + 1/2)*2^a, (n+1/2)*2^(-a-1)), the closest points to the origin are those corresponding to (m,m)=(1,1), (1,-1), (-1,1) or (-1,-1). All these points have a distance of at least max(2^a, 2^(-a-1)) from the origin.

Therefore for any point, let its distance from the origin be r. We can choose A such that if |a|>A then all the points of the form ((m + 1/2)*2^a, (n+1/2)*2^(-a-1)) will have distance at least 2*r from the origin. Then there are only finitely many grids left, (those for which |a| <= A), and so the point must be bounded away from these grids too.

>>5685634
yes, but as I prove here >>5685642 any finite region will only contain finitely many points. This is because as a line becomes more "dense" in the y direction (a goes to infinity), the displacement in the x-direction of the first line goes to infinity as well. Similarly as the line becomes dense in the x-direction (a goes to -infinity) the displacement in the y-direction of the first line goes to infinity.

So while what I plotted is incomplete, if I had restricted the axes to [0,6] it would actually be the true set (restricted to these axes).

I prove this here >>5685642

>>5685475
The value r is irrational.